free body diagram and equilibrium

Engineering Mechanics

For

ME/CE

By

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Contents

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Contents

Chapters Page No.

#1. Introduction 1

Introduction 1

#2. Free Body Diagram and Equilibrium 2 28

Introduction 2

Equivalent Force System 2 3

Newtons Laws of Motion 3

Equilibrium and Free Body Diagrams 3

Coplanar Concurrent Forces 4 6

Coplanar Non-Concurrent Forces 7

Condition for Body in Equilibrium 7 8

Friction 8

Solved Examples 9 24

Assignment 25 27

Answer Keys & Explanations 27 28

#3. Trusses and Frames 29 42

Trusses and Frames 29 31


Assignment 39 40


#4. Friction 43 51

Introduction 43

Dry Friction 43 44

Laws of Dry Friction 44 45

Rolling Resistance 45

Force of Friction on a Wheel 46 47

Assignment 48 49


Contents

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#5. Principle of Virtual Work 52 59

Principle of Virtual Work 52 54


#6. Kinematics and Dynamics of Particle 60 87

Introduction 60

Kinematics of Rectilinear Motion 60 65

Kinematics of Curvilinear Motion 65 66

Acceleration Analysis 66 77

Impulse and Momentum 77 79

Collision of Elastic Bodies 79 82

Assignment 83 85


#7. Work & Energy Methods 88 94

Work and Energy 88 89

Conservative/Non-Conservative Force Fields and Energy Balance 89 93

Assignment 94

Answer Keys & Explanations 94

#8. Kinematics and Dynamics of Rigid Body 95 107

Center of mass and Center of Gravity 95

Eulers Equation of Motion 95 96

Moment of Inertia 96 102

Conservation of Angular Momentum 103 104

Assignment 105 106


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"I am a slow walker ... but I

never walk backwards."

..Abraham Lincoln

Free Body Diagram

and Equilibrium Learning Objectives After reading this chapter, you will know:

1. Equivalent Force System, Newtons Law of Motion

2. Equilibrium and Free Body Diagrams, Type of Equilibrium

3. Static Friction, Virtual Work, Trusses and Frames, Statics Related Problems

Introduction Statics deals with system of forces that keeps a body in equilibrium. In other words the resultant of

force systems on the body are zero.

Force

A force is completely defined only when the following three characters are specified.

Magnitude

Point of Application

Line of action/Direction

Scalar and Vector

A quantity is said to be scalar if it is completely defined by its magnitude alone. e.g. length, energy,

work etc. A quantity is said to be vector if it is completely defined only when its magnitude and

direction is specified.

E.g.: Force, Acceleration.

Equivalent Force System Coplanar Force System: If all the forces in the system lie in a single plane, it is called coplanar force

system.

Concurrent Force System: If line of action of all the forces in a system passes through a single point it

is called concurrent force system.

Collinear Force System: In a system, all the forces parallel to each other, if line of action of all forces

lie along a single line then it is called a collinear force system.

CH

AP

TE

R

1

2

Free Body Diagram and Equilibrium


Force System Example Coplanar like parallel force is straight Weight of stationary train on rail off the track Coplanar concurrent force Forces on a rod resting against wall Coplanar non- concurrent force Forces on a ladder resting against a wall when a person

stands on a rung which is not at its center of gravity Non- coplanar parallel force The weight of benches in class room Non- coplanar concurrent force A tripod carrying camera Non- coplanar non-concurrent force Forces acting on moving bus

Newtons Laws of Motion First Law: Everybody continues in its state of rest or of uniform motion in a straight line unless it is

compelled to change that state by force acting on it.

Second Law: The rate of change of momentum of a body is directly proportional to the applied force

& it takes place in the direction in which the force acts.

F (mdv

dt)

Third Law: For every action, there is an equal and opposite reaction. Principle of Transmissibility of Forces: The state of rest or motion of rigid body is unaltered if a force action on a body is replaced by another force of the same magnitude and direction but acting anywhere on the body along the line of action of applied forces.

Parallelogram Law of Forces: If two forces acting simultaneously on a body at a point are

represented in magnitude and direction by the two adjacent sides of a parallelogram their resultant

is represented in magnitude and direction by the diagonal of the parallelogram which passes

through the point of intersection of the two sides representing the forces.

Equilibrium and Free Body Diagrams Equilibrium: Any system of forces which keeps the body at rest is said to be equilibrium, or when the

condition of the body is unaffected even though a number of forces acted upon it, is said to in

equilibrium.

