Ch3 Online Ppt

Ch. 3: Equilibrium

3.0 Outline 123 Mechanical System Isolation (FBD) 1252-D Systems Equilibrium Conditions 1443-D Systems Equilibrium Conditions 174

3.0 Outline

123

Ch. 3: Equilibrium

3.0 Outline

When a body is in equilibrium, the resultant on the body is zero.And if the resultant on a body is zero, the body is in equilibrium.

So,

is the necessary and sufficient conditions for equilibrium.

3.0 Outline

∑ ∑F = 0 M = 0

124

Ch. 3: Equilibrium

3.1 Mechanical System Isolation (FBD)

Free-Body Diagram (FBD) is the most important first step in the mechanics problems. It defines clearly the interested system to be analyzed. It represents all forces which act on the system. The system may be rigid, nonrigid, or their combinations. The system may be in fluid, gaseous, solid, or their combinations.

FBD represents the isolated / combination of bodies as a single body. Corresponding indicated forces may be

1. Contact force with other bodies that are removed virtually.

2. Body force such as gravitational or magnetic attraction forces.

3.1 FBD

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Ch. 3: Equilibrium

3.1 FBD

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Ch. 3: Equilibrium

3.1 FBD

Remarks1. Force by flexible cable is always a tension. Weight

of the cable may be significant.2. Smooth surface ideally cannot support the tangential

or frictional force. Contact force of the rough surfacemay not necessarily be normal to the tangential surface.

3. Roller, rocker, smooth guide, or slider ideallyeliminate the frictional force. That is the supportscannot provide the resistance to motionin the tangential direction.

4. Pin connection provides support force in any directionnormal to the pin axis. If the joint is not free to turn,a resisting couple may also be supported.

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Ch. 3: Equilibrium

3.1 FBD

128

Ch. 3: Equilibrium

3.1 FBD

Remarks5. The built-in / fixed support of the beam is capable of

supporting the axial force, the shear force, andthe bending moment.

6. Gravitational force is a kind of distributed non-contactforce. The resultant single force is the weight actedthrough C.M. towards the center of the earth.

7. Remote action force has the same overall effectson a rigid body as direct contact force of equalmagnitude and direction.

8. On the FBD, the force exerted on the bodyto be isolated by the body to be removed is indicated.

9. Sense of the force exerted on the FBD bythe removed bodies opposes the movement whichwould occur if those bodies were removed.

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Ch. 3: Equilibrium

3.1 FBD

Remarks10. If the correct sense cannot be known at first place,

the sense of the scalar component is arbitrarilyassigned. Upon computation, a negative algebraicsign indicates that the correct sense is oppositeto that assigned.

130

Ch. 3: Equilibrium

3.1 FBD

Construction of FBD1. Make decision which body or system is to be isolated.

That system will usually involve the unknown quantities.2. Draw complete external boundary of the system

to completely isolate it from all other contactingor attracting bodies.

3. All forces that act on the isolated body by the removedcontacting and attracting bodies are representedon the isolated body diagram. Forces should beindicated by vector arrows, each with its magnitude,direction, and sense. Consistency of the unknownsmust be carried throughout the calculation.

4. Assign the convenient coordinate axes.Only after the FBD is completed should the governingequations be applied.

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Ch. 3: Equilibrium

3.1 FBD

132

Ch. 3: Equilibrium

3.1 FBD

Note1. Include as much as possible the system in FBD

while the unknowns are still being revealed.2. Internal forces to a rigid assembly of members do not

influence the values of the external reactions. And sothe external response of the mechanism as a wholewould be unchanged.

3. Include the weights of the members on FBD.4. Try to get the correct sense of unknown vectors by

visualizing the motion of the whole system whenthe supports are pretended to disappear. The correctsense will oppose the motion’s direction.

5. Follow the action of force prototypes in determiningthe forces acted by the removed bodies.

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Ch. 3: Equilibrium

3.1 FBD

134

Ch. 3: Equilibrium

3.1 FBD

Ax Ay

MO

AxAy

Oy

Ox

Bx

135

Ch. 3: Equilibrium

3.1 FBD

136

Ch. 3: Equilibrium

3.1 FBD

Ax

MA

F

F

By

Ax

137

Ch. 3: Equilibrium

3.1 FBD

138

Ch. 3: Equilibrium

3.1 FBD

139

Ch. 3: Equilibrium

3.1 FBD

1. T

mg

x

NF

y

P

T

2.mg

R

On verge of being rolled overmeans the normal force N = 0

N = 0

y

x

140

Ch. 3: Equilibrium

3.1 FBD

y

x

3.

