Problems (Force Systems) - Dokuz Eylül Universitydebis.deu.edu.tr/userweb/binnur.goren/Statics2017/4P_Force Syste… · 8. In opening a door which is equipped with a heavy-duty return

1.

Problems (Force Systems)

2. Determine the x-y components of the tension T which is applied to point A of the

bar OA. Neglect the effects of the small pulley at B. Assume that r and q are

known. Also determine the n-t components of the tension T for T=100 N and

q=35o.


q

b

b

r + rcos q

r -

rsin

q

qqqq sin2cos23cossin22

-- rrrrrAB

qq

qb

qq

qb

sincos

sinTsinTT

sincos

cosTcosTT

y

x

223

1

223

1

-

--

-

x-y coordinates

n-t coordinates (for q=35o and T=100 N)

N..sinsinTT

N..coscosTT

.cos

sinarctan

t

n

o

5474191335100

6766191335100

1913351

351

-

bq

bq

b

q

b

T

T

qq

q

qq

qb

qq

q

qq

qb

sin2cos23

sin1

sin2cos23

sinsin

sin2cos23

cos1

sin2cos23

coscos

-

-

-

-

-

-

r

rr

r

rr

3. In the design of the robot to insert the small cylindrical part into a close-fitting circular hole, the robot

arm must exert a 90 N force P on the part parallel to the axis of the hole as shown. Determine the

components of the force which the part exerts on the robot along axes (a) parallel and perpendicular to the

arm AB, and (b) parallel and perpendicular to the arm BC.


(a) parallel and perpendicular to the arm AB(b) parallel and perpendicular to the arm BC.

force which the part exerts on the robot

vertical

P=90 N

vertical

horizontal

//ABP=90 N

30o

AB

45o P//AB

P AB

P//AB = PAB=90 cos45=63.64 N

vertical

horizontal

//BC

P=90 N

45o

BC

30oP//BC P BC

P//BC =90 cos30=77.94 N

P BC=90 sin30=45 N

45o

60o

15o

45o

15o

the robot arm must exert a 90 N force P on the part

4. The unstretched length of the spring is r.

When pin P is in an arbitrary position q,

determine the x- and y-components of the force

which the spring exerts on the pin. Evaluate your

answer for r=400 mm, k=1.4 kN/m and q =40°.


5. Three forces act on the bracket. Determine the magnitude and direction q of F2 so that the

resultant force is directed along the positive u axis and has a magnitude of 50 N.

F3=52 N

F1=80 N


if R=50 N q =? F2 =?

F3=52 N

F1=80 N

R

jiRjRiRR

25sin5025cos5025sin25cos --

Resultant

jiR

13.21315.45 -

iF

801

jFiFF

25sin25cos 222 - qq

jijiF

482013

1252

13

5523

jFiFRF yx

685.5425cos

315.452025cos80

2

2

-

q

q

F

FFx

13.6925sin

13.214825sin

2

2

--q

q

F

FFy

1

2

2

1

685.54

13.69

25cos

25sin

2

2

-

q

q

F

F

NF 14.8835.103

264.125tan

2

-

q

q

6. The turnbuckle T is tightened until the tension in cable OA is 5 kN. Express the

force Ԧ𝐹 acting on point O as a vector. Determine the projection of Ԧ𝐹 onto the y-axis

and onto line OB. Note that OB and OC lies in the x-y plane.


Ԧ𝐹 acting on point O

𝑭𝑭𝒛

𝑭𝒙𝒚

𝑷𝒓𝒐𝒋𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒏𝒕𝒐 𝒍𝒊𝒏𝒆 𝑶𝑩

𝒙

𝒚𝑶

𝑩30o

𝒏𝑶𝑩𝑼𝒏𝒊𝒕 𝒗𝒆𝒄𝒕𝒐𝒓 𝒐𝒇 𝒍𝒊𝒏𝒆 𝑶𝑩

jinOB

60cos30cos

kNF

F

jikjinFF

OB

OB

OBOB

63.2

5.091.2866.0358.1

5.0866.083.391.2358.1

F𝒛=5sin50=3.83 kN

F𝒙𝒚=5cos50=3.214 kN

𝒙𝒚 𝒑𝒍𝒂𝒏𝒆

𝒙

𝒚𝑶

𝑪

65o

𝑭𝒙𝒚

F𝑦= F𝑥𝑦sin65=3.214sin65=2.91 kN

F𝑥= F𝑥𝑦cos65=3.214cos65=1.358 kN

kjiF

83.391.2358.1

𝑷𝒓𝒐𝒋𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒏𝒕𝒐 𝒚 − 𝒂𝒙𝒊𝒔

kNjFFy 91.2

𝑻𝒉𝒆 𝒄𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔 𝒐𝒇 𝒑𝒐𝒊𝒏𝒕𝒔 𝑩 𝒂𝒏𝒅 𝑪 𝒂𝒓𝒆 𝐵 (1.6; −0.8𝑠𝑖𝑛30; 0.8𝑐𝑜𝑠30) 𝐵 (1.6;−0.4; 0.693) , 𝐶 (0; 0.7; 1.2)