Laws of Equilibrium

Force Law of Equilibrium: For any system of forces keeping a body in equilibrium, the algebraic sum of forces, in any direction is zero, ie. F = 0

Moment Law of Equilibrium: For any system of forces keeping a body in equilibrium, the algebraic sum of the moments of all the forces about any point in their plane is zero. i.e., M = 0 F d = 0 This law is applicable only to coplanar, non-concurrent force systems.

A

P

P



Coplanar Concurrent Forces Triangle Law of Forces

If two forces acting simultaneously on a body are represented by the sides of triangle taken in order,

their resultant is represented by the closing side of the triangle taken in the opposite order.

Polygon Law of Forces

If a number of forces acting at a point be represented in magnitude and direction by the sides of a

polygon in order, then the resultant of all these forces may be represented in magnitude and

direction by the closing side of the polygon taken in opposite order.

Resultant, (R) = P12 + P2

2 + 2P1P2cos

tan = (P2sin

P1+P2cos)

Where,

= Angle between two forces, = Inclination of resultant with force P1

When forces acting on a body are collinear, their resultant is equal to the algebraic sum of the forces.

Lamis Theorem: (Only three coplanar concurrent forces) If a body is in equilibrium under the action

of three forces, then each force is proportional to the sine of the angle between the other two forces.

P1

sin=

P2sin

=P3

sin

P1 P2

P3

P2

P1

P3

a

b

c

P2

B C

D E

P1

A

P1

P2 P3

P4 P2

P4

C

P1

P3

E

R

B A

R2

R1

D



Free Body Diagram: A free body diagram is a pictorial representation used to analyze the forces

acting on a free body. Once we decide which body or combination of bodies to analyze, we then treat

this body or combination as a single body isolated from all our surrounding bodies.

A free body diagram shows all contact and non-contact forces acting on the bodies.

Sample Free Body Diagrams

A Ladder Resting on Smooth Wall

A Cantilever Beam

A Block on a Ramp

In a free body diagram all the contacts/supports are replaced by reaction forces which will exert on

the structure. A mechanical system comprises of different types of contacts/supports.

mg m

Free Body Diagram of Just the Block

j

i

F3 F2 F1

V V V V

F

M W=m

g

y

x

P

600N

G

W 600N

R1

P R2



Types of Contacts/Supports

Following types of mechanical contacts can be found in various structures,

Flexible Cable, Belt, Chain or Rope

Force exerted by the cable is always a tension away from the body in the direction of the cables.

Smooth Surfaces

Contact force is compressive and is normal to the surfaces.

Rough Surfaces Rough surfaces are capable of supporting a tangential component F (frictional force as well as a normal component N of the resultant R.

Roller Support

Roller, rocker or ball support transmits a compressive force normal to supporting surface.

Freely Sliding Guide

Collar or slider support force normal to guide only. There is no tangential force as surfaces are considered to be smooth.

Pin Connection

A freely hinged pin supports a force in any direction in the plane normal to the axis; usually shown as two components Rx and Ry. A pin not free to turn also supports a couple M.

Built in or Fixed End

A built-in or fixed end supports an axial force F, a transverse force V, and a bending moment M.

A

M

F

V

A

Weld

O

r

A

R

y

Rx

Rx

Ry

M

N N

N

N

N

Weight of Cable Negligible

Weight of Cable not Negligible

T

T



Coplanar Non-Concurrent Forces Varignons Theorem: The algebraic sum of the moments of a system of coplanar forces about a

momentum center in their plane is equal to the moment of their resultant forces about the same

moment center.

R.d = P1.d1 +P2.d2

Effect of couple is unchanged if

Couple is rotated through any angle.

Couple is shifted to any position.

The couple is replaced by another pair of forces whose rotated effect is the same.

Couple is free vector.

Condition for Body in Equilibrium The algebraic sum of the components of the forces along each of the three mutually

perpendicular direction is zero.

The algebraic sum of the components of the moments acting on the body about each of the

three mutually perpendicular axis is zero.

When a body is in equilibrium, the resultant of all forces acting on it is zero. Thus, the resultant

force R and the resultant couple M are both zero and we have the equilibrium equations,

R = F = 0 & M = M = 0

For collinear force system

Fx = 0, Fy = 0 & Fz = 0

For non-collinear force system

MA = 0 , MB = 0 & MC = 0

These requirements are both necessary and sufficient conditions for equilibrium.