L

T

Rx

Ry

4.

y

xmg mOg

N

Ay

AX

141

Ch. 3: Equilibrium

OM 0=∑

3.1 FBD

5.

FN

R

mgT

y

x

O

6.y

xAy

AX

By

BX

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Ch. 3: Equilibrium

3.1 FBD

7.

y

x

8.mg

T

Ay

AX

y

x

L

T

Ay

AX

By

BX

143

Ch. 3: Equilibrium

3.2 2-D Eqilibrium Conditions

3.2 2-D Equilibrium Conditions

A body is in equilibrium if all forces and momentsapplied to it are in balance. In scalar form,

x y OF 0 F 0 M 0= = =∑ ∑ ∑• The x-y coordinate system and the moment point Ocan be chosen arbitrarily.• Complete equilibrium in 2-D motion must satisfy allthree equations. However, they are independentto each other. That is, equilibrium may only be satisfiedin some generalized coordinates.• System in equilibrium may stay still or move withconstant velocity. In both cases, the acceleration is zero.

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Ch. 3: Equilibrium


Categories of equilibrium

Some equations are automatically satisfied andso contribute nothing in solving the problems.

145

Ch. 3: Equilibrium


Equilibrium of a body under the action of two force only:The forces must be equal, opposite, and collinear.

Weights of the members negligible

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Ch. 3: Equilibrium


Equilibrium of a body underthe action of three force only:The lines of action of the three forces must be concurrent.The only exception is when the three forces are parallel.

The system may be reduced to the three-force member by successive addition of the known forces.

If all forces are concurrent, then the equilibrium statement calls for the closure of the polygon of forces.

147

Ch. 3: Equilibrium


Alternative Equilibrium Equations

( )x A B

Three independent equilibrium conditions:F 0 M 0 M 0

AB x-direction

= = =

¬ ⊥

∑ ∑ ∑

148

Ch. 3: Equilibrium

A B C

Three independent equilibrium conditions:M 0 M 0 M 0

A, B, and C are not on the same straight line= = =∑ ∑ ∑


Alternative Equilibrium Equations

149

Ch. 3: Equilibrium


Constraints and Statical Determinacy

The equilibrium equations may not always solve allunknowns in the problem. Simply put, if #unknowns(including geometrical variables) > #equations, then wecannot solve it. This is because the system has moreconstraints than necessary to maintain the equilibruim.This is call statically indeterminate system. Extraequations, from force-deformation material properties,must also be applied to solve the redundant constraints.

150

Ch. 3: Equilibrium



P Q

mg

#unknowns = 2#equilibrium eqs. = 2statically determinate

P

Ax

Ay

Bx

By

FBx

By

Ax

Ay

Cx

Cy

#unknowns = 4#equilibrium eqs. = 3statically indeterminate

#unknowns = 6#equilibrium eqs. = 3statically indeterminate

151

Ch. 3: Equilibrium


Adequacy of Constraints

152

Ch. 3: Equilibrium


Problem Solution1. List known – unknown quantities, and check the

number of unknowns and the number of availableindependent equations.

2. Determine the isolated system and draw FBD.3. Assign a convenient set of coordinate systems.

Choose suitable moment centers for calculation.4. Write down the governing equation, e.g. ,

before the calculation.5. Choose the suitable method in solving the problem:

scalar, vector, or geometric approach.

OM 0 = ∑

153

Ch. 3: Equilibrium


P. 3/1 In a procedure to evaluate the strength of thetriceps muscle, a person pushes down on aload cell with the palm of his hand as indicatedin the figure. If the load-cell reading is 160 N,determine the vertical tensile force F generatedby the triceps muscle. The mass of the lowerarm is 1.5 kg with mass center at G. Stateany assumptions.

