Ch. 2: Force Systems - Chulalongkorn University: Faculties ...pioneer.netserv.chula.ac.th/~pphongsa/mech1/statics/ch2.pdf · Ch. 2: Force Systems ... P. 2/21 It is desired to remove

Ch. 2: Force Systems

2.0 OutlineOverview of Forces

2-D Force SystemsRectangular Coordinate SystemsForce, Moment, and CoupleResultants

3-D Force SystemsRectangular Coordinate SystemsForce, Moment, and CoupleResultants

2.0 Outline


2.1 Overview of Forces

Force The measure of the attempt to move a body. It isa fixed vector. For the rigid body problems or only the external effects of the external force onto the objects are of interested, force can be treated as a sliding vector. Hence, the complete description must include magnitude, direction, and line of action. And the problem makes use of the principle of transmissibility.

Contact vs. Body Force

Concentrated vs. Distributed Force




Contact vs. Body Force

Concentrated vs. Distributed Force


A

P P

B



Force Measurement by comparison or by deformation of an elastic element (force sensor)

Action vs. Reaction Force Isolate the object and the force exerted on that body is represented FBD

Combining Force by parallelogram law and principle of moment

Force Components along the specified coordinate system to satisfy the parallelogram law (reverse step)




Orthogonal Projection along the specified direction. The components of a vector are not the same as the orthogonal projections onto the same coord. system, except the coordinate system is the orthogonal (rectangular) coordinate system.


R

F1

F2

Fa

Fb

a

b

R

Fa’

Fb’

a'

b'



Addition of Parallel Forces by graphic or algebraic (principle of moment) approach


F2

F1

(1)

F2

F1

F -F

R1 R2

(2)

R1

R2

R

(3)

Ch. 2: Force Systems2.2 2-D Simple Rectangular Coordinate Systems

2.2 2-D Rectangular Coord. Systems

y

xFx

FyF

θ

( )

x y

2 2x y

x y

y x

ˆ ˆF F

F = F F

F Fcos F Fsin

arctan 2 F ,F

θ θ

θ

+

+

= =

=

x yF = F F

F = i + j

Magnitude is always positiveScalar component includes sign information too!



2.2 2-D Arbitrary Rectangular Coordinate Systems

by convenience, right-handed, geometry of the problem

β

x

y

F

Fx = Fsinβ

Fy = Fcosβ

βx

yF

Fx = Fsin(π−β)

Fy = -Fcos(π−β)

x

Fx = Fcos(α−β)

Fy = Fsin(α−β)

β

y F

α


P. 2/12 If the two equal tension T in the pulley cabletogether produce a force of 5 kN on thepulley bearing, calculate T.

5 kN


P. 2/12

60゜

T

T 5 kN5 kN

By cosine law: 2 2 25 T T 2T Tcos60= + + ⋅ °

T = 2.89 kN


P. 2/14 While steadily pushing the machine upan incline, a person exerts a 180 N force Pas shown. Determine the components of Pwhich are parallel and perpendicular to theincline.


P. 2/14

t

n

180 N10゜

15゜15゜

( )( )

t

n

P 180cos 10 15 163.1 N

P 180sin 10 15 76.1 N

= + =

= − + = −


P. 2/16 Determine the resultant R of the two forcesapplied to the bracket. Write R in terms ofunit vectors along the x- and y-axes shown.


P. 2/16 clearly and carefully draw the picture!

10゜

20゜

15゜20゜

150 N200 N

y’ y

x

x’

( ) ( )( ) ( )

'

'

x

y

x

y

R 200cos 15 + 20 150sin 10 20 88.8 N

R 200sin 15 20 150cos 10 20 244.6 N

R 200cos15 150sin10 167.1 N

R 200sin15 150cos10 199.5 N

88.8 244.6 N 167.1 199.5 N

Non-Orthogonal Coordinate System x'-yby l

− + =

= + + + =

= − =

= + =

= +

=

R = i + j i' j'

( ) ( )

aw of sine and cosine200 N --> 174.34 55.1 N150 N --> -79.8 157.2 N

174.34-79.8 55.1+157.2 94.54 212.3 N

i' + ji' + j

R = i' + j = i' + j

200 N

110゜15゜

30゜

70゜80゜

150 N


P. 2/21 It is desired to remove the spike from thetimber by applying force along its horizontalaxis. An obstruction A prevents direct access,so that two forces, one 1.6 kN and the other P,are applied by cables as shown. Computethe magnitude of P necessary to ensureaxial tension T along the spike. Also find T.