Two forces can be in equilibrium only if they are equal in magnitude, opposite in direction, and

collinear in action. If a system is in equilibrium under the action of three forces, those three

forces must be concurrent.

A

B

R

P1

d1

d

d2

P2



Wrench: When the direction of resultant couple M and resultant force F are parallel, then it is

called wrench.

When direction of resultant couple & direction of resultant force is same then it is called

Positive wrench and when the direction opposite to each other it is called negative wrench.

Example of wrench is screw driver

Types of Equilibrium

There are three types of equilibrium as defined below,

Stable Equilibrium: A body is in stable equilibrium if it returns to its equilibrium position after it has

been displaced slightly.

Unstable Equilibrium: A body is in unstable equilibrium if it does not return to its equilibrium

position and does not remain in the displaced position after it has been displaced slightly.

Neutral Equilibrium: A body is in neutral equilibrium if it stays in the displaced position after if has

been displaced slightly.

Stable Equilibrium Unstable Equilibrium Neutral Equilibrium

Friction Friction is the force resisting the relative motion at solid surfaces, fluid layers and material elements

sliding against each other.

Types of Friction

1. Dry Friction: Friction between the contact surface.

2. Fluid Friction: Friction between the layers of fluid element.

3. Internal Friction: When cyclic load applied on the solid then, friction between the elements.

When applied force is, F

And static friction coefficient,

Fmax = N

FFmax

F = kN= Friction Force

k = Kinetic Friction Co-efficient



Solved examples Example 1

ABCD is a string suspended from points A and D and carries a weight of 5 N at B and a

weight of W N at C. The inclination to the vertical of AB and CD are 45 and 30

respectively and angle ABC is 165 . Find W and thetensions in the different parts of the

string.

Solution:

Let T1,T2 and T3 be the tensions in the parts AB, BC and CD respectively, as shown in figure.

For the equilibrium of point B, we have

T1sin 60

=T2

sin 135=

5

sin 165(From Lamis theorem)

T1 = 5sin 60

sin 165=

5 0.86602

0.25882= 16.73 N

T2 = 5sin 135

sin 165=

5 0.70710

0.25882= 13.66 N

For the equilibrium of point C, we have T2

sin 150=

T3sin 120

=W

sin 90 (From Lamis theorem)

T3 = T2 (sin 120

sin 150) = 13.66

3

2

2

1= 23.66 N

W = T2 sin 90

sin 150=

13.66 1

0.5 = 27.32 N

Example 2

A fine string ABCDE whose extremity A is fixed has weights W1 and W2 attached to it at B

and C and passes over a smooth pulley at D carrying a weight of 20 N at the free end E. If

in the position of equilibrium, BC is horizontal and AB, CD makes angles 60 and 30

respectively with the vertical, find

(A) Tensions in the portions AB, BC, CD and DE

(B) The value of the weights, W1 and W2

(C) The pressure on the pulley axis

Solution:

Since the string passes over a smooth pulley at D, the tension in CD portion of string is 20 N.

Let the tension in AB and BC be T1 and T2 respectively, as shown in figure.

For the equilibrium of point B, we have

T33

B

C

D

5 N

W

45

30 120

T2 T2

T11

A



T1sin 90

=T2

sin 120=

W1sin 150

And for the equilibrium of point C, T2

sin 150=

20

sin 90=

W

sin 120

Hence, W2 = 20 sin 120o

sin 90o = 20

3

2 = 17.32 N

T2 = 20 sin 150o

sin 90o = 20

1

2 = 10 N

Thus, T1 = T2 sin 90o

sin 120o =

10 2

3 = 11.55 N

W1 = T2 sin 150

sin 120o = 10

1

2

2

3 = 5.77 N

Pressure on the pulley

F = (20)2 + (20)2 + 2 20 20 cos 30o = 20 2 + 2 3

2 = 20 2 + 3 = 38.6 N

Example 3

A beam AB hinged at A and is supported at B by a vertical chord which passes over two

frictionless pulleys C and D. If the pulley D carries a vertical load W, find the position x of

the load P if the beam is to remain in equilibrium in the horizontal position.

x

P

A B

C

D

T1

W

l

T1

T1

A

D

E

W2 W1

B C

20 N 60

30

T1

T2 T2

20 N

20 N



Solution:

From pulley D

2T1 = W

T1 =W

2

Taking moments about A

MA = 0 =W

2 l p. x = 0

Wl

2= px

x =Wl

2P

Example 4

The wire passing round a telephone pole is horizontal and the two portions attached to

the pole are inclined at an angle of 60o to each other. The pole is supported by another

wire attached to the middle point of the pole and inclined at 60o to the horizontal. Show

that the tension in this wire is 43 times that of the telephone wire.