154

Ch. 3: Equilibrium


P. 3/1 Assumption: contraction force from biceps muscleacts at point O

1. Known: weight of lower hand,pushing forceUnknown: triceps force, biceps force

2. FBD: lower hand

y

x

O

y

M 0 -T 25-1.5g 150 160 300 0

T=1832 N

F 0 T-C-1.5g +160=0

C=1977 N

= × × + × =

=

∑

∑CT

160 N1.5g

155

Ch. 3: Equilibrium

equivalent tension forces at the middle pulleyF = 2Tcos30


P. 3/2 1. Unknown: l,RKnown: m, b, T

2. FBD: tensioning system with cut-cable

y

x

( )

O

22 2 2 2

Three-force member with m , , and For equilibrium, three lines of action must be concurrent.

2Tbcos30M 0 F b-mg l = 0 l = mg

R= F mg 3T m g

= × × ∴

+ = +

∑

∑

g F O

F = 0

F

RT

Tmg

156

Ch. 3: Equilibrium


P. 3/3 The exercise machine consists of a lightweight cart which ismounted on small rollers so that it is free to move alongthe inclined ramp. Two cables are attached to the cart –one for each hand. If the hands are together so that the cablesare parallel and if each cable lies essentially in a vertical plane,determine the force P which each hand must exert on its cablein order to maintain an equilibrium position. The mass of theperson is 70 kg, the ramp angle is 15°, and the angleβis 18°.In addition, calculate the force R which the ramp exertson the cart.

157

Ch. 3: Equilibrium


P. 3/3 Assumption: negligible rail friction

'

'

x

y

F 0 70gsin15 - Tcos9 = 0 T = 179.9 N

F 0 R-70gcos15-Tsin9 = 0 R = 691 N

= =

∑∑

T - 4Pcos9 = 0 P = 45.5 N

2P

2P

x’

T

T

70g

R

1. Unknown: P, T, R2. FBD: exercise machine, pulley

x’15° 9°9°

T

158

Ch. 3: Equilibrium


P. 3/4 A uniform ring of mass m and radius r carriesan eccentric mass mo at a radius b and is inan equilibrium position on the incline, whichmakes an angleαwith the horizontal. If thecontacting surfaces are rough enough toprevent slipping, write the expression for theangleθwhich defines the equilibrium position.

159

Ch. 3: Equilibrium


P. 3/4 1. Unknown: F, N, θ2. FBD: ring+eccentric mass

( )'

oO o

-1ox

o

m gbsinM 0 Fr - m gbsin = 0 F =r

r mF 0 F - m m gsin = 0 = sin 1 sinb m

θθ

α θ α

= ∴

= + ∴ +

∑

∑

x’

mg

mog

F

N

160

Ch. 3: Equilibrium


P. 3/5 The hook wrench or pin spanner is used toturn shafts and collars. If a moment of 80 Nmis required to turn the 200 mm diameter collarabout its center O under the action of the appliedforce P, determine the contact force R on thesmooth surface at A. Engagement of the pinat B may be considered to occur at the peripheryof the collar.

161

Ch. 3: Equilibrium

( )B A

A

Three - force member

M 0 N 0.1sin 60 P 0.375 + 0.1cos60 0

N 1047 N

= × − × = =

∑


P. 3/5

O

shaft & hook as one system

M 0 80-P 0.375 = 0 P = 213.3 N = × ∴ ∑y

x

NA

R B

80 Nm

162

Ch. 3: Equilibrium


P. 3/6 The small crane is mounted on one side of thebed of a pickup truck. For the positionθ=40°,determine the magnitude of the force supportedby the pin at O and the oil pressure p againstthe 50 mm-diameter piston of the hydrauliccylinder BC.

163

Ch. 3: Equilibrium

1

geometry at BCDO360 340sin 40 110cos 40tan 56.2

340cos 40 110sin 40d = 360cos 200 mm

α

α

− °+ − = = + =


P. 3/6

B

C O

D

α

αd

360

110 340

164

Ch. 3: Equilibrium


( )O

2

x x x

y y y

2 2x y

Three - force member

M 0 120g 785 + 340 cos 40 C d = 0 C = 5063 N

Fp = 2.58 MPar

F 0 O Ccos = 0 O 2820 N

F 0 - O 120g + Csin = 0 O 3030 N

O = O O

πα

α

= × − ×

=

= − = = − =

+

∑

∑∑

4140 N=

P. 3/6y

x

O

Oy

Ox

O

C120g

165

Ch. 3: Equilibrium


P. 3/7 The rubber-tired tractor shown has a mass of 13.5 Mg withthe C.M. at G and is used for pushing or pulling heavy loads.Determine the load P which the tractor can pull at a constantspeed of 5 km/h up the 15-percent grade if the driving forceexerted by the ground on each of its four wheels is 80 percentof the normal force under that wheel. Also find the total normalreaction NB under the rear pair of wheels at B.

