Beer11e Chapter 2 ISM - testbanklive.com · SOLUT Using th We hav Then And ION e force triang e: P So PR A. Q re (b le and the law 180 105 γ= = 2 (4 64 8.0 R R = = = 4kip sin(25

CHAPTER 2

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V e c t o r M e c h a n i c s f o r E n g i n e e r s S t a t i c s a n d D y n a m i c s 1 1 t h E d i t i o n B e e r S o l u t i o n s M a n u a lF u l l D o w n l o a d : h t t p : / / t e s t b a n k l i v e . c o m / d o w n l o a d / v e c t o r - m e c h a n i c s - f o r - e n g i n e e r s - s t a t i c s - a n d - d y n a m i c s - 1 1 t h - e d i t i o n - b e e r - s o l u t i o n s - m a n u a l /

F u l l d o w n l o a d a l l c h a p t e r s i n s t a n t l y p l e a s e g o t o S o l u t i o n s M a n u a l , T e s t B a n k s i t e : t e s t b a n k l i v e . c o m

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SOLUT

(a) P

(b) T

W

TION

arallelogram l

Triangle rule:

We measure:

PRO

Twomagn(b) th

law:

OBLEM 2.1

o forces are anitude and dirhe triangle rul

139R =

1

applied as shorection of theile.

1 kN, 47α =

own to a hooir resultant us

7.8°

ok. Determineing (a) the pa

1391 N=R

e graphically arallelogram la

N 47.8°

the aw,


SOLUT

(a) P

(b) T

W

TION

arallelogram l

Triangle rule:

We measure:

PR

Twograp(a)

law:

OBLEM 2.2

o forces are phically the the parallelo

90R =

2

applied as smagnitude

ogram law, (

06 lb, 2α =

shown to a band directio

(b) the triang

26.6°

bracket suppon of their rgle rule.

906R =

ort. Determiresultant usi

6 lb 26.6°

ine ing


PROBLEM 2.3

Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 10 kN and Q = 15 kN, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

(a) Parallelogram law:

(b) Triangle rule:

We measure: 20.1 kN,R = 21.2α = ° 20.1 kN=R 21.2°


PROBLEM 2.4

Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 6 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

(a) Parallelogram law:

(b) Triangle rule:

We measure: 8.03 kips, 3.8R α= = ° 8.03 kips=R 3.8°


PROBLEM 2.5

A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant.

SOLUTION

Using the triangle rule and the law of sines:

(a) 120 Nsin 30 sin 25

P=

° ° 101.4 NP =

(b) 30 25 180180 25 30125

ββ

° + + ° = °= ° − ° − °= °

120 Nsin 30 sin125

=° °

R 196.6 N=R


PROBLEM 2.6

A telephone cable is clamped at A to the pole AB. Knowing that the tension in the left-hand portion of the cable is T1 = 800 lb, determine by trigonometry (a) the required tension T2 in the right-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION


(a) 75 40 180180 75 4065

αα

° + ° + = °= ° − ° − °= °

2800 lbsin 65 sin 75

T=

° ° 2 853 lbT =

(b) 800 lbsin 65 sin 40

R=

° ° 567 lbR =


PROBLEM 2.7

A telephone cable is clamped at A to the pole AB. Knowing that the tension in the right-hand portion of the cable is T2 = 1000 lb, determine by trigonometry (a) the required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION


(a) 75 40 180180 75 4065

ββ

° + ° + = °= ° − ° − °= °

11000 lbsin 75° sin 65

T=

° 1 938 lbT =

(b) 1000 lbsin 75° sin 40

R=

° 665 lbR =


PROBLEM 2.8

A disabled automobile is pulled by means of two ropes as shown. The tension in rope AB is 2.2 kN, and the angle α is 25°. Knowing that the resultant of the two forces applied at A is directed along the axis of the automobile, determine by trigonometry (a) the tension in rope AC, (b) the magnitude of the resultant of the two forces applied at A.

SOLUTION

Using the law of sines:

2.2 kNsin 30° sin125 sin 25

ACT R= =

°

2.603 kN 4.264 kN

ACTR

=

=

(a) 2.60 kNACT =

(b) 4.26 kNR =


PROBLEM 2.9

A disabled automobile is pulled by means of two ropes as shown. Knowing that the tension in rope AB is 3 kN, determine by trigonometry the tension in rope AC and the value of α so that the resultant force exerted at A is a 4.8-kN force directed along the axis of the automobile.

SOLUTION

Using the law of cosines: 2 2 2(3 kN) (4.8 kN) 2(3 kN)(4.8 kN)cos 30°2.6643 kN

AC

AC

TT

= + −

=

Using the law of sines: sin sin 303 kN 2.6643 kN

34.3

α

α

°=

= °

2.66 kNAC =T 34.3°


SOLUT

Using th

(a)

(b)

TION

he triangle rule

e and law of s

si5s

α β+ +

sin117R

PROBLEM

Two forces amagnitude oangle α if thebe horizonta

ines:

in sin 2550 N 35 Nsin 0.6037

α

α

°=

=

37.138α =

25 180180117.862

β+ ° = °

= ° −=

35 N7.862 sin 2R

=°

M 2.10

are applied as of P is 35 N, de resultant R ol, (b) the corre

74

°

8°

25 37.1382

° − °°

N25°

shown to a hodetermine by of the two forcesponding mag

ook support. Ktrigonometry ces applied to gnitude of R.

Knowing that (a) the requirthe support is

37.1α = °

73.2 NR =

the red s to


SOLUT

Using th

(a)

(b)

TION

he triangle rulee and the law

50β + °

4s

4s

P

Athmfom

of sines:

60 18018070

β° + ° = °

= °= °

425 lbsin 70 sin 6

P=

°

425 lbsin 70 sin 5

R=

°

PROBLEM

A steel tank is hat α = 20°, d

magnitude of forces applied magnitude of R

50 60− ° − °

0°

0°

2.11

to be positiondetermine by the force P iat A is to be v

R.

ned in an excavtrigonometry if the resultanvertical, (b) th

vation. Knowi(a) the requir

nt R of the the correspondi

392 lbP =

346 lbR =

ing red two ing


SOLUT

Using th

(a)

(b)

TION

he triangle rulee and the law

( 3α +

sin

P

Athtrthc

of sines:

30 ) 60

sin (90 )425 lb

βββ

α

° + ° +

° −

90 α° −

(42.598 30R

° +

PROBLEM

A steel tank is hat the magrigonometry (he two force

corresponding

180180 (90sin 60500 lb

αα

= °= ° − += ° −

°=

47.402= °

500 lb0 ) sin 60

=° °

2.12

to be positiongnitude of Pa) the requires applied at magnitude of

30 ) 60° − °

ned in an excavP is 500 lb,d angle α if thA is to be v

f R.

vation. Knowi, determine he resultant Rvertical, (b)

42.6α = °

551 lbR =

ing by

R of the


SOLUT

The sma

(a) P

(b) R

TION

allest force P w

(425 lb)coP =

(425 lb)sinR =

will be perpen

os30°

n 30°

P

ADdoc

ndicular to R.

PROBLEM

A steel tank Determine bydirection of theof the two fcorresponding

2.13

is to be py trigonometre smallest forcforces appliedmagnitude of

positioned in ry (a) the ce P for whichd at A is vf R.

an excavatimagnitude a

h the resultantvertical, (b)

368 lb=P

21R =

on. and t R the

13 lb


SOLUT

The sma

(a) P

(b) R

TION

allest force P w

(50 N)sinP =

(50 N)cosR =

P

FmRc

will be perpen

25°

25°

PROBLEM

For the hook smagnitude andR of the twoorresponding

ndicular to R.

2.14

support of Prod direction of to forces applimagnitude of

ob. 2.10, deterthe smallest foied to the su

f R.

rmine by trigoforce P for whupport is hor

onometry (a) hich the resultrizontal, (b)

21.1 N=P

45.3 NR =

the tant the


PROBLEM 2.15

For the hook support shown, determine by trigonometry the magnitude and direction of the resultant of the two forces applied to the support.

SOLUTION

Using the law of cosines:

2 2 2(200 lb) (300 lb)

2(200 lb)(300 lb)cos (45 65 )413.57 lb

R

R

= +

− + °=

Using the law of sines: sin sin (45 65 )300 lb 413.57 lb

42.972

α

α

+ °=

= °

90 25 42.972β = + − ° 414 lb=R 72.0°


PROBLEM 2.16

Solve Prob. 2.1 by trigonometry.

PROBLEM 2.1

Two forces are applied as shown to a hook. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION


2 2 2(900 N) (600 N )2(900 N )(600 N)cos (135 )

1390.57N

R

R

= +− °

=

Using the law of sines: sin( 30 ) sin (135 )600N 1390.57N

30 17.7642

47.764

α

α

α

− °=

− = °

=

1391N=R 47.8°


SOLUT

Using th

We hav

Then

And

TION

he force triang

e:

P

So

PRA.Q re(b

gle and the law

180105

γ ==

2 (4

648.0

R

R

=

==

4 kipsin(25sin(25

25

°°°

ROBLEM 2

olve Problem 2

ROBLEM 2.4. Knowing tha

= 4 kips, desultant force

b) the triangle r

ws of cosines a

0 (50 255° − ° + °°

2

2

kips) (6 kip

.423 kips0264 kips

+

ps 8.0264) sin1) 0.4813

28.7753.775

αααα

=++ =° + =

= °

2.17

2.4 by trigono

4 Two structuat both memb

etermine graphexerted on thrule.

and sines:

)°

2ps) 2(4 kip−

4 kips05

375

°

°°

ometry.

ural members bers are in tenhically the ma

he bracket usin

s)(6 kips)cos1

B and C are bnsion and thatagnitude and ng (a) the par

105°

8.03 k=R

bolted to bract P = 6 kips adirection of rallelogram la

kips 3.8°

ket and the aw,


PROBLEM 2.18

For the stake of Prob. 2.5, knowing that the tension in one rope is 120 N, determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 160 N.

PROBLEM 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant.

SOLUTION

Using the laws of cosines and sines:

2 2 2(120 N) (160 N) 2(120 N)(160 N)cos 2572.096 N

PP

= + − °=

And sin sin 25120 N 72.096 N

sin 0.7034344.703

α

αα

°=

== °

72.1 N=P 44.7°


SOLUT

Using th

We have

Then

and

Hence:

TION

he force triang

e

gle and the law

R

48 sinsin

PROBLE

Two forces Knowing ththe magnitu

ws of cosines a

180 (2150

γ = ° −= °

2 2(48 N)2(48 N

104.366 N

R

R

= +−

=

N 104.366 Nsin150

n 0.2299613.2947

ααα

=°

== °

180180 1386.705

φ α= ° −= ° −= °

M 2.19

P and Q are ahat P = 48 N ude and directi

and sines:

20 10 )° + °

2(60 N)N)(60 N)cos15

N

+

N°

803.2947 80− °

° − °

applied to the and Q = 60 N

ion of the resu

50°

lid of a storagN, determine

ultant of the tw

104.=R

ge bin as showby trigonome

wo forces.

4 N 86.7°

wn. etry

°


SOLUT

Using th

We have

Then

and

Hence:

TION

he force triang

e

P

TKm

gle and the law

γ ==

2R

R

=

=

60 Nsinsin

ααα

=

==

18018083.2

φ = °= °=

PROBLEM

Two forces PKnowing thatmagnitude and

ws of cosines a

180 (20150

= ° − ° += °

2(60 N) (42(60 N)(48

104.366 N

= +−

=

104.366 Nsin150

0.2874516.7054

=°

== °

18016.7054

295

α° − − °° − ° −

°

M 2.20

P and Q are ap P = 60 N andd direction of

and sines:

10 )+ °

248 N)8 N)cos 150°

80°

pplied to the ld Q = 48 N, dthe resultant o

lid of a storagdetermine by of the two forc

104.4=R

ge bin as showtrigonometry ces.

4 N 83.3°

wn. the


SOLUT

Comput

29-lb Fo

50-lb Fo

51-lb Fo

TION

te the followin

orce:

orce:

orce:

ng distances:

PROB

Determi

2

2

2

(84)116 in.

(28)100 in.

(48)102 in.

OA

OB

OC

=

=

=

=

=

=

(29 lbxF = +

(29 lbyF = +

(50 lbxF = −

(50 lbyF = +

(51 lbxF = +

(51 lbyF = −

BLEM 2.21

ine the x and y

2 2

2 2

2 2

(80).

(96).

(90).

+

+

+

84b)116

80b)116

28b)100

96b)100

48b)102

90b)102

y components of each of the

F

F

F

F

e forces shown

21.0 lbxF = +

20.0 lbyF = +

14.00 lbxF = −

48.0 lbyF = +

24.0 lbxF = +

45.0 lbyF = −

n.


SOLUT

Comput

800-N F

424-N F

408-N F

TION

te the followin

Force:

Force:

Force:

ng distances:

PROBL

Determin

2

(600)1000 m

(560)1060 m

(480)1020 m

OA

OB

OC

=

=

=

=

=

=

(800 NxF = +

(800 NyF = +

(424 NxF = −

(424 NyF = −

(408 NxF = +

(408 NyF = −

LEM 2.22

ne the x and y

2 2

2 2

2 2

(800)mm

(900)mm

(900)mm

+

+

+

800N)1000

600N)1000

560N)1060

900N)1060

480N)1020

900N)1020

components oof each of the

F

forces shown.

640 NxF = +

480 NyF = +

224 NxF = −

360 NyF = −

192.0 NxF = +

360 NyF = −

.

N

N

N


PROBLEM 2.23

Determine the x and y components of each of the forces shown.

SOLUTION

80-N Force: (80 N)cos40xF = + ° 61.3 NxF =

(80 N)sin 40yF = + ° 51.4 NyF =

120-N Force: (120 N)cos70xF = + ° 41.0 NxF =

(120 N)sin 70yF = + ° 112.8 NyF =

150-N Force: (150 N)cos35xF = − ° 122. 9 NxF = −

(150 N)sin 35yF = + ° 86.0 NyF =


PROBLEM 2.24


SOLUTION

40-lb Force: (40 lb)cos60xF = + ° 20.0 lbxF =

(40 lb)sin 60yF = − ° 34.6 lbyF = −

50-lb Force: (50 lb)sin50xF = − ° 38.3 lbxF = −

(50 lb)cos50yF = − ° 32.1 lbyF = −

60-lb Force: (60 lb)cos25xF = + ° 54.4 lbxF =

(60 lb)sin 25yF = + ° 25.4 lbyF =


PROBLEM 2.25

Member BC exerts on member AC a force P directed along line BC. Knowing that P must have a 325-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component.

SOLUTION

2 2(650 mm) (720 mm)970 mm

BC = +

=

(a) 650970xP P⎛ ⎞= ⎜ ⎟

⎝ ⎠

or 970650

970325 N650

485 N

xP P ⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

485 NP =

(b) 720970

720485 N970

360 N

yP P ⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

970 NyP =


SOLUT

(a)

(b) V

TION

Vertical compo

PRO

MembKnowmagni

onent

OBLEM 2.26

ber BD exertswing that P muitude of the fo

sin35P ° =

P =

vP =

=

6

s on memberust have a 300-orce P, (b) its v

300 lb

300 lbsin 35°

cos35P °

(523 lb)cos35

r ABC a forc-lb horizontal vertical compo

5°

ce P directed component, d

onent.

along line Bdetermine (a) t

523 lbP =

428 lbvP =

BD. the


SOLUT

(a)

(b)

TION

PR

Thdircomma

180 4518015

α° = °

== °

cos

cos60cos621

xPPP

P

α =

=

=

=

tan

t

(60160

y

x

y x

PP

P P

α =

=

==

ROBLEM 2

he hydraulic cyrected along limponent perpagnitude of the

90 3045 90

α+ + ° +° − ° − ° −

s00 Ns151.17 N

xPα

°

tan

00 N) tan150.770 N

α

°

2.27

ylinder BC exine BC. Know

pendicular to me force P, (b)

030°

− °

xerts on membwing that P mumember AB, dits component

ber AB a forceust have a 600determine (a) t along line AB

621 NP =

160.8 NyP =

e P 0-N the B.