P. 2/21

x

y

No net force in y-direction

y

x

100 150R Psin atan 1.6sin atan 0200 200

P = 2.15 kN

100 150T = R Pcos atan 1.6cos atan200 200

= 3.20 kN

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠∴

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠


P. 2/24 As it inserts the small cylindrical part intoa close fitting circular hole, the robot armexerts a 90 N force P on the part parallel tothe axis of the hole as shown. Determinethe components of the force which the partexerts on the robot along axes (a) paralleland perpendicular to the arm AB, and (b)parallel and perpendicular to the arm BC.


P. 2/24Quasi-EquilibriumP is the force done by the robot on the part-P is the force done by the part on the robot

may be part of strength analysisof the robot arm

15゜45゜60゜

90 Nn1

t1n2

t2 -90cos45 90sin 45 63.6 63.6 N90cos60 90sin 60 45 77.9 N

− + = − +− + = +

1 1 1 1

2 2 2 2

P = n t n tP = n t n t


2.3 2-D Moment and Couple


Moment The measure of the attempt to rotate a body. It is induced by force. The moment vector’s direction is perpendicular to the plane established by the point and the line of action of the force. It is a fixed vector. For the rigid body problems or only the external effects of the external moment onto the objects are of interested, moment can be treated as a sliding vector. Hence, the complete description must include magnitude, direction, and line of action. And the problem makes use of the principle of transmissibility.


AM F


α

FA r

d

MA

For 2-D (in-plane rotation) problem, moment vector alwayspoints perpendicular to the plane. So it can be indicated bythe magnitude and the sense of rotation (CCW +, CW -)about the point.

Moment of about point A

( )AM F r sin Fdα= ×

= ⋅ =AM r F

*** Sign’s consistency throughout the problem ***



Varignon’s theorem The moment of a force about any point is equal to the sum of the moments of the components of the force about the same point

F

Ar

(MA ) FP

Q

(MA ) Q

(MA ) P ( )( )

( ) ( )

if

= ×

×

+

A F

A AP Q

M r F

= r P + Q

= M M

F = P + Q

Usage Calculate the moment of the force fromits components. Moments of some componentsmay be trivial to calculate.



Couple The measure of the attempt to purely rotate a body. It is produced by two equal, opposite, and non-collinear force. The couple vector’s direction is perpendicular to the plane established by those two lines of action of the force. It is a free vector and so no moment center. Only the magnitude and direction are enough to describe the couple.

α

-F

r

M

d

FrArB

( )

( ) M = F r sin Fdα

× ×

= ×

⋅ =

A BM = r F + r -Fr F

arbitrary chosen



M

d

F

-

F

M

d-F -F

M

dF

-2FM

d/2

2F

For rigid body, several pairs of equal & opposite forcescan determine the same couple. It is unique tocalculate the couple from the given pair of forces butit is non-unique to determine the pair of forces whichproduce that value of couple.

Usage Effect of the couple can be determined fromthe equivalent pair of forces. Effect from some specificpairs of forces may be trivial to calculate.

-F

F


P. 2/32 Calculate the moment of the 250 N force onthe handle of the monkey wrench about thecenter of the bolt.


P. 2/32

x

y

Varignon’s Theorem

OM 250cos15 0.2 250sin15 0.03 46.4 Nm CW

= − × + ×=


P. 2/33

x

y

Vector approach

0.03 0.35 m240cos10 - 240sin10 N

84.0 Nm= ×O

r = i + jF = i jM r F = - k


P. 2/42 The force exerted by the plunger of cylinderAB on the door is 40 N directed along the lineAB, and this force tends to keep the doorclosed. Compute the moment of this forceabout the hinge O. What force Fc normal tothe plane of the door must the door stop at Cexert on the door so that the combined momentabout O of the two forces is zero?