166

Ch. 3: Equilibrium

'

'

A Bx 2 2

A By 2 2

A B 2 2 2 2

A B

15F 0 P - 0.8N 0.8N 13500g 015 100

100F 0 N N 13500g 015 100

100 15M 0 N 1.8 P 0.6 -13500g 1.2 13500g 0.825 015 100 15 100

N 6.3 kN, N 124.7 kN, P = 8

= − + × = +

= + − × = +

= × − × × × − × × = + += =

∑

∑

∑5.1 kN


P. 3/7

'A B xM 0 M 0 F 0 = = = ∑ ∑ ∑

y’

x’

13500g

NA NB

0.8NA 0.8NB

alternative equations:

167

Ch. 3: Equilibrium


P. 3/8 Pulley A delivers a steady torque (moment) of100 Nm to a pump through its shaft at C. Thetension in the lower side of the belt is 600 N.The driving motor B has a mass of 100 kg androtates clockwise. Determine the magnitude Rof the force on the supporting pin at O.

168

Ch. 3: Equilibrium


P. 3/8

( )CM 0 600 - T 0.225 100 0 T = 155.6 N = × − = ∴ ∑

( )D y

y

x x x

M 0 O 0.25 600 0.2 0.075 100g 0.125

- T 0.075 - Tcos30 0.2 + Tsin30 0.125 = 0 O 906 N

F 0 Tcos30 + 600 - O 0 O 734.7 N

O = O

= × − × − − × × × ×

=

= = ∴ =

∑

∑2 2x y

y y

O 1.17 kN

F 0 Tsin30 -100g - P + O 0 P = 2.8 N

+ =

= = ∴ ∑

600 N

T 100 Nm by load

mg

y

x

OxOy

600 N

T

P spring compressed to resist rotation of the body

100g

D

D x OM 0 F 0 M 0 = = = ∑ ∑ ∑

169

Ch. 3: Equilibrium


P. 3/9 When setting the anchor so that it will dig into the sandy bottom,the engine of the 40 Mg cruiser with C.G. at G is run in reverseto produce a horizontal thrust T of 2 kN. If the anchor chainmakes an angle of 60°with the horizontal, determine theforward shift b of the center of buoyancy from its position whenthe boat is floating free. The center of buoyancy is the pointthrough which the resultant of the buoyant force passes.

170

Ch. 3: Equilibrium


P. 3/9 y

x

x

y

A

F 0 Acos60 - 2000 = 0 A = 4 kN

F 0 B - 40000g - Asin60 = 0 B = 395864 N

M 0 40000g 8 - 2000 3- Bx = 0 x = 7.915 m

b = 8- x = 85.2 mm

= ∴ = ∴ = × × ∴

∑∑∑

free floating (no thrust, tension): buoyancy force = weight, acting at C.G.backward motion: new buoyancy force acting at new position

to maintain equilibrium

40000g

ABxb

171

Ch. 3: Equilibrium


P. 3/10 A special jig for turning large concrete pipe sections (showndotted) consists of an 80 Mg sector mounted on a line of rollersat A and a line of rollers at B. One of the rollers at B is a gearwhich meshes with a ring of gear teeth on the sector so as toturn the sector about its geometric center O. When α= 0,a counterclockwise torque of 2460 Nm must be applied to thegear at B to keep the assembly from rotating. When α = 30,a clockwise torque of 4680 Nm is required to prevent rotation.Locate the mass center G of the jig by calculating r and θ.