SOLUT

(a)

(b)

TION

PRO

Cablethat Pof the

OBLEM 2.2

e AC exerts oP must have ae force P, (b)

cos 55yP

P =°

350 lbcos 55610.21

=°

=

sin 55xP P=

(610.21499.85

==

28

on beam AB aa 350-lb verticits horizontal

°

lb

5°

1 lb)sin 55lb

°

a force P direcal componentcomponent.

cted along lint, determine (a

ne AC. Knowia) the magnitu

610 lbP =

500 lbxP =

ing ude


SOLUT

(a)

(b)

TION

P

ThalpeP,

PROBLEM 2

he hydraulic long line BDerpendicular to, (b) its compo

750 N sinP=

2192P =

cosABCP P=

(219=

2.29

cylinder BD eD. Knowing o member ABonent parallel

n 20°

2.9 N

s20°

2.9 N)cos 20°

exerts on memthat P mustC, determine to ABC.

°

mber ABC a t have a 75(a) the magni

AP

force P direc0-N compontude of the fo

2190 NP =

2060 NABC =

ted ent rce


PROBLEM 2.30

The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P must have a 720-N component perpendicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC.

SOLUTION

(a) 371237 (720 N)122220 N

=

=

=

xP P

2.22 kNP =

(b) 351235 (720 N)122100 N

y xP P=

=

=

2.10 kN=yP


SOLUT

Compon

TION

nents of the fo

F

2

5

5

orces were dete

Force x

29 lb

50 lb

51 lb

R

tan

PROBLE

Determine t

PROBLEMforces show

ermined in Pro

x Comp. (lb)

+21.0

–14.00

+24.0

31.0xR = +

(31.0 lb)

n

23.031.036.573

23.0 lbsin (36.57

x y

y

x

R R

RR

R

α

α

= +

=

=

=

= °

=

i j

i

R

38.601 lb=

EM 2.31

the resultant o

M 2.21 Determwn.

oblem 2.21:

y Comp

+20

+4

–4

2yR = +

(23.0 lb)

b73 )

+

°

i j

b

of the three for

mine the x and

p. (lb)

0.0

8.0

5.0

3.0

rces of Problem

d y componen

38.6=R

m 2.21.

nts of each of

6 lb 36.6°

the


SOLUT

Compon

TION

nents of the fo

F

1

1

orces were det

orce x

80 N

20 N

50 N

R

PROBLE

Determine

PROBLEMforces show

termined in Pr

x Comp. (N)

+61.3

+41.0

–122.9

20.6xR = −

( 20.

tan

250.2tan20.6

tan 12.1485.29

250sin85

x

y

x

R

RR

R

α

α

αα

= +

= −

=

=

==

=

R i

EM 2.32

the resultant o

M 2.23 Determwn.

roblem 2.23:

y Com

+

+1

+

2yR = +

.6 N) (250.2

2 N6 N456930.2 N5.293

yR+

+

°

°

j

i

of the three for

mine the x and

mp. (N)

+51.4

112.8

+86.0

250.2

2 N)j

rces of Proble

d y componen

25=R

em 2.23.

nts of each of

1 N 85.3°

the

°


SOLUT

TION

F

4

5

6

orce x

40 lb

50 lb

60 lb

xR

tan

tan

tan

R

α

α

αα

R

PROBL

Determin

PROBLEthe forces

x Comp. (lb)

+20.00

–38.30

+54.38

36.08x = +

( 36.08 lb

41.42 lb36.08 lb1.1480048.942

41.42 lbsin 48.942

x y

y

x

R R

RR

R

α

α

αα

= +

= +

=

=

== °

=°

R i j

LEM 2.33

e the resultant

EM 2.24 Detes shown.

y Comp.

–34.6

–32.1

+25.3

41.4yR = −

) ( 41.42 lb+ −

°

i

t of the three f

ermine the x

(lb)

64

14

36

42

b)j

forces of Prob

and y compon

54.9=R

lem 2.24.

nents of each

9 lb 48.9°

h of


SOLUT

Compon

TION

nents of the fo

F

80

42

40

orces were det

orce x

00 lb

24 lb

08 lb

R

tan

PROBL

Determine

PROBLEforces sho

ermined in Pr

x Comp. (N)

+640

–224

+192

608xR = +

(608 lb)

n

24060821.541

240 Nsin(21.54653.65 N

x y

y

x

R R

RR

R

α

α

= +

=

=

=

= °

=

=

R i j

i

LEM 2.34

e the resultant

EM 2.22 Deterown.

oblem 2.22:

y Comp. (

+480

–360

–360

240yR = −

( 240 lb)

N41°)

N

+ −

j

j

t of the three f

rmine the x an

(N)

0

0

0

0

forces of Probl

nd y componen

654=R

lem 2.22.

nts of each of

4 N 21.5°

the


SOLUT

100-N F

150-N F

200-N F

TION

Force:

Force:

Force:

Force

100 N

150 N

200 N

PRO

Knowforces

(100 (100

x

y

FF

= +

= −

(150 (150

x

y

FF

= +

= −

(200 (200

x

y

FF

= −

= −

x Com

+

+

−1

xR = −

ta

OBLEM 2.35

wing that α =s shown.

N)cos35N)sin 35

° = +

° = −

N)cos65N)sin 65

° = +

° = −

N)cos35N)sin 35

° = −

° = −

mp. (N)

81.915

63.393

63.830

18.522

( 18.52

an

308.0218.52286.559

x y

y

x

R R

RR

α

α

= +

= −

=

=

= °

R i

308.02 sin86.5

R =

5

35°, determi

81.915 N57.358 N

+

−

63.393 N135.946 N

+

−

163.830 N114.715 N

−

−

y Comp. (N

−57.35

−135.94

−114.71

308.02yR = −

22 N) ( 308y

+ −

j

i

N59

ine the result

N)

58

46

15

2

8.02 N)j

309=R

tant of the th

9 N 86.6°

hree


SOLUT

Determi

Cable fo

500-N F

200-N F

and

Further:

TION

ine force comp

orce AC:

Force:

Force:

ponents:

(365 N

(365 N

= −

= −

x

y

F

F

(500 N)

(500 N)

x

y

F

F

=

=

(200 N)

(200 N

x

y

F

F

=

= −

2

(400 N474.37

= Σ = −

= Σ = −

= +

=

=

x x

y y

x

R FR F

R R R

255tan40032.5

α

α

=

= °

PRKnoresu

960N) 2414601100N) 271460

= −

= −

24) 480 N257) 140 N25

=

=

4) 160 N5

3N) 120 N5

=

= −

2

2

240 N 480275 N 140 N

N) ( 255 NN

− +

− +

+ −

yR

°

ROBLEM 2.owing that theultant of the th

40 N

75 N

N

2

N 160 N 4N 120 N

N)

+ =

− = −

36 tension in rop

hree forces exe

400 N255 N−

pe AC is 365 Nerted at point C

474=R

N, determine C of post BC.

4 N 32.5°

the


SOLUT

60-lb Fo

80-lb Fo

120-lb F

and

Further:

TION

orce:

orce:

Force:

(60 l(60 l

x

y

FF

=

=

(80 l(80 l

x

y

FF

=

=

(120(12

x

y

FF

=

= −

(20202.

x x

y y

R FR F

R

= Σ

= Σ

=

=

29.tan200

α =

tan

8.46

α −=

= °

PROKnowiforces

lb)cos 20 5lb)sin 20 20

° =

° =

lb)cos60 4lb)sin 60 69

° =

° =

0 lb)cos30 120 lb)sin 30

° =

° =

2

200.305 lb29.803 lb

00.305 lb) (.510 lb

=

=

+

803.305

1 29.803200.305

°

BLEM 2.37ing that α =shown.

6.382 lb0.521 lb

0.000 lb9.282 lb

103.923 lb60.000 lb= −

2(29.803 lb)

7 40°, determiine the result

203=R

ant of the th

3 lb 8.46°

hree


SOLUT

60-lb Fo

80-lb Fo

120-lb F

Then

and

TION

orce:

orce:

Force:

tan

tan

PROKnowforce

(60 lb) co(60 lb)si

x

y

FF

=

=

(80 lb) co(80 lb)si

x

y

FF

=

=

(120 lb) c(120 lb)s

x

y

FF

=

=

1611

x x

y y

R FR F

= Σ =

= Σ =

(168.95201.976

R =

=

110.676n168.953

n 0.6550733.228

α

αα

=

== °

OBLEM 2.3wing that α =es shown.

os 20 56.38in 20 20.52

° =

° =

os 95 6.97n 95 79.696

° = −

° =

cos 5 119.54sin 5 10.459

° =

° =

68.953 lb10.676 lb

253 lb) (110. lb

+

38 = 75°, determ

82 lb1 lb

725 lb6 lb

43 lb9 lb

2.676 lb)

mine the result

202=R

tant of the th

2 lb 33.2°

ree


PROBLEM 2.39 For the collar of Problem 2.35, determine (a) the required value of α if the resultant of the three forces shown is to be vertical, (b) the corresponding magnitude of the resultant.

SOLUTION

(100 N)cos (150 N)cos( 30 ) (200 N)cos

(100 N)cos (150 N)cos( 30 )

x x

x

R F

Rα α α

α α

= Σ

= + + ° −= − + + ° (1)

(100 N)sin (150 N)sin ( 30 ) (200 N)sin(300 N)sin (150 N)sin ( 30 )

y y

y

R F

Rα α αα α

= Σ

= − − + ° −= − − + ° (2)

(a) For R to be vertical, we must have 0.xR = We make 0xR = in Eq. (1):

100cos 150cos ( 30 ) 0

100cos 150(cos cos 30 sin sin 30 ) 029.904cos 75sin

α αα α α

α α

− + + ° =− + ° − ° =

=

29.904tan75

0.3987221.738

α

α

=

== ° 21.7α = °

(b) Substituting for α in Eq. (2):

300sin 21.738 150sin51.738

228.89 NyR = − ° − °

= −

| | 228.89 NyR R= = 229 NR =


PROBLEM 2.40 For the post of Prob. 2.36, determine (a) the required tension in rope AC if the resultant of the three forces exerted at point C is to be horizontal, (b) the corresponding magnitude of the resultant.

SOLUTION

960 24 4(500 N) (200 N)1460 25 5

48 640 N73

x x AC

x AC

R F T

R T

= Σ = − + +

= − + (1)

1100 7 3(500 N) (200 N)1460 25 5

55 20 N73

y y AC

y AC

R F T

R T

= Σ = − + −

= − + (2)

(a) For R to be horizontal, we must have 0.yR =

Set 0yR = in Eq. (2): 55 20 N 073 ACT− + =

26.545 NACT = 26.5 NACT =

(b) Substituting for ACT into Eq. (1) gives

48 (26.545 N) 640 N73

622.55 N623 N

= − +

=

= =

x

x

x

R

RR R 623 NR =


SOLUT

Using th

(a) Se

(b) Su

TION

he x and y axes

et 0yR = in E

ubstituting for

PRO

Deteof thBC,

s shown:

Eq. (2):

r ACT in Eq. (1

OBLEM 2.4

ermine (a) thehe three forces(b) the corresp

x xR F= Σ

93.6y y

y

R F

R

= Σ

=

93.631 lb −

1):

xR

R

41

e required tenss exerted at Pponding magn

sin10sin10

AC

AC

TT

= ° +

= ° +

(50 lb)sin 3

31 lb coACT

=

−

cos10AC

AC

TT

° =

=

(95.075 lb)94.968 lb

xR

=

==

sion in cable APoint C of boonitude of the r

(50 lb)cos3578.458 lb

+

+

5 (75 lb)sin

s10

° +

°

095.075 lb

sin10 78.45° +

AC, knowing tom BC must besultant.

5 (75 lb)co° +

n 60 coACT° −

58 lb

that the resultbe directed alo

os60°

s10°

95.1 lbACT =

95.0 lbR =

tant ong

(1)

(2)


SOLUT

Select th

Then

and

(a) Se

D

(b) Su

TION

he x axis to be

et 0yR = in E

Dividing each t

ubstituting for

R

e along a a′.

xR

yR

Eq. (2).

(80 lb

term by cosα

r α in Eq. (1)

60 lb (80xR = +

PROFor threquirto be pthe res

(60 lxF= Σ =

(80 lyF= Σ =

b)sin (120α −

gives:

(80

) gives:

0 lb)cos56.31°

OBLEM 2.42he block of Ped value of αparallel to the sultant.

lb) (80 lb)co+

lb)sin (120α −

lb)cos 0α =

0 lb) tan 1212tan8

56

α

α

α

=

=

=

(120lb)sin° +

2 Problems 2.37α if the resultan

incline, (b) th

os (120 lb)α +

0 lb)cosα

20 lb20 lb80 lb6.310°

n56.31 204.° =

7 and 2.38, dent of the threehe correspondi

sinα

.22 lb

etermine (a) e forces showning magnitude

56.3α = °

204 lbxR =

the n is e of

(1)

(2)


PROBLEM 2.43 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION

Free-Body Diagram Force Triangle

Law of sines:

400 lbsin 60 sin 40 sin80

AC BCT T= =

° ° °

(a) 400 lb (sin 60 )sin80ACT = °

° 352 lbACT =

(b) 400 lb (sin 40 )sin80BCT = °

° 261 lbBCT =


PROBLEM 2.44 Two cables are tied together at C and are loaded as shown. Knowing that α = 30°, determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION


Law of sines:

6 kN

sin 60 sin 35 sin85AC BCT T

= =° ° °

(a) 6 kN (sin 60 )

sin85ACT = °°

5.22 kNACT =

(b) 6 kN (sin 35 )

sin85BCT = °°

3.45 kNBCT =


PROBLEM 2.45

Two cables are tied together at C and loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION

Free-Body Diagram

1.4tan4.816.26021.6tan3

28.073

α

α

β

β

=

= °

=

= °

Force Triangle

Law of sines:

1.98 kN

sin 61.927 sin 73.740 sin 44.333AC BCT T

= =° ° °

(a) 1.98 kN sin 61.927

sin 44.333ACT = °°

2.50 kNACT =

(b) 1.98 kN sin 73.740

sin 44.333BCT = °°

2.72 kNBCT =


SOLUT

Law of s

(a)

(b)

TION

Free-Bod

sines:

dy Diagram

PROB

Two cabKnowing(a) in ca

sin 35 sACT

=°

5sACT =

5siBCT =

LEM 2.46

bles are tied g that P = 50

able AC, (b) in

For

500 Nsin 75 sin 70

BCT=

°

500 N sin 35in 70

°°

500 N sin 75in 70

°°

together at C00 N and α =n cable BC.

rce Triangle

N0°

C and are lo60°, determin

oaded as showne the tension

305 NACT =

514 NBCT =

wn. n in


SOLUT

Law of s

(a)

(b)

TION

Free-Bod

mg

(20196

W =

==

sines:

PRO

Twotensi

dy Diagram

00 kg)(9.81 m/62 N

sinAT

T

T

OBLEM 2.4

cables are tieion (a) in cabl

2/s )

15 sin 105AC BCT

=° °

(1962 NsinACT =

(1962 NsinBCT =

47

ed together at e AC, (b) in ca

F

1962 Nsin 60

=° °

N) sin 15n 60

°°

N)sin 105n 60

°°

C and are loaable BC.

Force Triang

aded as shown

gle

T

n. Determine

586 NACT =

2190 NBCT =

the


SOLUT

Law of s

(a)

(b)

TION

Free

sines:

e-Body Diagra

sin

PKAC

am

n 110 sin 5AC BCT T

=°

1200sin 6ACT =

1200sin 6BCT =

PROBLEM 2Knowing that α

C, (b) in rope

1200 lb5 sin 65

=° °

0 lb sin 11065

°°

lb sin 565

°°

2.48 20 ,α = ° dete

BC.