P. 2/42

( )O

C O

= atan 100/400 0.245 radM 40cos 0.075 40sin 0.425 7.03 Nm CWF M / 0.825 8.53 N

θθ θ

=

= − × − × =

= =

40 NFC

MO


P. 2/47 While inserting a cylindrical part into thecircular hole, the robot exerts the 90 N forceon the part as shown. Determine the momentabout points A, B, and C of the force whichthe part exerts on the robot.


P. 2/47

( ) ( )

at about

at about

0.55cos60 + 0.45cos45 0.55sin60 - 0.45sin45 0.593 0.158 m-90sin15 90cos15 23.29 86.93 N

90 0.15 13.5 Nm CCW55.23 Nm CCW

68.7 Nm CCW

= × =

= ×

= + =

AC

C

-P C A AC

A C -P C A

r = i + j = i + jF = -P = i + j = - i + jMM r F =M M M

-P

MC


P. 2/59 As part of a test, the two aircraft engines arerevved up and the propeller pitches areadjusted so as to result in the fore and aftthrusts shown. What force F must be exertedby the ground on each of the main brakedwheels at A and B to counteract the turningeffect of the two propeller thrusts? Necglectany effects of the nose wheel C, which isturned 90°and unbraked.


P. 2/59

CM 2 5 F 3 = 0F = 3.33 kN

= × − ×

no rotation resultant couple = 0

F


P. 2/63 A lug wrench is used to tighten a square-headbolt. If 250 N forces are applied to the wrenchas shown, determine the magnitude F of theequal forces exerted on the four contact pointson the 25 mm bolt head so that their externaleffect on the bolt is equivalent to that of thetwo 250 N forces. Assume that the forces areperpendicular to the flats of the bolt head.


P. 2/63

( )250 0.7 2 F 0.025F = 3500 N

× = ×

Equivalent couple system at bolt head

F


2.4 2-D Resultants

2.4 2-D Resultants

Force – push / pull body in the direction of force–- rotate the body about any axis except the

intersection line to the line of force

Dual effects : force and couple to separate the push / pulland rotate effect while maintaining theresultant force and moment (external effect)

FA

B

-F d

FA

B F

M=FdA

B F

Force-Couple System


2.4 2-D Resultants

Resultant is the simplest force combination which can replace the original system of forces, moments, and couples without altering the external effect of the system on the rigid body. The resultant force determination will be used in the Newton’s 2nd law :

m=∑F a


2.4 2-D Resultants

Resultant Determination• Force Polygon : head to tail of force vectorsNote: only magnitude and direction are ensured

i.e., line of action may be incorrect!

For the specified rectangular coordinate system,

( ) ( )( )

22

x x y y x y

y x

R F R F R= F F

arctan 2 R ,Rθ

+ + =

= = +

=

∑∑ ∑ ∑ ∑1 2 3R = F F F F


2.4 2-D Resultants

Resultant Determination• Prin. Transmissibility & Parallelogram Law :

graphical method quick and easy visualizable butlow accuracy

Note: magnitude, direction, and line of action are correct


2.4 2-D Resultants

Resultant Determination• Force-Couple Equivalent Method:

algebraic method high accuracyNote: magnitude, direction, and line of action are correct

1. Specify a convenient reference point2. Move all forces so the new lines of action

pass through point O Force-Couple Equivalence


2.4 2-D Resultants

Resultant Determination• Force-Couple Equivalent Method:

algebraic method high accuracyNote: magnitude, direction, and line of action are correct

3. Sum forces and couples to 4. Locate the correct line of action of

Force-Couple Equivalence

and OR MR

( )O i i ii i

M M Fd Rd Principle of Moment= = =∑∑ ∑

R = F


P. 2/61 In the design of the lifting hook the action ofthe applied force F at the critical section ofthe hook is a direct pull at B and a couple.If the magnitude of the couple is 4000 Nm,determine the magnitude of F.


P. 2/61Equivalent force-couple systemat the critical section

F 0.1 = 4000 F=40 kN× ∴


P. 2/68 Calculate the moment of the 1200 N forceabout pin A of the bracket. Begin by replacingthe 1200 N force by a force couple systemat point C. Calculate the moment of the1200 N force about the pin at B.