172

Ch. 3: Equilibrium

B 1 1

2 2

M 0 = 0 : 2460 - F 0.24 0, F 10250 N

= 30 : - 4680 + F 0.24 0, F 19500 N

α

α

= ° × = = ° × = =

∑


P. 3/10

y

x

( )OM 0 0 : 80000g rcos 10250 5 0

30 : -80000g rcos 180-30- 19500 5 0 r 367 mm, 79.8

α θ

α θθ

= = ° × − × = = ° × + × =

= = °

∑

2460 NmF1

4680 Nm

F2

NANBF

80000g

173

Ch. 3: Equilibrium


3.3 3-D Equilibrium Conditions

A body is in equilibrium if all forces and momentsapplied to it are in balance. In scalar form,

x y z

x y z

O O O

F 0 F 0 F 0

M 0 M 0 M 0

= = =

= = =∑ ∑ ∑∑ ∑ ∑

• The x-y-z coordinate system and the moment point Ocan be chosen arbitrarily.• Complete equilibrium in 3-D motion must satisfy allsix equations. However, they are independentto each other. That is, equilibrium may only be satisfiedin some generalized coordinates.• System in equilibrium may stay still or move withconstant velocity. In both cases, the acceleration is zero.

174

Ch. 3: Equilibrium


175

Ch. 3: Equilibrium


Categories of equilibrium

Some equations are automatically satisfied andso contribute nothing in solving the problems.

176

Ch. 3: Equilibrium



The equilibrium equations may not always solve allunknowns in the problem. Simply put, if #unknowns(including geometrical variables) > #equations, then wecannot solve it. This is because the system has moreconstraints than necessary to maintain the equilibrium.This is call statically indeterminate system. Extraequations, from force-deformation material properties,must also be applied to solve the redundant constraints.

177

Ch. 3: Equilibrium


Adequacy of Constraints

178

Ch. 3: Equilibrium


P. 3/11 The light right angle boom which supports the400 kg cylinder is supported by three cablesand a ball-and-socket joint at O attached to thevertical x-y surface. Determine the reactionsat O and the cable tensions.

179

Ch. 3: Equilibrium

AC

BD

BE OE

OB

OD

0.408 0.408 0.8160.707 0.707

, 0.6 0.80.6 0.8

= − −= −= − ===

n i + j kn j kn k n in i + kn i + j


P. 3/11

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

OB AC

AC AC OB AC

OD BE

AC AC BE BE OD BE

OE BD

BD BD

M 0 to find T

0.75 400g 2 T 0 T 4808.8 N

M 0 to find T

2 T 0.75 2 400g 1.5 T 0 T 654 N

M 0 to find T

2 T 0.75 2 40

= − × − + × = ⇒ =

= × + × − + × = ⇒ =

= × + × −

∑

∑

∑

i j k n n

k n i + k j i n n

j n i + k

( ) ( ) AC AC OE BD

x y z

0g 2 T 0 T 2775.1 N

O 1962 N, O 0 N, O 6540 N

+ × = ⇒ = = = = ∑

j k n n

F = 0

400g

OTBE

TBDTAC

180

Ch. 3: Equilibrium


P. 3/12 The 600 kg industrial door is a uniform rectangular panel whichrolls along the fixed rail D on its hanger-mounted wheels A and B.The door is maintained in a vertical plane by the floor-mountedguide roller C, which bears against the bottom edge. For theposition shown compute the horizontal side thrust on each ofthe wheels A and B, which must be accounted for in the designof the brackets.

181

Ch. 3: Equilibrium


P. 3/12

z

AB C C

A C x x

x x x C x

M 0 600g 0.15 N 3 0 N 294.3 N

M 0 N 0.6 B 3 0 B 58.86 N

F 0 A B N 0 A 235.44 N

= × − × = ⇒ = = × − × = ⇒ = = + − = ⇒ =

∑∑∑

NC

Ax

Az

Bz

Bx

600g

182

Ch. 3: Equilibrium


P. 3/13 The smooth homogeneous sphere rests in the120°groove and bears against the end platewhich is normal to the direction of the groove.Determine the angle θ, measured from thehorizontal, for which the reaction on each sideof the groove equals the force supported bythe end plate.

183

Ch. 3: Equilibrium


P. 3/13 Projection onto two orthogonal planes

y 1 2

z

x r

r

F 0 N N N

F 0 mgcos 2Ncos30

F 0 N mgsin

if N N, tan 1/ 2cos30 30 , N mg/2

θ

θ

θ θ

= = = = = = =

= = ⇒ = ° =

∑∑∑

x

yz

z mg

mgcosθ

θ

Nr

N2

N1

N1cos30+N2cos30

30°

184

Ch. 3: Equilibrium


P. 3/14 The mass center of the 30 kg door is in thecenter of the panel. If the weight of the door issupported entirely by the lower hinge A, calculatethe magnitude of the total force supported bythe hinge at B.