Force Triang

rmine the ten

gle

T

T

sion (a) in ca

1244 lbACT =

115.4 lbBCT =

ble


SOLUT

xFΣ

For P =

yFΣ

Solvin

TION

0 CAT= −

200N= we ha

0=

0.8

ng equations (

PRO

Two KnowBC.

sin30 CBT+

ve,

0.5 CAT−

coCAT

86603 0.8CAT +

1) and (2) sim

BLEM 2.49

cables are wing that P =

Free-B

sin30 cosP−

0.5 21A CBT+ +

os30 coCBT° −

86603 21CBT −

multaneously g

9

tied togeth= 300 N, de

Body Diagram

s 45 200N° − =

12.13 200− =

s30 sin 4P−

12.13 0= (2)

ives,

her at C anetermine the

m

0=

0 (1)

5 0=

)

nd are loadtension in c

T

T

ded as showcables AC a

134.6 NCAT =

110.4 NCBT =

wn. and


SOLUT

xFΣ

For CAT

yFΣ

Adding

Substitu

Andrange of

TION

0 CAT= −

0= we have,

0= cCAT

equations (1)

uting for CBT =

d 546.40CAT =f 179.315 N an

PRO

Two Detertaut.

sin30 CBT+

,

0.5 CBT +

cos30 cCBT° −

0.866

and (2) gives

0= into the eq

0N , 669P =nd 669.20 N.

BLEM 2.50

cables are rmine the ran

Free-B

sin30 cosP−

0.70711 2P+ −

cos30 sinP−

603 0.707CBT −

s, 1.36603 CT

quilibrium equ

0.50.86603

CAT−

9.20N Thus fo

0

tied togethnge of value

Body Diagram

s45 200N° − =

200 0= (1)

45 0= ; aga

711 0P = (2)

200CB = henc

uations and sol

0.707110.70711CA

PT+ −

−

for both cables

her at C anes of P for

m

0=

ain setting CAT

)

ce 146.4CBT =

lving simultan

200 01 0P− =

=

s to remain ta

nd are loadwhich both

0A = yields,

410N and P =

neously gives,

aut, load P mu

179.3 N

ded as showcables rema

179.315N=

ust be within

N < < 669 NP

wn. ain

the


SOLUT

Resolvin

Substitu

In the y-

Thus,

In the x-

Thus,

TION

ng the forces i

uting compone

-direction (one

-direction:

into x- and y-d

=R

ents: =R

e unknown for

−

(65

P

Twcoanm

directions:

A= + + +P Q F

(500 lb)[(650 lb)si

( cB AF F

= − +−+ −

j

i

rce):

500 lb (650− −

500 lbAF +

=

1302.70=

50 lb)cos50° +

cos50(1302.70

B AF F==

419.55 lb=

PROBLEM 2

wo forces P aonnection. Knnd that P =

magnitudes of t

0B+ =F

[(650 lb)cos5in 50 ]cos50 ) ( AF

°° +ji

lb)sin50 F° +

(650 lb)sin 50sin 50

+°

lb

cos5B AF F+ −

0 (650 lb)c0 lb)cos50

° −° −

b

2.51

and Q are apnowing that th

500= lb and the forces exer

50 ]

sin 50 ) 0A

°

° =

i

j

sin50 0AF ° =

0°

50 0° =

cos50(650 lb)cos50

°

pplied as showe connection i

650Q = lb, rted on the rod

Free

0°

wn to an aircris in equilibriu

determine ds A and B.

e-Body Diagr

1303 lbAF =

420 lbBF =

raft um the

ram


SOLUT

Resolvin

Substitu

In the x-

In the y-

TION

ng the forces i

uting compone

-direction (one

-direction:

into x- and y-d

=R P

ents: = −−+

R

e unknown for

cos 5Q

P− −

P ===

P

Tceod

directions:

A+ + +P Q F F

cos 50[(750 lb)cos[(750 lb)sin

P Q− +−+

j

rce):

50 [(750 lb)° −

(750 lb

127.710

Q =

=

sin 50Q− ° +

sin 50(127.710 lb

476.70 lb

Q= − ° += −=

PROBLEM

Two forces Pconnection. equilibrium anon rods A adetermine the

0B =F

0 sin 5050 ]50 ] (400 lb

Q° − °°° +

iij

cos 50 ] 400° +

b)cos 50 40cos 50

0 lb

° −°

(750 lb)sin 5

(750 lb)sin 5b)sin 50 (75+

° +

M 2.52

and Q are apKnowing th

nd that the maand B are Fmagnitudes o

b)

°j

i

0 lb 0=

00 lb

0 0° =

5050 lb)sin 50

°°

pplied as showhat the conagnitudes of th

750AF = lb anof P and Q.

Free-Bo

477 lb;P =

wn to an aircrnnection is he forces exernd 400BF =

ody Diagram

127.7 lbQ =

raft in

rted lb,


SOLUT

With

With FA

TION

A and FB as abo

F

FΣ

FF

F

FΣ

ove: F

PR

A wthe

BF =forc

Free-Body Di

30:5x BF F=

8 kN16 kN

A

B

FF

==

4 (16 kN)5CF =

0:y DF F= −

3 (16 kN)5DF =

ROBLEM 2.

welded connecfour forces 16= kN, dete

ces.

iagram of Co

35C AF F− − =

4) (8 kN)5

−

3 35 5B AF F+ −

3) (8 kN)5

−

53

ction is in equshown. Kn

ermine the m

nnection

0=

0A =

uilibrium undnowing that magnitudes of

F

F

der the action 8AF = kN a

f the other t

6.40 kNCF =

4.80 kNDF =

of and two


SOLUT

or

With

TION

F

yFΣ

BF

AF

BF

xFΣ

CF

PRO

A welfour fodeterm

Free-Body Di

0: DF= − −

35D AF F= +

5 kN, DF=

5 36 kN3 5

⎡= +⎢⎣

0: CF= − +

4 (54 (15 kN5

B AF F= −

= −

BLEM 2.54

lded connectioforces shown. mine the magn

iagram of Co

3 35 5A BF F+ =

8 kN=

3 (5 kN)5

⎤⎥⎦

4 45 5B AF F− =

)

5 kN)

4

on is in equiliKnowing tha

nitudes of the o

nnection

0=

0=

ibrium under tat 5AF = kN other two forc

F

F

the action of and 6DF = kes.

15.00 kNBF =

8.00 kNCF =

the kN,


SOLUT

(a) Su

(b) Fr

TION

ubstitute (1) in

rom (1):

0:xFΣ =

0:yFΣ =

nto (2): 0.6

AT

T

P

Athoncoanbode(b

Free-B

: cos 10ACBT

0.137CDT =

: sin 10ACBT °

0.67365 AT

67365 0ACBT +

269.46 lbCB =

0.137158CDT =

PROBLEM 2

A sailor is behat is suspenn the supponstant speend β = 10° oatswain’s etermine theb) in the trac

Body Diagram

0 cos 3ACBT° −

7158 ACBT

sin 30ACBT° +

0.5CB CDT+ =

0.5(0.137158T

b

8(269.46 lb)

2.55

eing rescued nded from a ort cable Ad by cable Cand that thechair and

e tension (a) ction cable C

m

30 cos 3CDT° −

0 sin 30CDT° +

200

) 200ACBT =

using a boapulley that

ACB and isCD. Knowine combined

the sailorin the suppo

CD.

30 0° =

0 200 0° − =

T

T

atswain’s chcan roll frees pulled at

ng that α = 3weight of t

r is 200 ort cable AC

269 lbACBT =

37.0 lbCDT =

hair ely

a 30° the lb,

CB,

(1)

(2)


SOLUT

TION

PAsucaKcathsu

Free-B

0:x AF TΣ =

AT

0: (30yFΣ =

+

PROBLEM A sailor is beinuspended fromable ACB and

Knowing that αable CD is 20he boatswain’support cable A

Body Diagram

cos 15ACB T° −

304.04 lACB =

04.04 lb)sin 1

(20 lb)sin 25+

2.56 ng rescued usim a pulley thatd is pulled at aα = 25° and β0 lb, determins chair and th

ACB.

m

cos 25ACBT ° −

lb

5 (304.04 l° +

5 0215.6

WW

° − ==

ing a boatswat can roll freela constant speβ = 15° and thne (a) the comhe sailor, (b) th

(20 lb)cos 25

lb)sin 25°

64 lb

(a)

(b) T

ain’s chair thatly on the supp

eed by cable Chat the tensionmbined weighthe tension in

5 0° =

216 lbW =

304 lbACBT =

t is port CD. n in t of the


PROBLEM 2.57

For the cables of prob. 2.44, find the value of α for which the tension is as small as possible (a) in cable bc, (b) in both cables simultaneously. In each case determine the tension in each cable.

SOLUTION


(a) For a minimum tension in cable BC, set angle between cables to 90 degrees. By inspection, 35.0α =

(6 kN)cos35ACT = 4.91 kNACT =

(6 kN)sin35BCT = 3.44 kNBCT =

(b) For equal tension in both cables, the force triangle will be an isosceles.

Therefore, by inspection, 55.0α =

6 kN(1 / 2)

cos35AC BCT T= =°

3.66 kNAC BCT T= =


SOLUT

(a) L

(b) L

TION

F

Law of cosines

Law of sines

Free-Body D

2P

P

sin600 N

β

β

PROB

For the allowablDetermi(b) the c

iagram

2 2(600) (7= +

784.02 NP =

sin (25 4N 784.02 N

° +=

46.0β = °

LEM 2.58

cables of Prole tension is 6ne (a) the ma

corresponding

2750) 2(600−

45 )N

°

46.0α∴ = °

oblem 2.46, it600 N in cablaximum forcevalue of α.

For

)(750)cos(25°

25° + °

t is known thae AC and 750e P that can b

rce Triangle

45 )° + °

at the maximu0 N in cable Bbe applied at

784 NP =

71.0α = °

um BC.

C,


SOLUT

To be sm

(a) T

(b)

TION

F

mallest, BCT m

Thus,

Free-Body D

must be perpen

α =

BCT =

P

Fthas

iagram

ndicular to the

5.00= °

(1200 lb)sin=

PROBLEM

or the situatiohe value of α fs possible, (b)

e direction of

n 5°

2.59

on described ifor which the ) the correspon

For

.ACT

n Figure P2.4tension in rop

nding value of

rce Triangle

α

T

48, determine pe BC is as smf the tension.

5.00α = °

104.6 lbBCT =

(a) mall


SOLUT

Free-Bo

TION

ody Diagram

PROBLE

Two cablesrange of vaeither cable

Requireme

From Eq. (

Requireme

From Eq. (

EM 2.60

s tied togetheralues of Q for e.

0:xFΣ =

0:yFΣ =

ent:

(2): sQ

ent:

(1): 75 lb −

r at C are loawhich the ten

coBCT Q− −

75 lbBCT =

sinACT Q−

sinACT Q=

60 lbACT =

sin 60 60 lb° =

69.3 Q =

60 lbBCT =

cos60 6Q− ° =

30Q =

aded as shownnsion will not

os60 75 lb° + =

b cos60Q− °

60 0° =

n60°

b:

lb

b:

0 lb

0.0 lb 30.0 lb

n. Determine t exceed 60 lb

0=

69.3 lbQ≤ ≤

the b in

(1)

(2)


PROBLEM 2.61

A movable bin and its contents have a combined weight of 2.8 kN. Determine the shortest chain sling ACB that can be used to lift the loaded bin if the tension in the chain is not to exceed 5 kN.

SOLUTION

Free-Body Diagram

tan0.6 m

α =h (1)

Isosceles Force Triangle

Law of sines: 12

12

(2.8 kN)sin

5 kN(2.8 kN)

sin5 kN

16.2602

AC

AC

TT

α

α

α

=

=

=

= °

From Eq. (1): tan16.2602 0.175000 m0.6 m

h h° = ∴ =

Half-length of chain 2 2(0.6 m) (0.175 m)0.625 m

AC= = +

=

Total length: 2 0.625 m= × 1.250 m


PROBLEM 2.62

For W = 800 N, P = 200 N, and d = 600 mm, determine the value of h consistent with equilibrium.

SOLUTION

Free-Body Diagram

800 NAC BCT T= =

( )2 2AC BC h d= = +

2 2

0: 2(800 N) 0yhF P

h dΣ = − =

+

2

800 12P d

h⎛ ⎞= + ⎜ ⎟⎝ ⎠

Data: 200 N, 600 mmP d= = and solving for h

2200 N 600 mm800 N 1

2 h⎛ ⎞= + ⎜ ⎟⎝ ⎠

75.6 mmh =

SOLUT

(a) F

(b) F

TION

Free Body: Co

Free Body: Co

ollar A

ollar A

PROBLE

Collar A is cslide on a magnitude equilibrium

15 in.x =

Forc

4.5P

Forc

15P

EM 2.63

connected as sfrictionless hof the force of the colla

ce Triangle

50 lb5 20.5

=

ce Triangle

50 lb25

=

shown to a 50horizontal rod

P required ar when (a)

0-lb load and cd. Determine

to maintain 4.5 in.,x =

10.98 lbP =

30.0 lbP =

can the the (b)


SOLUT

Similar

TION

Free Bo

Triangles

ody: Collar A

A

2 (5014

NN

==

4820 in. 14

x=

PROBLEM

Collar A is cslide on a fdistance x foP = 48 lb.

F

2 20) (48) 1.00 lb

− =

8 lb4 lb

M 2.64

connected as sfrictionless hoor which the

orce Triangle

196

shown to a 50orizontal rod.collar is in eq

e

0-lb load and c. Determine quilibrium wh

68.6 in.x =

can the hen


PROBLEM 2.65

Three forces are applied to a bracket as shown. The directions of the two 150-N forces may vary, but the angle between these forces is always 50°. Determine the range of values of α for which the magnitude of the resultant of the forces acting at A is less than 600 N.

SOLUTION

Combine the two 150-N forces into a resultant force Q:

2(150 N)cos25271.89 N

Q = °=

Equivalent loading at A:


2 2 2(600 N) (500 N) (271.89 N) 2(500 N)(271.89 N)cos(55 )cos(55 ) 0.132685

αα

= + + ° +° + =

Two values for :α 55 82.37527.4

αα

° + == °

or

55 82.37555 360 82.375

222.6

ααα

° + = − °° + = ° − °

= °

For 600 lb:R < 27.4 222.6α° < <


SOLUT

Free-Bo

TION

ody Diagram:

PROBL

A 200-kgDeterminethe free enthe same oCh. 4.)

: Pulley A

LEM 2.66

g crate is to bee the magnitund of the rope on each side o

cos

F

αα

Σ

For α = +

yFΣ =

For α = −

yFΣ =

e supported bude and direct

to maintain eof a simple pu

0: 2

0.5965553.377

xF P

αα

⎛= − ⎜

⎝== ± °

53.377 :+ °

160: 22

P⎛

= ⎜⎝

53.377 :− °

160: 22

P⎛

= ⎜⎝

by the rope-antion of the forquilibrium. (H

ulley. This can

5 c281

P⎛ ⎞

+⎟⎝ ⎠

6 sin 5381

P⎞

+⎟⎠

6 sin( 581

P⎞

+ −⎟⎠

nd-pulley arrarce P that muHint: The tensin be proved by

cos 0α =

3.377 1962 N° −

724=P

53.377 ) 196° −

17=P

ngement showust be exerted ion in the ropey the methods

N 0=

4 N 53.4°

62 N 0=

773 53.4°

wn. on

e is s of


SOLUT

Free-Bo (a)

(b)

(c)

(d )

(e)

TION

ody Diagram of Pulley

FΣ

FΣ

FΣ

FΣ

FΣ

0: 2yF T= −

0: 2yF T= −

0: 3yF T= −

0: 3yF T= −

0: 4yF T= −

PR

A 6andfor (Se

(600 lb) 0

12

T

− =

=

(600 lb) 0

12

T

− =

=

(600 lb) 0

1 (3

T

− =

=

(600 lb) 0

1 (3

T

− =

=

(600 lb) 0

14

T

− =

=

ROBLEM 2

600-lb crate id-pulley arrang

each arrangeee the hint for

(600 lb)

(600 lb)

(600 lb)

(600 lb)

(600 lb)

.67

is supported bgements as sh

ement the tensProblem 2.66

by several rophown. Determsion in the ro.)