𝑻𝒉𝒆 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 𝒗𝒆𝒄𝒕𝒐𝒓 𝑩𝑪 𝒊𝒔 mBCrkjir BCBC 2507.01.16.1507.01.16.1 222

// -

𝑻𝒉𝒆 𝒖𝒏𝒊𝒕 𝒗𝒆𝒄𝒕𝒐𝒓 𝒐𝒇 𝑻 𝒓𝑩𝑪 𝒊𝒔 kjikji

nnn TBCBC

253.0548.0797.0

2

507.01.16.1/ -

-

kjikjinTT BC

75.18941175.597253.0548.0797.0750 --

Tension 𝑻 acting on point B in vector form

𝑻𝑻 acting on point B BC

BCBC

r

rTnTT

/

/

7. The cable BC carries a tension of 750 N. Write this

tension as a force 𝑇 acting on point B in terms of the unit

vectors Ԧ𝑖, Ԧ𝑗 and 𝑘. The elbow at A forms a right angle.

8. In opening a door which is equipped with a heavy-duty return

mechanism, a person exerts a force P of magnitude 40 N as shown.

Force P and the normal n to the face of the door lie in a vertical plane.

Express P as a vector and determine the angles qx, qy and qz which the

line of action of P makes with the positive x-, y- and z-axes.

P𝒛=40sin30=20 N P𝒙𝒚=40cos30=34.64 N

P𝑥= P𝑥𝑦cos20=34.64cos20=32.55 N

kjiP

20848.1155.32

𝒙𝒚 𝒑𝒍𝒂𝒏𝒆

𝒙

𝒚

20o

𝒏

//𝒙

//𝒚

P𝒙𝒚

P𝑦= P𝑥𝑦sin20=34.64cos20=11.848 N

angles qx, qy and qz which the line of action of P makes with the positive x-, y- and z-axes

6040

20cos77.72

40

848.11cos536.35

40

55.32cos

aaa zyx qqq

9. The spring of constant k = 2.6 kN/m is attached to the disk at point A and to the end fitting at

point B as shown. The spring is unstretched when qA and qB are both zero. If the disk is rotated

15° clockwise and the end fitting is rotated 30° counterclockwise, determine a vector expression

for the force which the spring exerts at point A.


10. An overhead crane is used to reposition the boxcar within a railroad car-repair shop. If

the boxcar begins to move along the rails when the x-component of the cable tension

reaches 3 kN, calculate the necessary tension T in the cable. Determine the angle qxy

between the cable and the vertical x-y plane.


Tx=3 kN, calculate tension T, the angle qxy between the cable and the vertical x-y plane.

x-component of 𝑻

kjin

kjin

T

T

154.0617.077.0

145

45

222

kjiTT

154.0617.077.0

kNTT 896.3377.0 (magnitude of 𝑻)

kNTkNT zy 6.0896.3154.04.2896.3617.0

y and z-components of 𝑻

kjiT

6.04.23

Unit vector of 𝑻

𝑻 in vector form

kNTTT yxxy 84.34.23 2222

598.9896.089.3

84.3cos xy

xy

xyT

Tqq

11. The rectangular plate is supported by hinges along its side BC and by the cable AE.

If the cable tension is 300 N, determine the projection onto line BC of the force exerted

on the plate by the cable. Note that E is the midpoint of the horizontal upper edge of

the structural support.