185

Ch. 3: Equilibrium


P. 3/14

y

x

x A x x x

y A y y y

2 2x y

F 0 M 0 30g 0.36 B 1.5 0, B A 70.6 N

F 0 M 0 B 1.5 30g 0.9 0, B A 176.6 N

B B B 190.2 N

= = × − × = = = = = × − × = = =

= + =

∑ ∑∑ ∑

30g

30gx

y

z

Ay

By

Bx

Ax

186

Ch. 3: Equilibrium


P. 3/15 One of the three landing pads for the Mars Viking lander isshown in the figure with its approximate dimensions. The massof the lander is 600 kg. Compute the force in each leg whenthe lander is resting on a horizontal surface on Mars. Assumeequal support by the pads and consult Table D/2 in Appendix Das needed.

187

Ch. 3: Equilibrium


P. 3/15

( )

DC CA

BA DC

DC DC DC

x

DC DC CA CA CB

0.35 0.936 , 0.7664 0.418 0.4877

M 0 to find F

0.85 0.1 F 200g 0.55 0 F 1049.1 N

F 0 and symmetry about x-z plane

F 2T 0.7664 0 T T 239

= − = − + +

= × − × ⇒ =

= − × = ⇒ = =

∑

∑

n i k n i j k

k + i n j j =

n i

.5 N

g=3.73 m/s2

200g

FDC TCB

TCA

188

Ch. 3: Equilibrium


P. 3/16 The uniform 15 kg plate is welded to the vertical shaft, whichis supported by bearings A and B. Calculate the magnitudeof the force supported by bearing B during application of the120 Nm couple to the shaft. The cable from C to D preventsthe plate and shaft from turning, and the weight of theassembly is carried entirely by bearing A.

189

Ch. 3: Equilibrium

z

y

x

DC

O DC

A x x x

A y y y

2 2x y

0.95 0.316

M 0 120 0.6 T 0, T 632.9 N

M 0 B 0.2 15g 0.3 T 0.68 0, B 2265 N

M 0 B 0.2 T 0.68 0, B 680 N

B B B 2635 N

= − −

= + × = = = − × + × + × = = = × − × = =

= + =

∑∑∑

n i j

i n k


P. 3/16

x

y

z

T

15g

Bx

By

Ay

Ax15g

190

Ch. 3: Equilibrium


P. 3/17 The uniform 900x1200 mm trap door has amass of 200 kg and is propped open by the lightstrut AB at the angle θ= atan(4/3). Calculatethe compression FB in the strut and the forcesupported by the hinge D normal to the hingeaxis. Assume that the hinges act at the extremeends of the lower edge.

191

Ch. 3: Equilibrium

[ ]

( )

x

y

z

AB

C AB AB AB

C z z

C y AB AB y

2 2n y z

0.2857 0.4286 0.857

M 0 0.9 T 200g 0.45cos53.13 0, T 688 N

M 0 200g 0.6 D 1.2 0, D 981 N

M 0 D 1.2 T 0.9 0, D 147.4 N

D D D 992 N

= − − +

= × − × = = = − × + × = = = − × + − × = =

= + =

∑∑∑

n i j k

j n i

n i


P. 3/17

xy

z

DyDx

Dz

200gC

TAB

192

Ch. 3: Equilibrium


P. 3/18 The uniform rectangular panel ABCD has a mass of 40 kgand is hinged at its corners A and B to the fixed vertical surface.A wire from E to D keeps edges BC and AD horizontal. Hinge Acan support thrust along the hinge axis AB, whereas hinge Bsupports force normal to the hinge axis only. Find the tensionT in the wire and the magnitude B of the force supported byhinge B.

193

Ch. 3: Equilibrium


P. 3/18

( )[ ]

( )

x

y

DE

A

DE DE DE

A z z

AE y

0.35 0.707 0.61

M 0 0.6 40g cos30 sin 30

1.2 T 0 T 278.55 N

M 0 1.2 40g cos30 sin 30 2.4B 0, B 169.9 N

M 0 B 0 N

= − +

= × − − × ⇒ =

= × − − − = = = =

∑

∑∑

n i j k

j k i i

+ j n i =

i k i j

n B 169.9 N∴ =

40g

y

x

z

TDE

Az

Ay

Ax

Bz

By

194

Ch. 3: Equilibrium


P. 3/19 Under the action of the 40 Nm torque (couple) applied to thevertical shaft, the restraining cable AC limits the rotation of thearm OA and attached shaft to an angle of 60°measured fromthe y-axis. The collar D fastened to the shaft prevents downwardmotion of the shaft in its bearing. Calculate the bending momentM, the compression P, and the shear force V in the shaft atsection B. (note: Bending moment, expressed as a vector,is normal to the shaft axis, and shear force is also normal tothe shaft axis.)