300 lbT =

300 lbT =

200 lbT =

200 lbT =

150.0 lbT =

pe-ine pe.


SOLUT

Free-Bo

(b)

(d )

TION

ody Diagram of Pulley and

d Crate

0:yFΣ =

0:yFΣ =

PR

Solvthatcrat

PRby shotens2.66

3 (600 lbT

T

−

4 (600 lbT −

ROBLEM 2.

ve Parts b andt the free end te.

OBLEM 2.67several ropewn. Determinsion in the rop6.)

b) 0

1 (600 lb)3

T

=

=

b) 0

1 (600 lb)4

T

=

=

.68

d d of Problemof the rope is

7 A 600-lb cre-and-pulley ane for each ape. (See the h

)

m 2.67, assumis attached to

rate is supporarrangements arrangement hint for Probl

200 lbT =

150.0 lbT =

ing the

rted as

the em


SOLUT

Free-Bo

TION

ody Diagram:: Pulley C

PRO

A locableseconsuppo(a) th

(a) Σ

Henc

(b) FΣ

or

OBLEM 2.6

ad Q is applie ACB. The pnd cable CADorts a load he tension in c

0:x ACBF TΣ =

ce:

0: (

(1292.88 N)(y ACBF T=

69

ied to the pulpulley is heldD, which paP. Knowing

cable ACB, (b)

(cos25 cosB ° −

129ACBT =

(sin 25 sin 5

sin 25 sin 5

° +

° +

Q =

lley C, whichd in the positasses over the

that 75P =) the magnitud

s55 ) (750 N° −

92.88 N

AT

5 ) (750 N)

5 ) (750 N)

° +

° +

2219.8 N=

h can roll on ion shown bye pulley A a

50 N, determde of load Q.

N)cos55° 0=

1293 NACBT =

sin 55 0

sin 55 0

Q

Q

° − =

° − =

2220 NQ =

the y a and

mine

0

0


SOLUT

Free-Bo

TION

ody Diagram:: Pulley C

PRO

An 1on tha secsuppo(b) th

0xFΣ =

or

yFΣ =

or

(a) Substit

1.24

Hence:

(b) Using (

OBLEM 2.7

800-N load Qhe cable ACB. cond cable CAorts a load Phe magnitude o

0: (cos2ACBT

0: (sin 2ACBT

tute Equation (

4177 0.ACBT +

(1), 0.P =

70

Q is applied toThe pulley is

CAD, which p. Determine (of load P.

25 cos55 )° − ° −

25 sin55 )° + ° +

1.24177T

(1) into Equat

81915(0.5801

58010(1048.3

o the pulley Cheld in the po

passes over th(a) the tension

cos55 0P− ° =

0P =

sin55 1P+ ° −

0.81915ACBT +

tion (2):

10 ) 1800ACBT =

1048ACBT =

AT

37 N) 608.16=

C, which can rosition shown he pulley A an in cable AC

0

0.58010 ACBT

800 N 0=

5 1800 NP =

0 N

8.37 N

1048 NACBT =

6 N

608 NP =

roll by

and CB,

(1)

(2)


PROBLEM 2.71

Determine (a) the x, y, and z components of the 600-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

SOLUTION

(a) (600 N)sin 25 cos30xF = °

219.60 NxF = 220 NxF =

(600 N)cos 25°

543.78 Ny

y

F

F

=

= 544 NyF =

(380.36 N)sin 25 sin 30126.785 N

z

z

FF

= °= 126.8 NzF =

(b) 219.60 Ncos

600 Nx

xFF

θ = = 68.5xθ = °

543.78 Ncos600 N

yy

FF

θ = = 25.0yθ = °

126.785 Ncos

600 Nz

zFF

θ = = 77.8zθ = °


PROBLEM 2.72

Determine (a) the x, y, and z components of the 450-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

SOLUTION

(a) (450 N)cos 35 sin 40xF = − °

236.94 NxF = − 237 NxF = −

(450 N)sin 35°

258.11 Ny

y

F

F

=

= 258 NyF =

(450 N)cos35 cos40282.38 N

z

z

FF

= °=

282 NzF =

(b) -236.94 Ncos

450 Nx

xFF

θ = = 121.8xθ = °

258.11 Ncos450 N

yy

FF

θ = = 55.0yθ = °

282.38 Ncos

450 Nz

zFF

θ = = 51.1zθ = °

Note: From the given data, we could have computed directly 90 35 55 , which checks with the answer obtained.yθ = − =


PROBLEM 2.73

A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.)

SOLUTION

Recoil force 400 NF =

(400 N)cos40306.42 N

HF∴ = °=

(a) sin35

(306.42 N)sin35x HF F= − °

= − °

175.755 N= − 175.8 NxF = −

sin 40

(400 N)sin 40257.12 N

yF F= − °

= − °= − 257 NyF = −

cos35(306.42 N)cos35251.00 N

z HF F= + °= + °= + 251 NzF = +

(b) 175.755 Ncos400 N

xx

FF

θ −= = 116.1xθ = °

257.12 Ncos400 N

yy

FF

θ −= = 130.0yθ = °

251.00 Ncos

400 Nz

zFF

θ = = 51.1zθ = °


PROBLEM 2.74

Solve Problem 2.73, assuming that point A is located 15° north of west and that the barrel of the gun forms an angle of 25° with the horizontal.

PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.)

SOLUTION

Recoil force 400 NF =

(400 N)cos25362.52 N

HF∴ = °=

(a) cos15

(362.52 N)cos15x HF F= + °

= + °

350.17 N= + 350 NxF = +

sin 25

(400 N)sin 25169.047 N

yF F= − °

= − °= − 169.0 NyF = −

sin15(362.52 N)sin1593.827 N

z HF F= + °= + °= + 93.8 NzF = +

(b) 350.17 Ncos400 N

xx

FF

θ += = 28.9xθ = °

169.047 Ncos

400 Ny

yFF

θ −= = 115.0yθ = °

93.827 Ncos400 N

zz

FF

θ += = 76.4zθ = °


PROBLEM 2.75

The angle between spring AB and the post DA is 30°. Knowing that the tension in the spring is 50 lb, determine (a) the x, y, and z components of the force exerted on the circular plate at B, (b) the angles θx, θy, and θz defining the direction of the force at B.

SOLUTION

cos 60

(50 lb)cos 6025.0 lb

h

h

F F

F

= °

= °=

cos 35 sin60 sin35

( 25.0 lb)cos35 (50.0 lb)sin60 ( 25.0 lb)sin35

20.479 lb 43.301 lb 14.3394 l

x h y z h

x y z

x y z

F F F F F F

F F F

F F F

= − ° = = −

= − = = −

= − = = − b

(a)

20.5 lbxF = −

43.3 lbyF =

14.33 lbzF = −

(b) 20.479 lbcos

50 lbx

xFF

θ −= = 114.2xθ = °

43.301 lbcos50 lb

yy

FF

θ = = 30.0yθ = °

-14.3394 lbcos

50 lbz

zFF

θ = = 106.7zθ = °


PROBLEM 2.76

The angle between spring AC and the post DA is 30°. Knowing that the tension in the spring is 40 lb, determine (a) the x, y, and z components of the force exerted on the circular plate at C, (b) the angles θx, θy, and θz defining the direction of the force at C.

SOLUTION

cos 60

(40 lb)cos 6020.0 lb

h

h

F F

F

= °

= °=

(a)

cos 35x hF F= ° sin 60yF F= ° sin 35z hF F= − °

(20.0 lb)cos 35°= (40 lb)sin 60°= (20.0 lb)sin 35= − °

16.3830 lbxF = 34.641 lbyF = 11.4715 lbzF = −

16.38 lbxF =

34.6 lbyF =

11.47 lbzF = −

(b) 16.3830 lbcos

40 lbx

xFF

θ = = 65.8xθ = °

34.641 lbcos40 lb

yy

FF

θ = = 30.0yθ = °

-11.4715 lbcos40 lb

zz

FF

θ = = 106.7zθ = °


SOLUT

TION

P

CDcd

From trian

(a)

(b)

Fro

PROBLEM

Cable AB is Determine (a) cable on the adirection of tha

ngle AOB:

F

F

F

cosθ

om above:

2.77

65 ft long, athe x, y, and

anchor B, (b) at force.

cos y

y

θ

θ

sin

(3900 lbx yF F θ= −

= −

cosy yF F θ= +

(3900 lbzF = +

1839

xx

FF

θ = = −

30.5yθ =

cos zz

FF

θ =

and the tensiod z component

the angles θ

56 ft65 ft0.8615430.51

y

y

=

== °

cos 20

b)sin 30.51 co

°

°

(3900 lb)(0y =

b)sin 30.51° si

861 lb 0.47900 lb

= −

51°

677 lb3900 lb

= + =

on in that cats of the force

,xθ ,yθ and

os 20°

F

0.86154) F

in 20°

771

0.1736= +

able is 3900 e exerted by

zθ defining

1861 lbxF = −

3360 lbyF = +

677 lbzF = +

118.5xθ = °

30.5yθ = °

80.0zθ = °

lb. the the


PROBLEM 2.78

Cable AC is 70 ft long, and the tension in that cable is 5250 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor C, (b) the angles θx, θy, and θz defining the direction of that force.

SOLUTION

In triangle AOB: 70 ft56 ft5250 lb

ACOA

F

===

56 ftcos70 ft36.870

sin

(5250 lb)sin 36.8703150.0 lb

y

y

H yF F

θ

θ

θ

=

= °

=

= °=

(a) sin50 (3150.0 lb)sin50 2413.0 lbx HF F= − ° = − ° = −

2410 lbxF = −

cos (5250 lb)cos36.870 4200.0 lby yF F θ= + = + ° = +

4200 lbyF = +

cos50 3150cos50 2024.8 lbz HF F= − ° = − ° = − 2025 lbzF = −

(b) 2413.0 lbcos5250 lb

xx

FF

θ −= = 117.4xθ = °

From above: 36.870yθ = ° 36.9yθ = °

2024.8 lb5250 lb

zz

FF

θ −= = 112.7zθ = °


PROBLEM 2.79

Determine the magnitude and direction of the force F = (240 N)i – (270 N)j + (680 N)k.

SOLUTION

2 2 2

2 2 2(240 N) ( 270 N) ( 680 N)

x y zF F F F

F

= + +

= + − + − 770 NF =

240 Ncos770 N

xx

FF

θ = = 71.8xθ = °

270 Ncos770 N

yy

FF

θ −= = 110.5yθ = °

680 Ncos770 N

zy

FF

θ = = 28.0zθ = °


PROBLEM 2.80

Determine the magnitude and direction of the force F = (320 N)i + (400 N)j − (250 N)k.

SOLUTION

2 2 2

2 2 2(320 N) (400 N) ( 250 N)

x y zF F F F

F

= + +

= + + − 570 NF =

320 Ncos570 N

xx

FF

θ = = 55.8xθ = °

400 Ncos570 N

yy

FF

θ = = 45.4yθ = °

250 Ncos570 N

zy

FF

θ −= = 116.0zθ = °


PROBLEM 2.81

A force acts at the origin of a coordinate system in a direction defined by the angles θx = 69.3° and θz = 57.9°. Knowing that the y component of the force is –174.0 lb, determine (a) the angle θy, (b) the other components and the magnitude of the force.

SOLUTION

2 2 2

2 2 2

cos cos cos 1

cos (69.3 ) cos cos (57.9 ) 1

cos 0.7699

x y z

y

y

θ θ θ

θ

θ

+ + =

° + + ° =

= ±

(a) Since 0,yF < we choose cos 0.7699yθ = − 140.3yθ∴ = °

(b) cos

174.0 lb ( 0.7699)y yF F

F

θ=

− = −

226.0 lbF =

226 lbF =

cos (226.0 lb)cos69.3x xF F θ= = ° 79.9 lbxF =

cos (226.0 lb)cos57.9z zF F θ= = ° 120.1lbzF =


PROBLEM 2.82

A force acts at the origin of a coordinate system in a direction defined by the angles θx = 70.9° and θy = 144.9°. Knowing that the z component of the force is –52.0 lb, determine (a) the angle θz, (b) the other components and the magnitude of the force.

SOLUTION

2 2 2

2 2 2

cos cos cos 1

cos 70.9 cos 144.9 cos 1cos 0.47282

x y z

z

z

θ θ θ

θθ

+ + =

+ ° + ° == ±

(a) Since 0,zF < we choose cos 0.47282zθ = − 118.2zθ∴ = °

(b) cos52.0 ( 0.47282)

z zF Flb F

θ=− = −

110.0 lbF =

110.0 lbF =

cos (110.0 lb)cos70.9x xF F θ= = ° 36.0 lbxF =

cos (110.0 lb)cos144.9y yF F θ= = ° 90.0 lbyF = −


PROBLEM 2.83

A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, θz = 151.2°, and Fy < 0, determine (a) the components Fy and Fz, (b) the angles θx and θy.

SOLUTION

(a) cos (210 N)cos151.2z zF F θ= = °

184.024 N= − 184.0 NzF = −

Then: 2 2 2 2x y zF F F F= + +

So: 2 2 2 2(210 N) (80 N) ( ) (184.024 N)yF= + +

Hence: 2 2 2(210 N) (80 N) (184.024 N)yF = − − −

61.929 N= − 62.0 lbyF = −

(b) 80 Ncos 0.38095210 N

xx

FF

θ = = = 67.6xθ = °

61.929 Ncos 0.29490

210 Ny

yFF

θ = = = −

107.2yθ = °


PROBLEM 2.84

A force F of magnitude 1200 N acts at the origin of a coordinate system. Knowing that θx = 65°, θy = 40°, and Fz > 0, determine (a) the components of the force, (b) the angle θz.

SOLUTION

2 2 2

2 2 2

cos cos cos 1

cos 65 cos 40 cos 1cos 0.48432

x y z

z

z

θ θ θ

θθ

+ + =

+ ° + ° == ±

(b) Since 0,zF > we choose cos 0.48432, or 61.032z zθ θ= = 61.0zθ∴ = °

(a) 1200 NF =

cos (1200 N) cos65x xF F θ= =

507 NxF =

cos (1200 N)cos40y yF F θ= = ° 919 NyF =

cos (1200 N)cos61.032z zF F θ= = ° 582 NzF =


PROBLEM 2.85

A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

SOLUTION

2 2 2

(480 mm) (510 mm) (320 mm)

(480 mm) (510 mm ) (320 mm)770 mm

385 N [(480 mm) (510 mm) (320 mm) ]770 mm(240 N) (255 N) (160 N)

DB

DB

DB

F

DBFDB

= − +

= + +

==

=

= − +

= − +

i j k

F λ

i j k

i j k

240 N, 255 N, 160.0 Nx y zF F F= + = − = +


PROBLEM 2.86

For the frame and cable of Problem 2.85, determine the components of the force exerted by the cable on the support at E.

PROBLEM 2.85 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

SOLUTION

2 2 2

(270 mm) (400 mm) (600 mm)

(270 mm) (400 mm) (600 mm)770 mm

385 N [(270 mm) (400 mm) (600 mm) ]770 mm(135 N) (200 N) (300 N)

EB

EB

EB

F

EBFEB

= − +

= + +

==

=

= − +

= − +

i j k

F λ

i j k

F i j k

135.0 N, 200 N, 300 Nx y zF F F= + = − = +


PROBLEM 2.87

In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AB is 2 kips, determine the components of the force exerted at A by the cable.

SOLUTION

Cable AB: ( 46.765 ft) (45 ft) (36 ft)74.216 ft

46.765 45 3674.216

AB

AB AB AB

ABAB

T

− + += =

− + += =

i j kλ

i j kT λ

( ) 1.260 kipsAB xT = −

( ) 1.213 kipsAB yT = +

( ) 0.970 kipsAB zT = +


PROBLEM 2.88

In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AC is 1.5 kips, determine the components of the force exerted at A by the cable.