P. 2/68

C

A C

B A

M 1200 0.2 240 Nm CCW1M M 1200 0.6 562 Nm CCW5

2M M 1200 0.5 1099 Nm CCW5

= × =

= + × × =

= + × × =

Force-Couple equivalent systemmakes the moment calculation intuitive


P. 2/71 The combined drive wheels of a front-wheel-drive automobile are acted on by a 7000 Nnormal reaction force and a friction force F,both of which are exerted by the road surface.If it is known that the resultant of these twoforces makes a 15°angle with the vertical,determine the equivalent force-couple systemat the car mass center G. Treat this as a2D problem.


P. 2/71

G

Rcos15 = 7000 R = 7246.9 NM 7000 1 7246.9sin15 0.5 7937.8 Nm CW

∴= × + × =

R


P. 2/76 Determine and locate the resultant R of thetwo forces and one couple acting on the I-beam.


O

R = 8-5 = 3 kN downwardM 25 5 2 8 2 1 kNm CW= − × − × =

P. 2/76 typical step in strength analysis

First, find the equivalent force-couple at point O

Then, locate the correct line of action by prin. of moment13d = 1 d = 1/3 m & x = 4 m3

∴

O


P. 2/77 If the resultant of the two forces and couple Mpasses through point O, determine M.


P. 2/77 Resultant passes through point Omeans there is no moment at point O

OM M - 400 0.15cos30 -320 0.3 = 0M = 148 Nm CCW

= × ×


P. 2/80 The directions of the two thrust vectors of anexperimental aircraft can be independentlychanged from the conventional forward directionwithin limits. For the thrust configuration shown,determine the equivalent force-couple systemat point O. Then replace this force-couplesystem by a single force and specify the pointon the x-axis through which the line of actionof this resultant passes.


( ) ( )O

T + Tcos15 Tsin15 1.966T 0.259T NM Tcos15 3- T 3- Tsin15 10 = 2.69T Nm CW= × × ×

R = i + j = i + j P. 2/80

0.259T x = -2.69T x = -10.4 m× ∴

R

MO

Rx

Ry

Since Rx of the new force systemdoes not contribute moment about O,only Ry can be used in calculation.

MO


P. 2/84 Two integral pulleys are subjected to the belttensions shown. If the resultant R of theseforces passes through the center O, determineT and the magnitude of R and the CCW angleθit makes with the x-axis.


P. 2/84 Resultant force passes O MO = 0

( ) ( )( ) ( )

160 T 100 150 200 200 0 T = 60 N

200 +150 -160cos30 - 60cos30 160sin30 + 60sin30159.5 110 N

R = 193.7 N = 34.6θ °

− × + − × = ∴

R = i + j = i + j


P. 2/90 A rear-wheel-drive car is stuck in the snow between otherparked cars as shown. In an attempt to free the car, threestudents exert forces on the car at points A, B, and Cwhile the driver’s actions result in a forward thrust of 200 Nacting parallel to the plane of rotation of each rear wheel.Treating the problem as 2D, determine the equivalentforce-couple system at the car center of mass G and locatethe position x of the point on the car centerline throughwhich the resultant passes. Neglect all forces not shown.


( ) ( )

G

200 + 400 + 200 + 250sin30 250cos30 350925 + 566.5 N

M 350 1.65 250sin 30 0.9 690 Nm CCW

+

= × + × =

R = i + j = i j

P. 2/90• 400 N and y-direction of 250 Ncause no moment about O.• Moments by thrust force 200 Ncancel each other.

566.5 x = 690 x = 1.218 m× ∴

Rx

Ry

Since Rx of the new force systemdoes not contribute moment about G,only Ry can be used in calculation.

MO


P. 2/91 An exhaust system for a pickup truck is shown in the figure.The weights Wh, Wm, and Wt of the headpipe, muffler, andtailpipe are 10, 100, and 50 N, respectively, and act at theindicated points. If the exhaust pipe hanger at point A isadjusted so that its tension FA is 50 N, determine the requiredforces in the hangers at points B, C, and D so that theforce-couple system at point O is zero. Why is a zeroforce-couple system at O desirable.