195

Ch. 3: Equilibrium


P. 3/19

section the shaft at B revealing the reaction force and moment

[ ]

( )x y

x y

AC

z AC AC AC

z AC AC

2 2 2 2x y AC

B B B AC AC

B B

0.53 0.38 0.758

M 0 40 0.18 T 0 T 419.3 N

F 0 P T 0 P 317.8 N

V V V T P 273.5 N

M M 40 0.09 0.18 T

M 42.87 Nm, M 20.0

= −

= + × = ⇒ = = + ⇒ =

= + = − =

= + − + × =

= =

∑∑

∑

n i + j k

j n k

n k =

M 0 i + j+ k j n 0

x y

2 2b B B Nm M M M 47.3 Nm∴ = + =

xy

TAC

MBx

MBy40 NmP

Vx

Vy

196

Ch. 3: Equilibrium


P. 3/20

197

Ch. 3: Equilibrium

yOM 0 100 0.15 P 0.3 0, P 50 N = × − × = = ∑

x

y

B

C C

B A C

M 0 200 0.2 100sin15 0.35

Psin30 0.075 N 0.5 0, N 58.13 N

M 0 N 0.525 N 0.2625 200 0.2625

100cos15 0.52 100sin15 0.2237

= × − × − × − × = =

= − × − × + × + × − ×

∑

∑

A

z A B C B

Pcos30 0.635 Psin30 0.1125 0, N 108.56 N

F 0 N N 100sin15 N Psin30 200 0, N 32.44 N

− × + × = =

= + + + − − = = ∑


P. 3/20

FBD of reel only

x y

z

NA

NBNC

198

Ch. 3: Equilibrium


P. 3/21 The drum and shaft are welded together andhave a mass of 50 kg with mass center at G.The shaft is subjected to a torque (couple) of120 Nm, and the drum is prevented fromrotating by the cord wrapped securely aroundit and attached to point C. Calculate themagnitudes of the forces supported bybearings A and B.

199

Ch. 3: Equilibrium

y

z

x

B

B x x

B z z

x x x x

z

M 0 T 0.15 120 0, T 800 N

M 0 Tcos66.87 0.36 A 0.7 0, A 161.6 N

M 0 50g 0.3 Tsin66.87 0.36 A 0.7 0, A 588.6 N

F 0 A B Tcos66.87 0, B 152.6 N

F 0

= × − = = = × − × = = = × + × − × = = = + − = = =

∑∑∑∑∑ z z z

2 2 2 2x z x z

A B 50g Tsin66.87 0, B 637.6 N

A A A 610.4 N, B B B 655.6 N

+ − − = =

= + = = + =


P. 3/21

0.180.24

0.15

66.87°

T

T

50g

Az

Ax

BxBzx

y

z

200

Ch. 3: Equilibrium


P. 3/22

201

Ch. 3: Equilibrium

( ) ( )( ) ( )

BC AD

AB z

z BC BC AD AD

x BC BC AD AD

BC AD

y

0.13 0.91 0.39 , 0.48 0.84 0.241

M 0 O 0 N

M 0 1.8 T 2.1 T 0

M 0 2.1 T 2.1 T 50g 2.1 0

T 625 N, T 1024 N

M 0 M 50gx

= − + = − − +

= = = × + × = × + × − × =

= =

= + +

∑∑∑

∑

n i j k n i j k

i n k j n k =

j n i j n i

( )BC BC

y y BC BC AD AD y

x x BC BC AD AD x

2 2 2x y z

1.5 T 0

M 365.66 490.5

F 0 O T T 0, O 1429 N

F 0 O T T 0, O 410 N

O O O O 1487 N

x

×

= −

= + + = = = + + = =

= + + =

∑∑

i n j =

n j n j

n i n i


P. 3/22 double U-joint

x

z

M

50g

TADTBC

Oz

OyOx

202

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