SOLUTION

Cable AB: ( 46.765 ft) (55.8 ft) ( 45 ft)85.590 ft

46.765 55.8 45(1.5 kips)85.590

AC

AC AC AC

ACAC

T

− + + −= =

− + −= =

i j kλ

i j kT λ

( ) 0.820 kipsAC xT = −

( ) 0.978 kipsAC yT = +

( ) 0.789 kips= −AC zT


PROBLEM 2.89

A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AB is 408 N, determine the components of the force exerted on the plate at B.

SOLUTION

We have:

(320 mm) (480 mm) (360 mm) 680 mmBA BA= + =i + j - k

Thus:

8 12 9F

17 17 17B BA BA BA BABAT T TBA

⎛ ⎞= = = ⎜ ⎟⎝ ⎠

λ i + j - k

8 12 9 0

17 17 17BA BA BAT T T⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

i j k

Setting 408 NBAT = yields,

192.0 N, 288 N, 216 Nx y zF F F= + = + = −


PROBLEM 2.90

A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 429 N, determine the components of the force exerted on the plate at D.

SOLUTION

We have:

(250 mm) (480 mm) (360 mm) 650 mmDA DA= − =i + j + k

Thus:

5 48 36F

13 65 65D DA DA DA DADAT T TDA

⎛ ⎞= = = −⎜ ⎟⎝ ⎠

λ i + j + k

5 48 36 0

13 65 65DA DA DAT T T⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

i j k

Setting 429 NDAT = yields,

165.0 N, 317 N, 238 Nx y zF F F= − = + = +


PROBLEM 2.91

Find the magnitude and direction of the resultant of the two forces shown knowing that P = 300 N and Q = 400 N.

SOLUTION

(300 N)[ cos30 sin15 sin 30 cos30 cos15 ](67.243 N) (150 N) (250.95 N)

(400 N)[cos50 cos 20 sin 50 cos50 sin 20 ](400 N)[0.60402 0.76604 0.21985](241.61 N) (306.42 N) (87.939 N)

(174.

= − ° ° + ° + ° °= − + += ° ° + ° − ° °= + −= + −= +=

P i j ki j k

Q i j ki j

i j kR P Q

2 2 2

367 N) (456.42 N) (163.011 N)

(174.367 N) (456.42 N) (163.011 N)515.07 N

R

+ +

= + +

=

i j k

515 NR =

174.367 Ncos 0.33853515.07 N

xx

RR

θ = = = 70.2xθ = °

456.42 Ncos 0.88613515.07 N

yy

RR

θ = = = 27.6yθ = °

163.011 Ncos 0.31648515.07 N

zz

RR

θ = = = 71.5zθ = °


PROBLEM 2.92


SOLUTION

(400 N)[ cos30 sin15 sin 30 cos30 cos15 ](89.678 N) (200 N) (334.61 N)

(300 N)[cos50 cos 20 sin 50 cos50 sin 20 ](181.21 N) (229.81 N) (65.954 N)

(91.532 N) (429.81 N) (268.66 N)

(91.5R

= − ° ° + ° + ° °= − + += ° ° + ° − ° °= + −= += + +

=

P i j ki j k

Q i j ki j k

R P Qi j k

2 2 232 N) (429.81 N) (268.66 N)515.07 N

+ +

= 515 NR =

91.532 Ncos 0.177708515.07 N

xx

RR

θ = = = 79.8xθ = °

429.81 Ncos 0.83447515.07 N

yy

RR

θ = = = 33.4yθ = °

268.66 Ncos 0.52160515.07 N

zz

RR

θ = = = 58.6zθ = °


PROBLEM 2.93

Knowing that the tension is 425 lb in cable AB and 510 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

SOLUTION

2 2 2

2 2 2

(40 in.) (45 in.) (60 in.)

(40 in.) (45 in.) (60 in.) 85 in.

(100 in.) (45 in.) (60 in.)

(100 in.) (45 in.) (60 in.) 125 in.

(40 in.) (45 in.) (60 in.)(425 lb)85 in.AB AB AB AB

AB

AB

AC

AC

ABT TAB

= − +

= + + =

= − +

= + + =

− += = =

i j k

i j k

i j kT λ

(200 lb) (225 lb) (300 lb)

(100 in.) (45 in.) (60 in.)(510 lb)125 in.

(408 lb) (183.6 lb) (244.8 lb)(608) (408.6 lb) (544.8 lb)

AB

AC AC AC AC

AC

AB AC

ACT TAC

⎡ ⎤⎢ ⎥⎣ ⎦

= − +

⎡ ⎤− += = = ⎢ ⎥

⎣ ⎦= − +

= + = − +

T i j k

i j kT λ

T i j kR T T i j k

Then: 912.92 lbR = 913 lbR =

and 608 lbcos 0.66599912.92 lbxθ = = 48.2xθ = °

408.6 lbcos 0.44757912.92 lbyθ = = − 116.6yθ = °

544.8 lbcos 0.59677912.92 lbzθ = = 53.4zθ = °


PROBLEM 2.94

Knowing that the tension is 510 lb in cable AB and 425 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

SOLUTION

2 2 2

2 2 2

(40 in.) (45 in.) (60 in.)

(40 in.) (45 in.) (60 in.) 85 in.

(100 in.) (45 in.) (60 in.)

(100 in.) (45 in.) (60 in.) 125 in.

(40 in.) (45 in.) (60 in.)(510 lb)85 in.AB AB AB AB

AB

AB

AC

AC

ABT TAB

= − +

= + + =

= − +

= + + =

− += = =

i j k

i j k

i j kT λ

(240 lb) (270 lb) (360 lb)

(100 in.) (45 in.) (60 in.)(425 lb)125 in.

(340 lb) (153 lb) (204 lb)(580 lb) (423 lb) (564 lb)

AB

AC AC AC AC

AC

AB AC

ACT TAC

⎡ ⎤⎢ ⎥⎣ ⎦

= − +

⎡ ⎤− += = = ⎢ ⎥

⎣ ⎦= − +

= + = − +

T i j k

i j kT λ

T i j kR T T i j k

Then: 912.92 lbR = 913 lbR =

and 580 lbcos 0.63532912.92 lbxθ = = 50.6xθ = °

423 lbcos 0.46335912.92 lbyθ −

= = − 117.6yθ = °

564 lbcos 0.61780912.92 lbzθ = = 51.8zθ = °


PROBLEM 2.95

For the frame of Problem 2.85, determine the magnitude and direction of the resultant of the forces exerted by the cable at B knowing that the tension in the cable is 385 N.

PROBLEM 2.85 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

SOLUTION

2 2 2

(480 mm) (510 mm) (320 mm)

(480 mm) (510 mm) (320 mm) 770 mm

BD

BD

= − + −

= + + =

i j k

(385 N) [ (480 mm) (510 mm) (320 mm) ](770 mm)(240 N) (255 N) (160 N)

BD BD BD BDBDT TBD

= =

= − + −

= − + −

F λ

i j k

i j k

2 2 2

(270 mm) (400 mm) (600 mm)

(270 mm) (400 mm) (600 mm) 770 mm

BE

BE

= − + −

= + + =

i j k

(385 N) [ (270 mm) (400 mm) (600 mm) ](770 mm)(135 N) (200 N) (300 N)

BE BE BE BEBET TBE

= =

= − + −

= − + −

F λ

i j k

i j k

(375 N) (455 N) (460 N)BD BE= + = − + −R F F i j k

2 2 2(375 N) (455 N) (460 N) 747.83 NR = + + = 748 NR =

375 Ncos

747.83 Nxθ−

= 120.1xθ = °

455 Ncos

747.83 Nyθ = 52.5yθ = °

460 Ncos

747.83 Nzθ−

= 128.0zθ = °


PROBLEM 2.96

For the plate of Prob. 2.89, determine the tensions in cables AB and AD knowing that the tension in cable AC is 54 N and that the resultant of the forces exerted by the three cables at A must be vertical.

SOLUTION

We have:

( )

(320 mm) (480 mm) (360 mm) 680 mm

(450 mm) (480 mm) (360 mm) 750 mm

(250 mm) (480 mm) 360 mm 650 mm

AB AB

AC AC

AD AD

= − − + =

= − + =

= − − =

i j k

i j k

i j k

Thus:

( )

( )

( )

320 480 360680

54 450 480 360750

250 480 360650

ABAB AB AB AB

AC AC AC AC

ADAD AD AD AD

TABT TABACT TAC

TADT TAD

= = = − − +

= = = − +

= = = − −

T λ i j k

T λ i j k

T λ i j k

Substituting into the Eq. = ΣR F and factoring , , :i j k

320 25032.40680 650

480 48034.560680 650

360 36025.920680 650

AB AD

AB AD

AB AD

T T

T T

T T

⎛ ⎞= − + +⎜ ⎟⎝ ⎠

⎛ ⎞+ − − −⎜ ⎟⎝ ⎠⎛ ⎞+ + −⎜ ⎟⎝ ⎠

R i

j

k


PROBLEM 2.96 (Continued)

Since R is vertical, the coefficients of i and k are zero:

:i 320 25032.40 0680 650AB ADT T− + + = (1)

:k 360 36025.920 0680 650AB ADT T+ − = (2)

Multiply (1) by 3.6 and (2) by 2.5 then add:

252 181.440 0680 ABT− + =

489.60 NABT =

490 NABT =

Substitute into (2) and solve for :ADT

360 360(489.60 N) 25.920 0680 650 ADT+ − =

514.80 NADT =

515 NADT =


PROBLEM 2.97

The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC.

SOLUTION

Cable AB: 183 lbABT =

( 48 in.) (29 in.) (24 in.)(183 lb)61in.

(144 lb) (87 lb) (72 lb)

AB AB AB AB

AB

ABT TAB

− + += = =

= − + +

i j kT

T i j k

λ

Cable AC: ( 48 in.) (25 in.) ( 36 in.)65 in.

48 25 3665 65 65

AC AC AC AC AC

AC AC AC AC

ACT T TAC

T T T

− + + −= = =

= − + −

i j kT

T i j k

λ

Load P: P=P j

For resultant to be directed along OA, i.e., x-axis

360: (72 lb) 065z z ACR F T ′= Σ = − = 130.0 lbACT =


PROBLEM 2.98

For the boom and loading of Problem. 2.97, determine the magnitude of the load P.

PROBLEM 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC.

SOLUTION

See Problem 2.97. Since resultant must be directed along OA, i.e., the x-axis, we write

250: (87 lb) 065y y ACR F T P= Σ = + − =

130.0 lbACT = from Problem 2.97.

Then 25(87 lb) (130.0 lb) 065

P+ − = 137.0 lbP =


PROBLEM 2.99

A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AB is 6 kN.

SOLUTION

Free-Body Diagram at A:

The forces applied at A are: , , , andAB AC ADT T T W

where .W=W j To express the other forces in terms of the unit vectors i, j, k, we write

(450 mm) (600 mm) 750 mm(600 mm) (320 mm) 680 mm(500 mm) (600 mm) (360 mm) 860 mm

AB ABAC ACAD AD

= − + =

= + − =

= + + + =

i jj ki j k

and ( 450 mm) (600 mm)

750 mmAB AB AB AB ABABT T TAB

− += = =

i jT λ

45 6075 75 ABT⎛ ⎞= − +⎜ ⎟

⎝ ⎠i j

(600 mm) (320 mm)

680 mm60 3268 68

(500 mm) (600 mm) (360 mm)860 mm

50 60 3686 86 86

−= = =

⎛ ⎞= −⎜ ⎟⎝ ⎠

+ += = =

⎛ ⎞= + +⎜ ⎟⎝ ⎠

AC AC AC AC AC

AC

AD AD AD AD AD

AD

ACT T TAC

T

ADT T TAD

T

i jT λ

j k

i j kT λ

i j k



Equilibrium condition: 0: 0AB AC ADFΣ = ∴ + + + =T T T W

Substituting the expressions obtained for , , and ;AB AC ADT T T factoring i, j, and k; and equating each of the coefficients to zero gives the following equations:

From i: 45 50 075 86AB ADT T− + = (1)

From j: 60 60 60 075 68 86AB AC ADT T T W+ + − = (2)

From k: 32 36 068 86AC ADT T− + = (3)

Setting 6 kNABT = in (1) and (2), and solving the resulting set of equations gives

6.1920 kN5.5080 kN

AC

AC

TT

=

= 13.98 kNW =


PROBLEM 2.100

A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AD is 4.3 kN.

SOLUTION

See Problem 2.99 for the figure and analysis leading to the following set of linear algebraic equations:

45 50 075 86AB ADT T− + = (1)

60 60 60 075 68 86AB AC ADT T T W+ + − = (2)

32 36 068 86AC ADT T− + = (3)

Setting 4.3 kNADT = into the above equations gives

4.1667 kN3.8250 kN

AB

AC

TT

== 9.71 kNW =


SOLUT

The forc

where P

and

TION

ces applied at A

.P=P j To exp

A are:

press the other

ABACAD

=

=

=

AB

AC

AD

=

=

=

T

T

T

PROB

Three cthe verttension

FREE-BOD

,ABT

r forces in term

(4.20 m)(2.40 m) (

(5.60 m)

= − −

= −

= − −

ii

j

AB AB AB

AC AC A

AD AD A

T T

T T

T T

= =

= =

= =

λ

λ

λ

BLEM 2.101

cables are usetical force P ein cable AD i

DY DIAGRA

, , andAC ADT T

ms of the unit

(5.60 m)(5.60 m) (4

(3.30 m)

−

+

−

jj

k

( 0.6

(0.32

( 0.8

B

AC

AD

ABABACACADAD

= −

=

= −

1

ed to tether a exerted by the s 481 N.

AM AT A

d P

vectors i, j, k

.20 m)ABACAD

k

0.8 )

2432 0.7567

86154 0.507

ABT−

−

−

i j

i

j

balloon as shballoon at A k

, we write

7.00 m7.40 m6.50 mD

=

=

=

76 0.56757

769 ) ADT

+j k

k

hown. Determknowing that

) ACTk

ine the


Equilibr

Substitu

Equating

Setting

rium condition

uting the expre

(−

g to zero the c

481 NADT =

P

n:

essions obtaine

0.6 0.3ABT− +

coefficients of

0.8 ABT− −

in (2) and (3),

PROBLEM 2

0: ABFΣ = T

ed for ,AB AT T

2432 ) ((0.56757

ACT +

+

i

f i, j, k:

0.6 AT−

0.75676 ACT −

0.56757T

, and solving t

4323

AC

AD

TT

=

=

2.101 (Con

B AC AD+ +T T

, andAC ADT an

( 0.8 0.77 0.50769

AB

AC

TT− −

−

0.32432ABT T+

0.86154 ADT− +

0.50769ACT T−

the resulting se

30.26 N32.57 N

ntinued)

0P+ =j

nd factoring i,

5676 0.89 ) 0

AC

AD

TT

−

=k

0ACT =

0P+ =

0ADT =

et of equation

, j, and k:

86154 )ADT P+

s gives

)j

926 N=P

(1)

(2)

(3)


PROBLEM 2.102

Three cables are used to tether a balloon as shown. Knowing that the balloon exerts an 800-N vertical force at A, determine the tension in each cable.

SOLUTION

See Problem 2.101 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3).

0.6 0.32432 0AB ACT T− + = (1)

0.8 0.75676 0.86154 0AB AC ADT T T P− − − + = (2)

0.56757 0.50769 0AC ADT T− = (3)

From Eq. (1): 0.54053AB ACT T=

From Eq. (3): 1.11795AD ACT T=

Substituting for ABT and ADT in terms of ACT into Eq. (2) gives

0.8(0.54053 ) 0.75676 0.86154(1.11795 ) 0AC AC ACT T T P− − − + =

2.1523 ; 800 N800 N2.1523371.69 N

AC

AC

T P P

T

= =

=

=

Substituting into expressions for ABT and ADT gives

0.54053(371.69 N)1.11795(371.69 N)

AB

AD

TT

==

201 N, 372 N, 416 NAB AC ADT T T= = =


PROBLEM 2.103

A 36-lb triangular plate is supported by three wires as shown. Determine the tension in each wire, knowing that a = 6 in.