P. 2/91

So the pipe is in equilibrium w/ no external reaction force at support O.Therefore stress at O is zero no breakage

Force-couple at point O is zero force-couple at any point is zero too!At point E,

E

( ) ( ) ( )h m t A B

B

W 0.2 1.3 0.9 W 0.65 0.9 W 0.4 F 1.3 0.9 F 0.9 0F 98.9 N

× + + + × + + × − × + − × =

=

A B C D h m t

D C

C D

F F F cos30 F cos30 W W W 0F sin 30 F sin 30 0F F 6.415 N

+ + + − − − =− =

= =

Force components in horizontal and vertical direction = 0


2 2 2x y z x y z

x x y y z z

F x y z

F

F F F F = F F F

F Fcos F Fcos F Fcos

directional unit vector cos cos cos

F

θ θ θ

θ θ θ

+ +

= = =

=

F = i + j+ k

n i + j+ k

F = n


2.5 3-D Rectangular Coordinate Systems

Force Vector description : directional unit vectorand magnitude

xz y xy z yz xF Fsin F Fsin F Fsinθ θ θ= = =



Direction of Force Vector by Two Points

( ) ( ) ( )( ) ( ) ( )

2 1 2 1 2 1F 2 2 2

2 1 2 1 2 1

FF

x x y y z zABAB x x y y z z

− − −= =

− + − + −

i + j+ kn

F = n



Direction of Force Vector by Two Angles

( ) ( ) ( )F

F

F

cos cos cos sin sin

1F

φ θ φ θ φ=

=

n i + j+ k

nF = n


F n


Orthogonal Projection of the Force Vector may not be equal to its component. They are equal when the rectangular coordinate system is used.

Orthogonal Projection of in the - direction

θ

Fn

( )

F

n F

n n F

F

F F cosF

F F

θ

=

= =

=

F nF nF n = n n

F n = n n n

ii i

i

magnitude = dot product of with F n


P. 2/102 In opening a door which is equipped witha heavy duty return mechanism, a personexerts a force P of magnitude 32 N as shown.Force P and the normal n to the face ofthe door lie in a vertical plane. Express P asa vector and determine the angles θx θy θzwhich the line of action P makes with thepositive x-, y-, and z-axes.


P. 2/102

x

y

z

Pcos30cos20 Pcos30sin20 Psin3026.0 + 9.48 +16 N

acos 35.5P

acos 72.8P

acos 60P

θ

θ

θ

°

°

°

= =

= =

= =

P = i + j+ k = i j k

P i

P j

P k

i

i

i

top view

20゜

Pxy=Pcos30

angle description


P. 2/104 The rectangular plate is supported by hingesalong its side BC and by the cable AE. If thecable tension is 300 N, determine the projectiononto line BC of the force exerted on the plateby the cable. Note that E is the midpoint of thehorizontal upper edge of the structural support.


P. 2/104 ( )( )( )( )( )

BC

BC BC

0.4,0,1.2sin 25

0,0,1.2sin 25

0,1.2cos 25,0

0.4,1.2cos 25,0

0,0.6cos 25,0

T 142.1 193.2 180.2

0.9063 0.4226

T 251.2 N

A

B

C

D

E

AEAE

BCBC

= −

=

=

= −

=

= + −

= = −

= =

T = i j k

n j k

T ni

Orthogonal projection in a direction: magnitude = dot product

2-point description


P. 2/107 The power line is strung from the power-polearm at A to point B on the same horizontalplane. Because of the sag of the cable in thevertical plane, the cable makes an angle of 15°with the horizontal where it attaches to A.If the cable tension at A is 800 N, write T asa vector and determine the magnitude of itsprojection onto the x-z plane.