Solution:

Let the tension in the two portions of the telephone wire be T1each and the tension in

another wire be T2, as shown in figure.

Then T = 2 T1cos 30o = 3T1

Let AC = BC = l

Taking moments about B, we get

T 2l = T2 cos 60 l

T2 = 2T1 3 2

= 4 3T1 T2T1

= 43

T1

T A

B

C

T1

T2

60

D 60

A

P

x

T1 =W

2

l



Example 5

Two halves of a round homogeneous cylinder is held together by a thread wrapped round

the cylinder with two equal weights, P attached to its ends, as shown in figure. The

complete cylinder weighs, W Newton. The plane of contact of both of its halves is vertical.

Determine the minimum value of P for which both halves of the cylinder will be in

equilibrium on a horizontal plane.

Solution:

Given the problem as shown in below figure. We draw the free body diagram as follows.

Note the following salient points

In the F.B.D. As the question is to find the minimum value of force F on rope for which the two halves

just remain in contact, we see that the limiting case is that the two halves are just about to touch. In this case, the two halves rotate about point of contact C. So, the point of contact as acts as revolute joint about which the two semi-circular cylinders rotate and hence has two normal reactions Nx & Ny as shown in FBD.

In case of a half turn rope (rope which goes around the cylinder just half a turn around

the top half once), to split the two halves as we have to cut the rope once. On cutting

the rope, the rope tension force P is exposed once on the top tip each half, which is why

it is marked on top.

The left two force, are gravity of each half which is 1

2 acting on Center of Gravity of each

semi cylindrical half. To calculate the CG location, we know that

CG = xdA

A

dA

A

Now, employing polar coordinates and taking a infinitesimal element as shown in figure, we get as follows.

1

2W

1

2W

Nx Nx

Ny Ny

P

P

P

P

P P

W

C

N

4r

3

W/2

r

W

P P

P

G

(a) (b)

A



CG = xdA

A

dA

A

= r

R

0sin rdrd

0

(rdr)dR

0

0

=

23 R

3

12 R

2=

4R

3& CGy =

yda

A

dA

A

= r

R

0cos (rdr)d

0

(rdr)dR

0

0

(which is clear from symmetry of the body about X axis)

Now, as the body is to be in static equilibrium in the limiting condition, we get by zero moment sum about point P as follows.

(W

2) (

4R

3) + (P)(R) (P)(2R) = 0 P =

2W

3

Modification/Extension for Multiple Turns: When the number of turns on the cylinder

increases, by physical intuition, clearly, the minimum value of P required to just hold the

two halves together must be lesser, right? Lets check if the solution gives this analytically.

Assume that the rope turns n full turns around the cylinder. In this case, when we try to

draw FBD of two halves separately, we have to cut the ropes n+1 times on top edge of

cylinder which means a force of (n+1)P acts on top edge. In addition, we have to cut rope

n times at bottom edge ie at point C, which means force acting at bottom point is nP. With

these modifications, we get the new FBD as follows. Now,

Again taking moment about point C, we get (W

2) (

4R

3) + (P)(R) ((n + 1)P)(2R) = 0

P =1

2n + 1

2W

3

Clearly, as number of turns n 1

2n + 1 The minimum force to hold the halves together P

FBD for the General n Rope Turn Case

1

2w

P P

(n + 1)P (n + 1)P

nP nP Ny Ny

1

2w

Nx Nx

Semi-Cylinder COG Calculation Diagram

dA = (r d)dr

x = r sin

y = r cos

d dr r

Y

X



Example 6

A smooth circular cylinder of radius 2 m is lying in a triangular groove, one side of which

makes an angle of 10 and the other an angle of 30 with the horizontal, as shown in

figure. Find the reactions at the surface of contact if there is no friction and the weight of

the cylinder is 150 N.

Solution:

Let R1 and R2 be the reaction of the 10 and 30 planes respectively.