SOLUTION

By Symmetry DB DCT T= Free-Body Diagram of Point D:

The forces applied at D are: , , , andDB DC DAT T T P

where (36 lb) .P= =P j j To express the other forces in terms of the unit vectors i, j, k, we write

(16 in.) (24 in.) 28.844 in.

(8 in.) (24 in.) (6 in.) 26.0 in.(8 in.) (24 in.) (6 in.) 26.0 in.

DA DADB DBDC DC

= − =

= − − + =

= − − − =

i ji j ki j k

and (0.55471 0.83206 )

( 0.30769 0.92308 0.23077 )

( 0.30769 0.92308 0.23077 )

DA DA DA DA DA

DB DB DB DB DB

DC DC DC DC DC

DAT T TDADBT T TDBDCT T TDC

= = = −

= = = − − +

= = = − − −

T λ i j

T λ i j k

T λ i j k



Equilibrium condition: 0: (36 lb) 0DA DB DCFΣ = + + + =T T T j

Substituting the expressions obtained for , , andDA DB DCT T T and factoring i, j, and k:

(0.55471 0.30769 0.30769 ) ( 0.83206 0.92308 0.92308 36 lb)

(0.23077 0.23077 ) 0DA DB DC DA DB DC

DB DC

T T T T T TT T

− − + − − − +

+ − =

i jk

Equating to zero the coefficients of i, j, k:

0.55471 0.30769 0.30769 0DA DB DCT T T− − = (1)

0.83206 0.92308 0.92308 36 lb 0DA DB DCT T T− − − + = (2)

0.23077 0.23077 0DB DCT T− = (3)

Equation (3) confirms that DB DCT T= . Solving simultaneously gives,

14.42 lb; 13.00 lbDA DB DCT T T= = =


PROBLEM 2.104

Solve Prob. 2.103, assuming that a = 8 in.

PROBLEM 2.103 A 36-lb triangular plate is supported by three wires as shown. Determine the tension in each wire, knowing that a = 6 in.

SOLUTION

By Symmetry DB DCT T= Free-Body Diagram of Point D:

The forces applied at D are: , , , andDB DC DAT T T P

where (36 lb) .P= =P j j To express the other forces in terms of the unit vectors i, j, k, we write

(16 in.) (24 in.) 28.844 in.

(8 in.) (24 in.) (8 in.) 26.533 in.(8 in.) (24 in.) (8 in.) 26.533 in.

DA DADB DBDC DC

= − =

= − − + =

= − − − =

i ji j ki j k

and (0.55471 0.83206 )

( 0.30151 0.90453 0.30151 )

( 0.30151 0.90453 0.30151 )

DA DA DA DA DA

DB DB DB DB DB

DC DC DC DC DC

DAT T TDADBT T TDBDCT T TDC

= = = −

= = = − − +

= = = − − −

T λ i j

T λ i j k

T λ i j k



Equilibrium condition: 0: (36 lb) 0DA DB DCFΣ = + + + =T T T j


(0.55471 0.30151 0.30151 ) ( 0.83206 0.90453 0.90453 36 lb)

(0.30151 0.30151 ) 0DA DB DC DA DB DC

DB DC

T T T T T TT T

− − + − − − +

+ − =

i jk


0.55471 0.30151 0.30151 0DA DB DCT T T− − = (1)

0.83206 0.90453 0.90453 36 lb 0DA DB DCT T T− − − + = (2)

0.30151 0.30151 0DB DCT T− = (3)

Equation (3) confirms that DB DCT T= . Solving simultaneously gives,

14.42 lb; 13.27 lbDA DB DCT T T= = =


Solution

where P

and

Equilibr

n The forces ap

.P=P j To exp

rium Condition

pplied at A are

press the other

ABABACACADAD

==

==

==

AB

AC

AD

=

=

=

=

=

=

T

T

T

n with =W

FΣ =

PR

A theis 5

e:

, ,AB AC ADT T T

r forces in term

(36 in.)75 in.(60 in.) (368 in.(40 in.) (677 in.

= − +=

= +=

= +=

i

j

i

( 0.48 0.8

(0.88235

(0.51948

AB AB A

AC AC A

AD AD A

T T

T T

T T

= =

= − +

= =

= +

= =

= +

λ

i

λ

j

λ

i

W= − j

0: AB A= +T T

ROBLEM 2

crate is suppoe weight of the544 lb.

andD W

ms of the unit

(60 in.) (27

32 in.)

60 in.) (27 in

−

−

j

k

j

8 0.36 )

0.47059 )

0.77922 0.3

B

AB

AC

AC

AD

ABAB

TACAC

TADAD

−

−

j k

k

j

AC AD W+ −T j

2.105

orted by threee crate knowin

vectors i, j, k

in.)

n.)

k

k

35065 )

B

C

ADTk

0=

e cables as shng that the ten

, we write

hown. Determnsion in cable A

ine AC



Substituting the expressions obtained for , , andAB AC ADT T T and factoring i, j, and k:

( 0.48 0.51948 ) (0.8 0.88235 0.77922 )

( 0.36 0.47059 0.35065 ) 0AB AD AB AC AD

AB AC AD

T T T T T WT T T

− + + + + −

+ − + − =

i jk


0.48 0.51948 0AB ADT T− + = (1)

0.8 0.88235 0.77922 0AB AC ADT T T W+ + − = (2)

0.36 0.47059 0.35065 0AB AC ADT T T− + − = (3)

Substituting 544 lbACT = in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives:

374.27 lb345.82 lb

AB

AD

TT

== 1049 lbW =


SOLUT

The forc

where P

and

Equilibr

TION

ces applied at A

.P=P j To exp

rium Condition

A are:

press the other

ABABACACADAD

==

==

==

AB

AC

AD

=

=

=

=

=

=

T

T

T

n with =W

FΣ =

P

AD

, ,AB AC ADT T T

r forces in term

(36 in.)75 in.(60 in.) (368 in.(40 in.) (677 in.

= − +=

= +=

= +=

i

j

i

( 0.48 0.8

(0.88235

(0.51948

AB AB A

AC AC A

AD AD A

T T

T T

T T

= =

= − +

= =

= +

= =

= +

λ

i

λ

j

λ

i

W= − j

0: AB A= +T T

PROBLEM 2

A 1600-lb cratetermine the t

andD W

ms of the unit

(60 in.) (27

32 in.)

60 in.) (27 in

−

−

j

k

j

8 0.36 )

0.47059 )

0.77922 0.3

B

AB

AC

AC

AD

ABAB

TACAC

TADAD

−

−

j k

k

j

AC AD W+ −T j

2.106

te is supportetension in each

vectors i, j, k

in.)

n.)

k

k

35065 )

B

C

ADTk

0=

ed by three ch cable.

, we write

ables as showwn.



Substituting the expressions obtained for , , andAB AC ADT T T and factoring i, j, and k:

( 0.48 0.51948 ) (0.8 0.88235 0.77922 )

( 0.36 0.47059 0.35065 ) 0AB AD AB AC AD

AB AC AD

T T T T T WT T T

− + + + + −

+ − + − =

i jk


0.48 0.51948 0AB ADT T− + = (1)

0.8 0.88235 0.77922 0AB AC ADT T T W+ + − = (2)

0.36 0.47059 0.35065 0AB AC ADT T T− + − = (3)

Substituting 1600 lbW = in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives,

571 lbABT =

830 lbACT =

528 lbADT =


PROBLEM 2.107

Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that 0,Q = find the value of P for which the tension in cable AD is 305 N.

SOLUTION

0: 0A AB AC ADΣ = + + + =F T T T P where P=P i

(960 mm) (240 mm) (380 mm) 1060 mm

(960 mm) (240 mm) (320 mm) 1040 mm

(960 mm) (720 mm) (220 mm) 1220 mm

AB AB

AC AC

AD AD

= − − + =

= − − − =

= − + − =

i j k

i j k

i j k

48 12 1953 53 53

12 3 413 13 13

305 N [( 960 mm) (720 mm) (220 mm) ]1220 mm

(240 N) (180 N) (55 N)

AB AB AB AB AB

AC AC AC AC AC

AD AD AD

ABT T TAB

ACT T TAC

T

⎛ ⎞= = = − − +⎜ ⎟⎝ ⎠

⎛ ⎞= = = − − −⎜ ⎟⎝ ⎠

= = − + −

= − + −

T λ i j k

T λ i j k

T λ i j k

i j k

Substituting into 0,AΣ =F factoring , , ,i j k and setting each coefficient equal to φ gives:

48 12: 240 N53 13AB ACP T T= + +i (1)

:j 12 3 180 N53 13AB ACT T+ = (2)

:k 19 4 55 N53 13AB ACT T− = (3)

Solving the system of linear equations using conventional algorithms gives:

446.71 N341.71 N

AB

AC

TT

== 960 NP =


PROBLEM 2.108

Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that 1200 N,P = determine the values of Q for which cable AD is taut.

SOLUTION

We assume that 0ADT = and write 0: (1200 N) 0A AB AC QΣ = + + + =F T T j i

(960 mm) (240 mm) (380 mm) 1060 mm

(960 mm) (240 mm) (320 mm) 1040 mm

AB AB

AC AC

= − − + =

= − − − =

i j k

i j k

48 12 1953 53 53

12 3 413 13 13

AB AB AB AB AB

AC AC AC AC AC

ABT T TAB

ACT T TAC

⎛ ⎞= = = − − +⎜ ⎟⎝ ⎠

⎛ ⎞= = = − − −⎜ ⎟⎝ ⎠

T λ i j k

T λ i j k

Substituting into 0,AΣ =F factoring , , ,i j k and setting each coefficient equal to φ gives:

48 12: 1200 N 053 13AB ACT T− − + =i (1)

12 3: 053 13AB ACT T Q− − + =j (2)

19 4: 053 13AB ACT T− =k (3)

Solving the resulting system of linear equations using conventional algorithms gives:

605.71 N705.71 N300.00 N

AB

AC

TT

Q

==

= 0 300 NQ≤ <

Note: This solution assumes that Q is directed upward as shown ( 0),Q ≥ if negative values of Q are considered, cable AD remains taut, but AC becomes slack for 460 N.Q = −


SOLUT

We notePoint A.

We have

Thus:

Substitu

TION

e that the we

e:

(4

(2

AB

AC

AD

= −

=

=

uting into the E

ight of the pl

FΣ

(320 mm)

450 mm) (4

250 mm) (4

− −

−

−

i

i

i

AB

AC

AD

=

=

=

T

T

T

Eq. 0FΣ = an

PROB

A rectaKnowinweight

late is equal

0: ABF = +T

(

(480 mm) (

480 mm) (3

480 mm) 3

+

+

−

j

j

j

AB AB A

AC AC A

AD AD A

T T

T T

T T

= =

= =

= =

λ

λ

λ

nd factoring ,i

817

1217

917

AB

AB

AB

T

T

T

⎛ − +⎜⎝

⎛+ −⎜⎝⎛+ +⎜⎝

BLEM 2.10

angular plate ng that the teof the plate.

in magnitude

AC AD+ + +T T

)

(360 mm)

360 mm)

360 mm

A

A

A

k

k

k

(

817

0.6

513

AB

AC

AD

ABAB

ACACADAD

⎛= −⎜⎝

=

⎛= ⎜⎝

i

i

, , :j k

50.613

0.64

7.0.4813

AC A

B AC

AC

T T

T

T

+

− −

+ −

9

is supportedension in cabl

to the force

0P =j

680 mm

750 mm

650 mm

AB

AC

AD

=

=

=

12 917 17

0.64 0.48

9.6 7.213 13

− +

− +

− −

i j k

j

i j k

9.613

.2 03

AD

AD

AD

T

T P

T

⎞⎟⎠

⎞+ ⎟⎠

⎞ =⎟⎠

i

j

k

d by three cale AC is 60 N

P exerted by

Free

m

m

m

)8

AB

AC

AD

T

T

T

⎞⎟⎠

⎞⎟⎠

k

k

k

j

Dim

ables as showN, determine

y the support

e Body A :

mensions in m

wn. the

on

mm



Setting the coefficient of i, j, k equal to zero:

:i 8 50.6 017 13AB AC ADT T T− + + = (1)

:j 12 9.60.64 07 13AB AC ADT T T P− − − + = (2)

:k 9 7.20.48 017 13AB AC ADT T T+ − = (3)

Making 60 NACT = in (1) and (3):

8 536 N 017 13AB ADT T− + + = (1′)

9 7.228.8 N 017 13AB ADT T+ − = (3′)

Multiply (1′) by 9, (3′) by 8, and add:

12.6554.4 N 0 572.0 N13 AD ADT T− = =

Substitute into (1′) and solve for :ABT

17 536 572 544.0 N8 13AB ABT T⎛ ⎞= + × =⎜ ⎟

⎝ ⎠

Substitute for the tensions in Eq. (2) and solve for P :

12 9.6(544 N) 0.64(60 N) (572 N)17 13844.8 N

P = + +

= Weight of plate 845 NP= =


PROBLEM 2.110

A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 520 N, determine the weight of the plate.

SOLUTION

See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below:

8 50.6 017 13AB AC ADT T T− + + = (1)

12 9.60.64 017 13AB AC ADT T T P− + − + = (2)

9 7.20.48 017 13AB AC ADT T T+ − = (3)

Making 520 NADT = in Eqs. (1) and (3):

8 0.6 200 N 017 AB ACT T− + + = (1′)

9 0.48 288 N 017 AB ACT T+ − = (3′)

Multiply (1′) by 9, (3′) by 8, and add:

9.24 504 N 0 54.5455 NAC ACT T− = =

Substitute into (1′) and solve for :ABT

17 (0.6 54.5455 200) 494.545 N8AB ABT T= × + =

Substitute for the tensions in Eq. (2) and solve for P:

12 9.6(494.545 N) 0.64(54.5455 N) (520 N)17 13768.00 N

P = + +

= Weight of plate 768 NP= =


PROBLEM 2.111

A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AB is 840 lb, determine the vertical force P exerted by the tower on the pin at A.

SOLUTION

0: 0AB AC AD PΣ = + + + =F T T T j Free-Body Diagram at A:

= 20 100 25 105 ft

60 100 18 118 ft

20 100 74 126 ft

AB AB

AC AC

AD AD

− − + =

= − + =

= − − − =

i j k

i j k

i j k

We write

4 20 521 21 21

AB AB AB AB

AB

ABT TAB

T

= =

⎛ ⎞= − − +⎜ ⎟⎝ ⎠

T λ

i j k

30 50 959 59 59

AC AC AC AC

AC

ACT TAC

T

= =

⎛ ⎞= − +⎜ ⎟⎝ ⎠

T λ

i j k

10 50 3763 63 63

AD AD AD AD

AD

ADT TAD

T

= =

⎛ ⎞= − − −⎜ ⎟⎝ ⎠

T λ

i j k

Substituting into the Eq. 0Σ =F and factoring , , :i j k



4 30 1021 59 63

20 50 5021 59 63

5 9 37 021 59 63

AB AC AD

AB AC AD

AB AC AD

T T T

T T T P

T T T

⎛ ⎞− + −⎜ ⎟⎝ ⎠

⎛ ⎞+ − − − +⎜ ⎟⎝ ⎠⎛ ⎞+ + − =⎜ ⎟⎝ ⎠

i

j

k

Setting the coefficients of , , ,i j k equal to zero:

:i 4 30 10 021 59 63AB AC ADT T T− + − = (1)

:j 20 50 50 021 59 63AB AC ADT T T P− − − + = (2)

:k 5 9 37 021 59 63AB AC ADT T T+ − = (3)

Set 840 lbABT = in Eqs. (1) – (3):

30 10160 lb 059 63AC ADT T− + − = (1′)

50 50800 lb 059 63AC ADT T P− − − + = (2′)

9 37200 lb 059 63AC ADT T+ − = (3′)

Solving, 458.12 lb 459.53 lb 1552.94 lbAC ADT T P= = = 1553 lbP =


PROBLEM 2.112

A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 590 lb, determine the vertical force P exerted by the tower on the pin at A.