2 2xz x z

y

xz y

1.5atan 8.5310

Tcos15cos + Tcos15sin Tsin15764.2 +114.6 207

T T T 792 N

or acos 81.76T

T Tsin 792 N

θ

θ θ

θ

θ

°

°

⎛ ⎞ =⎜ ⎟⎝ ⎠

−−

= + =

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

=

T = i j k = i j k

T ji

θ

P. 2/107




Scalar approach in 3-D is more difficult than vector approach

Moment of about point

and normal to the plane and through

= ×

⊥ ⊥

O

O

O O

O

M F OM r FM r M FM O



Vector Cross Product

( ) ( ) ( )y z z y z x x z x y y x

x y z

x y z

r F r F r F r F r F r F

r r r for remembranceF F F

× − − −r F = i + j+ k

i j k

Proof check & Visualizationby Prin. of Moment

x y z z y

y z x x z

z x y y x

M r F r F

M r F r F

M r F r F

= −

= −

= −



Moment of about axis through point F λ

OM

λM

1. Find moment of about point F O= ×OM r F

OMλ

2. Orthogonally project in the-direction along axis n

( ) ( )λ = ×OM M n n = r F n ni i

x y z

x y z

x y z

r r rM F F F

n n nλ =

* Point O can be any point on axis λ



3-D Couple

Couple as free vector



3-D Equivalent Force-Couple System


P. 2/114 The helicopter is drawn here with certain3-D geometry given. During a ground test,a 400 N aerodynamic force is applied tothe tail rotor at P as shown. Determine themoment of this force about point O ofthe airframe.


P. 2/114Force P does not cause moment in y-direction

( ) ( )O 400 1.2 400 6 480 + 2400 N= × ×M i + k = i k

For this simple force P, we can determinethe moment component-wise


P. 2/128 In picking up a load from position A, a cabletension T of magnitude 21 kN is developed.Calculate the moment that T produces aboutthe base O of the construction crane.


P. 2/128

Vectorial approach

( ) ( )0,18,30 6,13,0

T 4.06 3.39 20.32 kN

18 30 m-264.2 +121.9 - 73.2 kNm

A B =

ABAB

OA

=

= − −

== ×O

T = i j k

r = j+ k M r T = i j k


P. 2/128Algebraic approach

( ) ( )

( )

( )

( )

x

y

z

x

T

y

T

z

T

0,18,30 6,13,0

T 4.06 3.39 20.32 kN

translate force to , moment at O by T4.06 13

moment at O by T

3.39 6

moment at O by T20.32 13 20.32 6

264.16 121.92 7

A B =

ABAB

B

=

= − −

= − ×

= − ×

= − × + ×

∴ = − + −

O

O

O

O

T = i j k

M k

M k

M i j

M i j 3.12 kNmk


P. 2/130 The special-purpose milling cutter is subjectedto the force of 1200 N and a couple of 240 Nmas shown. Determine the moment of thissystem about point O.


P. 2/130MO = moment induced by force + free vector couple

1200cos30 -1200sin30 1039 - 600 N0.2 + 0.25 m

240cos30 - 240sin30 -259.8 + 327.8 +87.8 Nm= ×O

R = j k = j k r = i k M r R + j k = i j k


P. 2/133 A 5 N vertical force is applied to the knobof the window-opener mechanism when thecrank BC is horizontal. Determine the momentof the force about point A and about line AB.


P. 2/133

( )

( )

75cos30 + 75 + 75sin30 mm-5 375 325 Nmm

cos30 sin 30281 162.4 Nmm

= × = − +

= +

= = − −

A

AB

AB A AB AB

r = i j kM r k i jn i kM M n n i ki


and R MR

2.7 3-D Resultants

2.7 3-D Resultants

Resultant is the simplest force combination which can replace the original system of forces, moments, and couples without altering the external effect of the system on the rigid body.

Vectorial approach is more suitable in 3-D problems.1. Define the suitable rectangular coord. System and

specify a convenient point O2. Move all forces so the new lines of action pass

through point O force-couple equivalence3. Sum forces and couples to 4. Locate the correct line of action of

force-couple equivalence solving piercing point(2 unknowns: ) rank-2 degenerated×r R = M