If T=300 N, determine the projection onto line BC of the force exerted on the plate by the cable.

x

y

z

𝑪𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔 𝒐𝒇 𝒑𝒐𝒊𝒏𝒕𝒔 𝑨,𝑩, 𝑪 𝒂𝒏𝒅 𝑬 𝒘𝒊𝒕𝒉 𝒓𝒆𝒔𝒑𝒆𝒄𝒕 𝒕𝒐 𝒄𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆 𝒔𝒚𝒔𝒕𝒆𝒎

𝑨 400, 0, 0 B (0, 0, 0)

𝑪 0, 1200𝑠𝑖𝑛25,−1200𝑐𝑜𝑠25C (0, 507.14, -1087.57)

𝑬 0, 1200𝑠𝑖𝑛25,−600𝑐𝑜𝑠25E (0, 507.14, -543.78)

kjiT

kjinTT T

21.19319.18012.142

33.844

78.54314.507400300

--

--

kjkjinTT BCBC

906.0423.021.19319.18012.142 ---

Unit vector of 𝒍𝒊𝒏𝒆 𝑩𝑪

z

y

25o

25o

BCn

kjjknBC

906.0423.025sin25cos --

Projection of T onto line BC

NTBC 26.251

12. The y and z scalar components of a force are 100 N and 200 N, respectively. If

the direction cosine l=cosqx of the line of action of the force is -0.5, write Ԧ𝐹 as a

vector.

5.0200100 - lNFNF zy

75.05.011 22222222 - nmnmnml

yzzy FNFF 61.22322

FnFFFmFF zzyy qq coscos

NFF 2.25861.22375.0

NFF xx 1.129cos - q

kjiF

2001001.129 -


13. Determine the parallel and normal components of force Ԧ𝐹 in vector form with

respect to a line passing through points A and B.

35

50°x

y

z

A

B

(3, 2, 5) m

(6, 4, 8) m

F=75 kN

F

kNFxy 311.6435

575

22

kNFF xyx 34.4150cos

kNFF xyy 265.4950sin

kNFz 59.3835

375

22

kjiF

59.38265.4934.41

Cartesian components of 𝑭


35

50°x

y

z

A

B

(3, 2, 5) m

(6, 4, 8) m

F=75 kN

F

kjiF

59.38265.4934.41

Unit vector of line AB

kjikji

nAB

639.0426.0639.0

323

323

222

Parallel component of 𝑭 to line AB (its scalar value) :

kNF

kjikjiF

nFF AB

14.72639.059.38426.0265.49639.034.41

639.0426.0639.059.38265.4934.41

//

//

//

Parallel component of 𝑭 to line AB (in vector form):

kjikjinFF AB

14.4676.3014.46639.0426.0639.014.72////

Normal component of 𝑭 to line AB (in vector form):

kjiFFF

55.7505.188.4// ---

ABn

14. Determine the magnitude and

direction angles of the resultant force

acting on the bracket.

kjiF

xyxyFF

45sin45030cos45cos45030sin45cos450

11

1 -

21 FFFR

kjiF

2.31857.2751.1591 -

Resultant


kjiF

120cos60060cos60045cos6002

kjiF

30030026.4242 -

Direction angles for 𝑭 𝟐 906045 zyx qqq

1coscoscos 222 zyx qqq

1222 nml

Direction cosines

1cos60cos45cos 222 zq

5.0cos25.0cos2 zz qq

1205.0cos90 - zzz qqq

kjiF

30030026.4242 -

Direction angles for 𝑹

029.097.633

2.18cos907.0

97.633

57.575cos418.0

97.633

16.265cos

coscoscos

zyx

zz

y

yx

xR

R

R

R

R

R

qqq

qqq

Direction Cosines of Resultant Force

3.88)029.0arccos(9.24)907.0arccos(3.65)418.0arccos( zyx qqq

kjiF

2.31857.2751.1591 -

Resultant 21 FFFR

kjiR

3002.31830057.27526.4241.159 --

kjiR

2.1857.57516.265

Magnitude of Resultant Force

NRR 97.6332.1857.57516.265 222

15. Express the force Ԧ𝐹 as a vector in terms of unit vectors Ԧ𝑖, Ԧ𝑗 and 𝑘.

Determine the direction angles qx, qy and qz which Ԧ𝐹 makes with the

positive x-, y-, and z-axes.


Position vector

kjiABr

kjiABr

AB

AB

657040

254020501525

/

/

-

------

Unit vector

kjin

kji

r

rn

AB

AB

627.0676.0386.0

657040

657040

56.103

222/

/

-

-

kjikjinFF

25.4705075.289627.0676.0386.0750 --

1627.0676.0386.0 222 - nmlnmlDirection cosines

Direction angles

-

17.51627.0cos

47.47676.0cos

7.112386.0cos

zz

yy

xx

n

m

l

qq

qq

qq

Documents

Problems (Force Systems) - Dokuz Eylül Universitydebis.deu.edu.tr/userweb/binnur.goren/Statics2017/4P_Force Syste… · 8. In opening a door which is equipped with a heavy-duty return