SOLUTION

0: 0AB AC AD PΣ = + + + =F T T T j Free-Body Diagram at A:

= 20 100 25 105 ft

60 100 18 118 ft

20 100 74 126 ft

AB AB

AC AC

AD AD

− − + =

= − + =

= − − − =

i j k

i j k

i j k

We write

4 20 521 21 21

AB AB AB AB

AB

ABT TAB

T

= =

⎛ ⎞= − − +⎜ ⎟⎝ ⎠

T λ

i j k

30 50 959 59 59

AC AC AC AC

AC

ACT TAC

T

= =

⎛ ⎞= − +⎜ ⎟⎝ ⎠

T λ

i j k

10 50 3763 63 63

AD AD AD AD

AD

ADT TAD

T

= =

⎛ ⎞= − − −⎜ ⎟⎝ ⎠

T λ

i j k

Substituting into the Eq. 0Σ =F and factoring , , :i j k



4 30 1021 59 63

20 50 5021 59 63

5 9 37 021 59 63

AB AC AD

AB AC AD

AB AC AD

T T T

T T T P

T T T

⎛ ⎞− + −⎜ ⎟⎝ ⎠

⎛ ⎞+ − − − +⎜ ⎟⎝ ⎠⎛ ⎞+ + − =⎜ ⎟⎝ ⎠

i

j

k

Setting the coefficients of , , ,i j k equal to zero:

:i 4 30 10 021 59 63AB AC ADT T T− + − = (1)

:j 20 50 50 021 59 63AB AC ADT T T P− − − + = (2)

:k 5 9 37 021 59 63AB AC ADT T T+ − = (3)

Set 590 lbACT = in Eqs. (1) – (3):

4 10300 lb 021 63AB ADT T− + − = (1′)

20 50500 lb 021 63AB ADT T P− − − + = (2′)

5 3790 lb 021 63AB ADT T+ − = (3′)

Solving, 1081.82 lb 591.82 lbAB ADT T= = 2000 lbP =


PROBLEM 2.113

In trying to move across a slippery icy surface, a 175-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope.

SOLUTION

Free-Body Diagram at A

16 3034 34

N ⎛ ⎞= +⎜ ⎟⎝ ⎠

N i j

and (175 lb)W= = −W j j

( 30 ft) (20 ft) (12 ft)38 ft

15 10 619 19 19

AC AC AC AC AC

AC

ACT T TAC

T

− + −= = =

⎛ ⎞= − + −⎜ ⎟⎝ ⎠

i j kT λ

i j k

( 30 ft) (24 ft) (32 ft)50 ft

15 12 1625 25 25

AB AB AB AB AB

AB

ABT T TAB

T

− + += = =

⎛ ⎞= − + +⎜ ⎟⎝ ⎠

i j kT λ

i j k

Equilibrium condition: 0Σ =F

0AB AC+ + + =T T N W



Substituting the expressions obtained for , , ,AB ACT T N and W; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations:

From i: 15 15 16 025 19 34AB ACT T N− − + = (1)

From j: 12 10 30 (175 lb) 025 19 34AB ACT T N+ + − = (2)

From k: 16 6 025 19AB ACT T− = (3)

Solving the resulting set of equations gives:

30.8 lb; 62.5 lbAB ACT T= =


PROBLEM 2.114

Solve Problem 2.113, assuming that a friend is helping the man at A by pulling on him with a force P = −(45 lb)k.

PROBLEM 2.113 In trying to move across a slippery icy surface, a 175-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope.

SOLUTION

Refer to Problem 2.113 for the figure and analysis leading to the following set of equations, Equation (3) being modified to include the additional force ( 45 lb) .= −P k

15 15 16 025 19 34AB ACT T N− − + = (1)

12 10 30 (175 lb) 025 19 34AB ACT T N+ + − = (2)

16 6 (45 lb) 025 19AB ACT T− − = (3)

Solving the resulting set of equations simultaneously gives:

81.3 lbABT =

22.2 lbACT =


PROBLEM 2.115

For the rectangular plate of Problems 2.109 and 2.110, determine the tension in each of the three cables knowing that the weight of the plate is 792 N.

SOLUTION

See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below. Setting 792 NP = gives:

8 50.6 017 13AB AC ADT T T− + + = (1)

12 9.60.64 792 N 017 13AB AC ADT T T− − − + = (2)

9 7.20.48 017 13AB AC ADT T T+ − = (3)

Solving Equations (1), (2), and (3) by conventional algorithms gives

510.00 NABT = 510 NABT =

56.250 NACT = 56.2 NACT =

536.25 NADT = 536 NADT =


PROBLEM 2.116

For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that

2880 NP = and 0.Q =

SOLUTION

0: 0A AB AC ADΣ = + + + + =F T T T P Q

Where P=P i and Q=Q j

(960 mm) (240 mm) (380 mm) 1060 mm

(960 mm) (240 mm) (320 mm) 1040 mm

(960 mm) (720 mm) (220 mm) 1220 mm

AB AB

AC AC

AD AD

= − − + =

= − − − =

= − + − =

i j k

i j k

i j k

48 12 1953 53 53

12 3 413 13 13

48 36 1161 61 61

AB AB AB AB AB

AC AC AC AC AC

AD AD AD AD AD

ABT T TAB

ACT T TAC

ADT T TAD

⎛ ⎞= = = − − +⎜ ⎟⎝ ⎠

⎛ ⎞= = = − − −⎜ ⎟⎝ ⎠

⎛ ⎞= = = − + −⎜ ⎟⎝ ⎠

T λ i j k

T λ i j k

T λ i j k

Substituting into 0,AΣ =F setting (2880 N)P = i and 0,Q = and setting the coefficients of , , i j k equal to 0, we obtain the following three equilibrium equations:

48 12 48: 2880 N 053 13 61AB AC ADT T T− − − + =i (1)

12 3 36: 053 13 61AB AC ADT T T− − + =j (2)

19 4 11: 053 13 61AB AC ADT T T− − =k (3)




1340.14 N1025.12 N915.03 N

AB

AC

AD

TTT

==

= 1340 NABT =

1025 NACT =

915 NADT =


PROBLEM 2.117


2880 NP = and 576 N.Q =

SOLUTION

See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:

48 12 48 053 13 61AB AC ADT T T P− − − + = (1)

12 3 36 053 13 61AB AC ADT T T Q− − + + = (2)

19 4 11 053 13 61AB AC ADT T T− − = (3)

Setting 2880 NP = and 576 NQ = gives:

48 12 48 2880 N 053 13 61AB AC ADT T T− − − + = (1′)

12 3 36 576 N 053 13 61AB AC ADT T T− − + + = (2′)

19 4 11 053 13 61AB AC ADT T T− − = (3′)

Solving the resulting set of equations using conventional algorithms gives:

1431.00 N1560.00 N183.010 N

AB

AC

AD

TTT

=== 1431 NABT =

1560 NACT =

183.0 NADT =


PROBLEM 2.118


2880 NP = and 576Q = − N. (Q is directed downward).

SOLUTION

See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:

48 12 48 053 13 61AB AC ADT T T P− − − + = (1)

12 3 36 053 13 61AB AC ADT T T Q− − + + = (2)

19 4 11 053 13 61AB AC ADT T T− − = (3)

Setting 2880 NP = and 576 NQ = − gives:

48 12 48 2880 N 053 13 61AB AC ADT T T− − − + = (1′)

12 3 36 576 N 053 13 61AB AC ADT T T− − + − = (2′)

19 4 11 053 13 61AB AC ADT T T− − = (3′)

Solving the resulting set of equations using conventional algorithms gives:

1249.29 N490.31 N1646.97 N

AB

AC

AD

TTT

=== 1249 NABT =

490 NACT =

1647 NADT =


PROBLEM 2.119

For the transmission tower of Probs. 2.111 and 2.112, determine the tension in each guy wire knowing that the tower exerts on the pin at A an upward vertical force of 1800 lb.

PROBLEM 2.111 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AB is 840 lb, determine the vertical force P exerted by the tower on the pin at A.

SOLUTION

See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below:

:i 4 30 10 021 59 63AB AC ADT T T− + − = (1)

:j 20 50 50 021 59 63AB AC ADT T T P− − − + = (2)

:k 5 9 37 021 59 63AB AC ADT T T+ − = (3)

Substituting for 1800 lbP = in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives:

4 30 10 021 59 63AB AC ADT T T− + − = (1′)

20 50 50 1800 lb 021 59 63AB AC ADT T T− − − + = (2′)

5 9 37 021 59 63AB AC ADT T T+ − = (3′)

973.64 lb531.00 lb532.64 lb

AB

AC

AD

TTT

==

= 974 lbABT =

531 lbACT =

533 lbADT =


PROBLEM 2.120

Three wires are connected at point D, which is located 18 in. below the T-shaped pipe support ABC. Determine the tension in each wire when a 180-lb cylinder is suspended from point D as shown.

SOLUTION

Free-Body Diagram of Point D:

The forces applied at D are:

, , andDA DB DCT T T W

where 180.0 lb .= −W j To express the other forces in terms of the unit vectors i, j, k, we write

(18 in.) (22 in.)28.425 in.

(24 in.) (18 in.) (16 in.)34.0 in.(24 in.) (18 in.) (16 in.)34.0 in.

DADADBDBDCDC

= +=

= − + −=

= + −=

j k

i j k

i j k



and

(0.63324 0.77397 )

( 0.70588 0.52941 0.47059 )

(0.70588 0.52941 0.47059 )

DA Da DA Da

DA

DB DB DB DB

DB

DC DC DC DC

DC

DAT TDA

TDBT TDB

TDCT TDC

T

= =

= +

= =

= − + −

= =

= + −

T λ

j k

T λ

i j k

T λ

i j k

Equilibrium Condition with W= −W j

0: 0DA DB DCF WΣ = + + − =T T T j


( 0.70588 0.70588 )(0.63324 0.52941 0.52941 ) (0.77397 0.47059 0.47059 )

DB DC

DA DB DC

DA DB DC

T TT T T W

T T T

− +

+ + −

− −

ij

k


0.70588 0.70588 0DB DCT T− + = (1)

0.63324 0.52941 0.52941 0DA DB DCT T T W+ + − = (2)

0.77397 0.47059 0.47059 0DA DB DCT T T− − = (3)

Substituting 180 lbW = in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives,

119.7 lbDAT =

98.4 lbDBT =

98.4 lbDCT =


PROBLEM 2.121

A container of weight W is suspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable that passes over a pulley at B and through ring A and that is attached to a support at D. Knowing that 1000 N,W = determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.)

SOLUTION

The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector along the cable. That is, with

2 2 2

(0.78 m) (1.6 m) (0 m)

( 0.78 m) (1.6 m) (0)1.78 m

[ (0.78 m) (1.6 m) (0 m) ]1.78 m

( 0.4382 0.8989 0 )

AB AB AB

AB

AB AB

AB

AB

ABT TAB

T

T

= − + +

= − + +

=

= =

= − + +

= − + +

i j k

T λ

i j k

T i j k

and 2 2 2

(0) (1.6 m) (1.2 m)

(0 m) (1.6 m) (1.2 m) 2 m

[(0) (1.6 m) (1.2 m) ]2 m

(0.8 0.6 )

ACAC AC AC

AC AC

AC

AC

TACT TAC

T

= + +

= + + =

= = = + +

= +

i j k

T λ i j k

T j k

and 2 2 2

(1.3 m) (1.6 m) (0.4 m)

(1.3 m) (1.6 m) (0.4 m) 2.1 m

[(1.3 m) (1.6 m) (0.4 m) ]2.1 m

(0.6190 0.7619 0.1905 )

ADAD AD AD

AD AD

AD

AD

TADT TAD

T

= + +

= + + =

= = = + +

= + +

i j k

T λ i j k

T i j k



Finally, 2 2 2

(0.4 m) (1.6 m) (0.86 m)

( 0.4 m) (1.6 m) ( 0.86 m) 1.86 m

[ (0.4 m) (1.6 m) (0.86 m) ]1.86 m

( 0.2151 0.8602 0.4624 )

AE AE AE

AE

AE AE

AE

AE

AET TAE

T

T

= − + −

= − + + − =

= =

= − + −

= − + −

i j k

T λ

i j k

T i j k

With the weight of the container ,W= −W j at A we have:

0: 0AB AC AD WΣ = + + − =F T T T j

Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations:

0.4382 0.6190 0.2151 0AB AD AET T T− + − = (1)

0.8989 0.8 0.7619 0.8602 0AB AC AD AET T T T W+ + + − = (2)

0.6 0.1905 0.4624 0AC AD AET T T+ − = (3)

Knowing that 1000 NW = and that because of the pulley system at B ,AB ADT T P= = where P is the externally applied (unknown) force, we can solve the system of linear Equations (1), (2) and (3) uniquely for P.

378 NP =


PROBLEM 2.122

Knowing that the tension in cable AC of the system described in Problem 2.121 is 150 N, determine (a) the magnitude of the force P, (b) the weight W of the container.

PROBLEM 2.121 A container of weight W is suspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable that passes over a pulley at B and through ring A and that is attached to a support at D. Knowing that 1000 N,W = determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.)

SOLUTION

Here, as in Problem 2.121, the support of the container consists of the four cables AE, AC, AD, and AB, with the condition that the force in cables AB and AD is equal to the externally applied force P. Thus, with the condition

AB ADT T P= =

and using the linear algebraic equations of Problem 2.131 with 150 N,ACT = we obtain

(a) 454 NP =

(b) 1202 NW =


PROBLEM 2.123

Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables.

SOLUTION

Free Body Diagram at A:

Since tension inBACT = cable BAC, it follows that

AB AC BACT T T= =

( 17.5 in.) (60 in.) 17.5 6062.5 in. 62.5 62.5

(60 in.) (25 in.) 60 2565 in. 65 65

(80 in.) (60 in.) 4 3100 in. 5 5

AB BAC AB BAC BAC

AC BAC AC BAC BAC

AD AD AD AD AD

AE AE AE AE

T T T

T T T

T T T

T T

− + −⎛ ⎞= = = +⎜ ⎟⎝ ⎠

+ ⎛ ⎞= = = +⎜ ⎟⎝ ⎠

+ ⎛ ⎞= = = +⎜ ⎟⎝ ⎠

= =

i jT λ i j

i kT λ j k

i jT λ i j

T λ (60 in.) (45 in.) 4 375 in. 5 5AET− ⎛ ⎞= −⎜ ⎟

⎝ ⎠

j k j k



Substituting into 0,AΣ =F setting ( 200 lb) ,= −P j and setting the coefficients of i, j, k equal to ,φ we obtain the following three equilibrium equations:

From 17.5 4: 062.5 5BAC ADT T− + =i (1)

From 60 60 3 4: 200 lb 062.5 65 5 5BAC AD AET T T⎛ ⎞+ + + − =⎜ ⎟

⎝ ⎠j (2)

From 25 3: 065 5BAC AET T− =k (3)


76.7 lb; 26.9 lb; 49.2 lbBAC AD AET T T= = =


PROBLEM 2.124

Knowing that the tension in cable AE of Prob. 2.123 is 75 lb, determine (a) the magnitude of the load P, (b) the tension in cables BAC and AD.

PROBLEM 2.123 Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables.

SOLUTION

Refer to the solution to Problem 2.123 for the figure and analysis leading to the following set of equilibrium equations, Equation (2) being modified to include Pj as an unknown quantity:

17.5 4 062.5 5BAC ADT T− + = (1)

60 60 3 4 062.5 65 5 5BAC AD AET T T P⎛ ⎞+ + + − =⎜ ⎟

⎝ ⎠ (2)

25 3 065 5BAC AET T− = (3)

Substituting for 75 lbAET = and solving simultaneously gives:

(a) 305 lbP =

(b) 117.0 lb; 40.9 lbBAC ADT T= =


SOLUT

For both

Here

or

Thus, w

Now

Where y

From th

Setting t

With

Now, fr

Setting t

And usi

TION

h Problems 2.

when y given, z

y and z are in u

he F.B. Diagra

the j coefficie

rom the free bo

the k coefficie

ing the above r

125 and 2.126

(0

z is determined

λ

units of meter

am of collar A:

ent to zero give

ody diagram o

ent to zero giv

result for ABT

P

CcaPteofsy

6: 2 2( )AB x=

20.525 m) (0=

2 2 0.y z+ =

d,

10.525 m0.38095

ABABAB

=

=

=

λ

i

rs, m.