( )go together to determine the resultant

Principle of Moment

⎧⎪⎨

×⎪⎩

∑∑

R = F

M = r F

M

2.7 3-D Resultants

The selected point O specifies the couple

Dynamics: calculate the resultantsat C.M.G

m

I θ

=

=

∑∑

G

G

F x

M

Statics: calculate the resultants at any point0

0

=

=∑∑

F

M


( ) 0× =∑ ∑R = F M = r F

= F = = ×∑ ∑OR M M r R

R

2.7 3-D Resultants

Resultants of Special Force Systems

Concurrent Forces No moment about the pointof concurrency

Parallel Forces Magnitude of = magnitude ofalgebraic sum of the given forces

Wrench Resultant as the resultantof screwdriver

R M


2.7 3-D Resultants

Wrench Resultant – Force-Couple Equivalence

a) Determine the force-couple resultantat convenient point O

b) Orthogonally project along and perp. to

c) Transform couple into equiv. pair of with applied at O to cancel

d) Resultant with correct line of action andremains wrench resultant

and R M

M Rn

R

and -R R

( ) = = =R 1 R R 2 1Rn M M n n M M - MR

i

2M-R R

1M R

Wrench resultant is the simplest form to visualizethe effect of general force system on to the object :translate and rotate about the unique axis – screw axis


2.7 3-D Resultants

axis of the wrench, which is , lies in a planethrough O and plane defined by⊥

Rand R M


P. 2/139 The pulley and gear are subjected to the loadsshown. For these forces, determine theequivalent force-couple system at point O.


P. 2/139

( )800 + 200 -1200sin10 1200cos10792 +1182 N

800 N : 800 0.55 -800 0.1200 N : 200 0.55 200 0.11200 N : 1200sin10 0.22 1200cos10 0.075 1200cos10 0.22

260 504 28.6 Nm

+

= − × ×= − × ×= × × ×

= + = − +

1

2

3

O 1 2 3

R = i j = i j

M j kM j+ kM j+ k + i

M M + M M i j k

typical problem in shaft analysis


P. 2/141 Two upward loads are exerted on the small 3Dtruss. Reduce these two loads to a singleforce-couple system at point O. Show that Ris perpendicular to Mo. Then determine thepoint in the x-z plane through which theresultant passes.


P. 2/141

2400 N800 2.4 1600 2.4 1600 0.91440 + 5760 Nm

determine line of action of must be x m far from yz plane to produce 5760 Nm

5760 = 2400 x x = 2.4 m must be - z m far from xy plane to

= × × ×

× ∴

O

R = jM k + k + i = i k

RR k

R

( ) produce 1440 Nm

1440 = 2400 -z z = -0.6 m× ∴

i


P. 2/148 Replace the two forces acting on the blockby a wrench. Write the moment M associatedwith the wrench as a vector and specify thecoordinates of the point P in the x-y planethrough which the line of action of the wrenchpasses.


( )F F

F a + c Fb=O

R = i - kM j- k

P. 2/148 a) Determine force-couple resultant at O

b) Project MO || and ┴ nR

( )

( )

1 12 2

Fb Fb2 2

Fb FbF a + c2 2⊥

=

= = −

= − = − −

R

O R R

O

n i - k

M M n n i k

M M M i + j k

i


x y

bx=a+c, y=2

⊥ = ×r = i + jM r R

P. 2/148c) Transform couple M┴ into pair of force R and –R

If r points to the piercing point of the xy plane,

d) Wrench consisting of and actsb through xy plane at x = a+c, y = 2

R M


P. 2/149 The resultant of the two forces and couplemay be represented by a wrench. Determinethe vector expression for the moment M ofthe wrench and find the coordinates of the pointP in the x-z plane through which the resultantforce of the wrench passes.


P. 2/149

( )( ) ( )

( ) ( )

100 100 NLet point P in xz plane, where the wrench passes, has the coordinate x, 0, z .

Moment about point P = 100 z 100 0.4 x 100 0.4 z 100 0.3 20

100z 20-100z 10-100x NmThis moment at po

× × − × − ×

=P

R = i + j

i + k + j- k - j

M i + j+ k int P must equal to the couple of the wrench passing through point P.

And since it is the wrench, .x = 0.1 m, z = 0.1 m

10 10 Nm∴

= +

P

P

M R

M i j

Documents

Ch. 2: Force Systems - Chulalongkorn University: Faculties ...pioneer.netserv.chula.ac.th/~pphongsa/mech1/statics/ch2.pdf · Ch. 2: Force Systems ... P. 2/21 It is desired to remove