: 0:Σ =F

es (1.9P −

AB

P

T

of collar B:

ves

, we have

PROBLEM 2

ollars A and Ban slide fre

(341 N)=P j iension in the wf the force Qystem.

2 2 2y z+ +

2 20.20 m) y+

2.23563 m

(0.20

1.90476

y z

y

− +

− +

i j

i j

: x zN N+ +i k

90476 ) ABy T =

341 N341 N

1.90476

P

y

=

=

0: NΣ =F

(1.9ABQ T−

(ABQ T z= =

2.125

B are connecteeely on fricis applied to wire when y =required to m

Free

2z+

)m

1.90476

z

z+

k

k

AB ABP T λ+ +j

0

x yN N Q+ +i j k

0476 ) 0z =

341 N (1.(1.90476)y

ed by a 525-mctionless rod

collar A, de155 mm,= (b

maintain the eq

e-Body Diagr

0=

0AB ABT− =k λ

(34.90476 )z =

mm-long wire ads. If a foetermine (a) b) the magnituquilibrium of

rams of Colla

0

41 N)( )zy

and rce the ude the

rs:



Then from the specifications of the problem, 155 mm 0.155 my = =

2 2 20.23563 m (0.155 m)

0.46 mzz

= −=

and

(a) 341 N0.155(1.90476)1155.00 N

ABT =

=

or 1155 N=ABT

and

(b) 341 N(0.46 m)(0.866)(0.155 m)

(1012.00 N)

Q =

=

or 1012 N=Q


PROBLEM 2.126

Solve Problem 2.125 assuming that 275 mm.y =

PROBLEM 2.125 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force (341 N)=P j is applied to collar A, determine (a) the tension in the wire when y 155 mm,= (b) the magnitude of the force Q required to maintain the equilibrium of the system.

SOLUTION

From the analysis of Problem 2.125, particularly the results:

2 2 20.23563 m341 N

1.90476341 N

AB

y z

Ty

Q zy

+ =

=

=

With 275 mm 0.275 m,y = = we obtain:

2 2 20.23563 m (0.275 m)

0.40 mzz

= −=

and

(a) 341 N 651.00(1.90476)(0.275 m)ABT = =

or 651 NABT =

and

(b) 341 N(0.40 m)(0.275 m)

Q =

or 496 NQ =


SOLUT

Using th

we have

Then

and

Hence:

TION

he force triang

e

gle and the law

γ

R

R

10 kNsin

sin

α

α

α

φ

PROB

Two stshown.that thedetermresultanB.

ws of cosines a

180 (40120

γ = ° −= °

2 2

2

(15 kN)2(15 kN

475 kN21.794 kN

R

R

= +−

==

N 21.794 kNsin120

10 kN21.794 k

0.3973723.414

α

α

α

=°

⎛= ⎜⎝

==

50φ α= + ° =

BLEM 2.12

tructural mem. Knowing the force is 15

mine by trigonnt of the force

and sines,

0 20 )° + °

2(10 kN)N)(10 kN)cos1

N

+

N

N sin120kN

⎞ °⎟⎠

73.414

27

mbers A and hat both memkN in membeometry the mes applied to t

20°

B are bolted mbers are in cer A and 10 k

magnitude and the bracket by

21.8=R

to a bracket compression akN in member

direction of y members A a

8 kN 73.4°

as and r B, the and


PROBLEM 2.128


SOLUTION

Compute the following distances:

2 2

2 2

2 2

(24 in.) (45 in.)51.0 in.

(28 in.) (45 in.)53.0 in.

(40 in.) (30 in.)50.0 in.

OA

OB

OC

= +

=

= +

=

= +

=

102-lb Force: 24 in.102 lb51.0 in.xF = − 48.0 lbxF = −

45 in.102 lb51.0 in.yF = + 90.0 lbyF = +

106-lb Force: 28 in.106 lb53.0 in.

= +xF 56.0 lbxF = +

45 in.106 lb53.0 in.yF = + 90.0 lbyF = +

200-lb Force: 40 in.200 lb50.0 in.xF = − 160.0 lbxF = −

30 in.200 lb50.0 in.yF = − 120.0 lbyF = −


SOLUT

(a) Fo

S

(b) Si

TION

or R to be ver

et

ince R is to be

rtical, we must

e vertical:

PROBLEMA hoist trollα = 40°, detresultant of magnitude o

xR = Σ1xR P= −

yR = FΣ

410.yR =

t have 0.xR =

0xR = in E

0 17177.86P

P= −=

=R

M 2.129 ley is subjecteermine (a) ththe three forcf the resultant

(20xF PΣ = +177.860 lb

(200 lb)cyF =

32 lb

.

Eq. (1)

77.860 lb60 lb

410 lb= =yR

ed to the threehe required maces is to be vt.

0 lb)sin 40° −

cos40 (400° +

e forces showagnitude of th

vertical, (b) th

(400 lb)cos 4

lb)sin 40°

wn. Knowing the force P if he correspondi

40°

177.9 lbP =

410 lbR =

that the ing

(1)

(2)


SOLUT

Law of s

(a)

(b)

TION

Free

sines:

PRO

Knowalongcable

e-Body Diagra

sinF

OBLEM 2.1

wing that α =g line AC, dete BC.

am

n 35 sin 50AC BCF T

=°

300 lbsin 95ACF =

300 lbsin 95BCT =

30

55° and that ermine (a) the

300 lb0 sin 95

=° °

b sin 355

°°

b sin 505

°°

boom AC exe magnitude o

Force Triang

erts on pin C of that force, (

gle

F

a force direc(b) the tension

172.7 lbACF =

231 lbBCT =

ted n in


SOLUT

(a)

(b)

TION

0:xΣ = −F

0:1yΣ =F

T

PR

TwoKnoAC,

Fr

12 4 (313 5ACT− +

5 (312 N)13

T+

480 NBCT = −

ROBLEM 2.

o cables are owing that P =

(b) in cable B

ee Body: C

360 N) 0=

3 (360 N5BCT +

120 N 216 N−

131

tied together 360 N,= determ

BC.

N) 480 N 0− =

N

at C and lomine the tens

0

T

oaded as showsion (a) in ca

312 NACT =

144.0 NBCT =

wn. able


SOLUT

Force tri

(a)

S

(b)

TION

iangle is isosc

ince 0,P > th

Free-Body D

celes with

he solution is c

PROB

Two cabthe maxdeterminapplied a

Diagram: C

2 184

ββ

==

2P =

correct.

1α =

LEM 2.132

bles tied togethximum allowne (a) the maat C, (b) the co

80 857.5

° − °°

2(800 N)cos 47

80 50 47° − ° −

2

her at C are lowable tensionagnitude of thorresponding

7.5° 1081 N=

7.5 82.5° = °

oaded as shown in each cahe largest forcvalue of α.

Force Trian

wn. Knowing table is 800 ce P that can

ngle

1081 NP =

82.5α = °

that N, be


PROBLEM 2.133

The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AC is 120 lb, determine (a) the components of the force exerted by this wire on the pole, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

SOLUTION

(a) (120 lb)cos 60 cos 20xF = ° °

56.382 lbxF = 56.4 lbxF = +

(120 lb)sin 60

103.923 lby

y

F

F

= − °

= − 103.9 lbyF = −

(120 lb)cos 60 sin 2020.521 lb

z

z

FF

= − ° °= − 20.5 lbzF = −

(b) 56.382 lbcos120 lb

xx

FF

θ = = 62.0xθ = °

103.923 lbcos120 lb

yy

FF

θ −= = 150.0yθ = °

20.52 lbcos120 lb

zz

FF

θ −= = 99.8zθ = °


PROBLEM 2.134

Knowing that the tension in cable AC is 2130 N, determine the components of the force exerted on the plate at C.

SOLUTION

2 2 2

(900 mm) (600 mm) (920 mm)

(900 mm) (600 mm) (920 mm)1420 mm

2130 N [ (900 mm) (600 mm) (920 mm) ]1420 mm

(1350 N) (900 N) (1380 N)

CA CA CA

CA

CA

CA

CA

T

CATCA

= − + −

= + +

==

=

= − + −

= − + −

i j k

T λ

T i j k

i j k

( ) 1350 N, ( ) 900 N, ( ) 1380 NCA x CA y CA zT T T= − = = −


PROBLEM 2.135


SOLUTION

(600 N)[sin 40 sin 25 cos 40 sin 40 cos 25 ](162.992 N) (459.63 N) (349.54 N)(450 N)[cos55 cos30 sin 55 cos55 sin 30 ](223.53 N) (368.62 N) (129.055 N)

(386.52 N) (828.25 N) (220.49 N)

(3R

= ° ° + ° + ° °= + += ° ° + ° − ° °= + −= += + +

=

P i j ki j k

Q i j ki j k

R P Qi j k

2 2 286.52 N) (828.25 N) (220.49 N)940.22 N

+ +

= 940 NR =

386.52 Ncos940.22 N

xx

RR

θ = = 65.7xθ = °

828.25 Ncos940.22 N

yy

RR

θ = = 28.2yθ = °

220.49 Ncos940.22 N

zz

RR

θ = = 76.4zθ = °


SOLUT

Setting

TION

coefficients of

( 13

1345

AB ABT

ABTAB

T

T

=

=

−=

⎛= −⎜⎝

T λ

( 15

1549

0:

AC AC

AB

T

ACTAC

T

T

F

=

=

−=

⎛= −⎜⎝

Σ =

T λ

T

f i, j, k equal t

13:45

T− −i

40:45

T+ +j

16:45

T+ −k

0 mm) (400450

3 40 165 45 45

+

+ +

i

i j

0 mm) (40049

5 40 249 49 49

B AC

+

+ −

+ + +

i

i j

T Q

to zero:

15 049

T P− + =

40 049

T W− =

24 049

T Q− + =

PROBLE

A containeCable BACfixed suppQ Q= k acontainer i

376 N,=the same in

0 mm) (1600 mm

+

⎞⎟⎠

j

k

0 mm) ( 2490 mm

0

+ −

⎞⎟⎠

+ + =

j

k

P W

0.595

1.705

0 0.13424

EM 2.136

er of weight C passes throuports at B andare applied tin the positiodetermine P

n both portions

0 mm)k

40 mm)k

01T P=

21T W=

40T Q=

W is suspendugh the ring and C. Two forto the ring ton shown. Kand Q. (Hints of cable BAC

Free-Bod

ded from ring nd is attachedrces P=P i ato maintain

Knowing that t: The tensionC.)

dy A:

A. d to and the W

n is

(1)

(2)

(3)



Data: 376 N 1.70521 376 N 220.50 NW T T= = =

0.59501(220.50 N) P= 131.2 NP =

0.134240(220.50 N) Q= 29.6 NQ =


SOLUT

A:

Collar A

Substitu

Collar B

Substitu

(a)

Fr

(b) F

TION

A:

ute for ABλ an

B:

ute for ABλ and

rom Eq. (2):

From Eq. (1):

Aλ

Σ

nd set coefficie

0: (6Σ =F

d set coefficien

9 in.x =

PROB

Collars Afreely onB as sh

9 in.x =required

Free-Body D

BAB xAB

−= =

i

0: P NΣ = +F i

ent of i equal t

60 lb) xN ′+ +k i

nt of k equal to

60 lb2T

−

2(9 in.) +

60 lb25

AT−

(125P =

LEM 2.137

A and B are con frictionless hown, determ., (b) the co

d to maintain th

Diagrams of C

B

(20 in.)25 in.

z− +i j

y zN N T+ +j k

to zero:

25 inABT xP −

y AB ABN T′+ −j λ

o zero:

05 in.ABT z

=

2 2(20 in.) zz

+

(12 in.)in.

AB

5.0 lb)(9 in.)25 in.

7

onnected by a rods. If a 60-l

mine (a) the orresponding he equilibrium

Collars:

B:

zk

0AB ABT =λ

0.

=

0B =

2(25 in.)12 in.z

==

25-in.-long wlb force Q is

tension in magnitude o

m of the system

T

wire and can slapplied to colthe wire wh

of the force m.

125.0 lbABT =

45.0 lbP =

ide llar hen

P

(1)

(2)


PROBLEM 2.138

Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. Determine the distances x and z for which the equilibrium of the system is maintained when

120 lbP = and 60 lb.Q =

SOLUTION

See Problem 2.137 for the diagrams and analysis leading to Equations (1) and (2) below:

025 in.

ABT xP = = (1)

60 lb 025 in.

ABT z− = (2)

For 120 lb,P = Eq. (1) yields (25 in.)(20 lb)ABT x = (1′)

From Eq. (2): (25 in.)(60 lb)ABT z = (2′)

Dividing Eq. (1′) by (2′), 2xz

= (3)

Now write 2 2 2 2(20 in.) (25 in.)x z+ + = (4)

Solving (3) and (4) simultaneously,

2 2

2

4 400 625

456.7082 in.

z z

zz

+ + =

==

From Eq. (3): 2 2(6.7082 in.)13.4164 in.

x z= ==

13.42 in., 6.71 in.x z= =


PROBLEM 2F1

Two cables are tied together at C and loaded as shown. Draw the free-body diagram needed to determine the tension in AC and BC.

SOLUTION

Free-Body Diagram of Point C:

2

3

(1600 kg)(9.81 m/s )

15.6960(10 ) N=15.696 kN

W

WW

=

=


PROBLEM 2.F2

Two forces of magnitude TA = 8 kips and TB = 15 kips are applied as shown to a welded connection. Knowing that the connection is in equilibrium, draw the free-body diagram needed to determine the magnitudes of the forces TC and TD.

SOLUTION

Free-Body Diagram of Point E:


PROBLEM 2.F3

The 60-lb collar A can slide on a frictionless vertical rod and is connected as shown to a 65-lb counterweight C. Draw the free-body diagram needed to determine the value of h for which the system is in equilibrium.

SOLUTION

Free-Body Diagram of Point A:


SOLUT

Free-Bo

Free-Bo

TION

ody Diagram

ody Diagram

of Point B:

of Point C:

P

AshNdrde

1

1

250 N8tan

1tan2

E

AB

BC

W

θ

θ

−

−

=

=

=

Use this free

1 1tanCDθ −=

Use this free

Then weight

PROBLEM 2

A chairlift has hown. Knowin

N and that the sraw the freeetermine the w

765 N 1018.25 30.51014

10 22.62024

+ =

= °

= °

body to determ

1.1 10.38896

= °

body to determ

of skier WS is

2.F4

been stoppedng that each cskier in chair Ee-body diagraweight of the s

15 N

°

mine TAB and

mine TCD and

found by

SW =

d in the positichair weighs 2E weighs 765 ams needed skier in chair F

TBC.

WF.

250 NFW= −

ion 250

N, to

F.


PROBLEM 2.F5

Three cables are used to tether a balloon as shown. Knowing that the tension in cable AC is 444 N, draw the free-body diagram needed to determine the vertical force P exerted by the balloon at A.

SOLUTION



PROBLEM 2.F6

A container of mass m = 120 kg is supported by three cables as shown. Draw the free-body diagram needed to determine the tension in each cable

SOLUTION


2(120 kg)(9.81 m/s )

1177.2 NW =

=


PROBLEM 2.F7

A 150-lb cylinder is supported by two cables AC and BC that are attached to the top of vertical posts. A horizontal force P, which is perpendicular to the plane containing the posts, holds the cylinder in the position shown. Draw the free-body diagram needed to determine the magnitude of P and the force in each cable.

SOLUTION

Free-Body Diagram of Point C:


PROBLEM 2.F8

A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. Knowing that the tension in wire AB is 630 lb, draw the free-body diagram needed to determine the vertical force P exerted by the tower on the pin at A.

SOLUTION

Free-Body Diagram of point A:


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Beer11e Chapter 2 ISM - testbanklive.com · SOLUT Using th We hav Then And ION e force triang e: P So PR A. Q re (b le and the law 180 105 γ= = 2 (4 64 8.0 R R = = = 4kip sin(25