Using Lamis theorem, we get W

sin 40 =

R1sin 150

= R2

sin 170

R1 = W sin 150

sin 40= W

0.5

0.64278 = 0.778 W

= 0.778 150 = 116.6 N

R2 = W sin 170

sin 40= 150

0.17365

0.64278 = 40.52 N

Example 7

Two smooth spheres of weight, W and radius, r each are in equilibrium in a horizontal

channel of width (b



Solution:

Let R1R2 and R3 be the reactions at C, E and D respectively. Also let P be the force exerted

by one sphere on the other at the point of contact O. Then,

cos = b 2r

2r

The forces acting at the point A are (R3 W), R1 and P. Using Lamis theorem, we get R3 W

sin =

R1sin(90 )

= P

sin 90

P = R3 W

sin

R1 = (R3 W) cot

The forces acting at the point B are W, R2 and P. Again using Lamis theorem. R2

sin(90 ) =

P

sin 90 =

W

sin ; R2 = W cot ; P =

W

sin

For r = 25 cm and b = 90 cm

cos =90 (2 25)

2 25=

40

50= 0.8

= 36.87 R2 = 100 cot 36.87 = 133.3 N

P = 100

sin 36.87 = 166.66 N

R3 = P sin + W = 166.66 sin 36.87 + 100 = 200 N

R1 = (200 100) cot 36.87 = 133.33 N.

Example 8

A uniform wheel of 0.5 m diameter and weighing 1.5 kN rests against a rectangular block

0.2 m hight lying on a horizontal plane, as shown in figure. It is to be pulled over this

block by a horizontal force, P applied to the end of a string around the circumference of

the wheel. Find the force, P when the wheel is just about to roll over the block.

Solution:

Let W = weight of wheel, RA = reaction on the wheel at A

The three forces P, W and RA are in equilibrium. Since P and W meet at D, therefore RA

must pass through D. Using Lamis theorem, we have P

sin(180 ) =

W

sin(90 + )

0.2m 0.2m

1.5 kN

0.25 m

A

B

C

B

C

D D P

P

(b) (a)

W

E

A

RA



P

sin = W cos

P = W tan

tan = AE

DE

AE = (AC)2 (CE)2 = (0.25)2 (0.25 0.2)2

0.0625 0.0025 = 0.06 = 0.245 m

tan = 0.245

0.3 = 0.8165

= 39.23

P = 1.5 0.8165 = 1.225 kN

Example 9

Two rollers of weights W1 and W2 are connected by a flexible string AB. The rollers rest

on two mutually perpendicular planes DE and EF, as shown in figure. Find the tension in

the string and the angle, that it makes with the horizontal when the system is in

equilibrium. Take, W1 = 60 N, W2 = 120 N and = 30

Solution:

Let RA and RB be the reaction on the planes at A and B respectively and T the tension in

the string AB. These forces are shown in figure.

Roller A[Fig (b)] Applying Lamis theorem at A, we have

T

sin(90 + )=

W1sin{180 ( + )}

T

cos =

W1sin( + )

. (1)

Roller B [Fig. (c)] T

sin(180 ) =

W2sin{90 + ( + )}

(a)

A

B T T

(90 )

W1

W2

RB

RA

B A

D

E

F

(90 )

W1

W2



T

sin =

W2cos ( + )

. (2)

sin( + ) = W1 cos

T; cos ( + ) =

W2 sin

T

tan( + ) = W1W2

cot = 60

120 cot 30 = 0.866

+ = 40.89; = 40.89 30 = 10.89

T =W1co s

sin ( + )=

60co s 30

si n 40.89= 79.38 N

Example 10

Three cables are joined at the junction ring C. Determine the tensions in cables AC and BC

caused by the weight of the 30 kg cylinder.

Solution:

Let T1 and T2 be the tension in the string AC and BC respectively.

60o

T = 294.3 N T1

T2

45o 15o

120o 135o

45o

30o

30o

A

45o

B

C 15o

30kg

D

T

T

W1

W2

(90 ) (90 )

(90 )

(90+ )

(90+ + )

()

RA RB

(b)

(c)



3

4

36.87o

294.3

sin 105o=

T1sin 135o

=T2

sin 120o

T1 = 304.68 sin 135o = 215.44 N

T2 = 304.68 sin 120o = 263.86 N

T = 294.3 N T1 = 215.44 N T2 = 263.80 N

Example 11

The flanged steel cantilever beam with riveted bracket is subjected to the couple and two

forces shown and their effect on the design of the attachment at A must be determined.

Replace the two forces and couple by an equivalent couple M and resultant force, R at A.

Solution:

Fx = 2 cos 70 + 1.2 cos 36. 87o

Fx = 1.644 kN

Fy = 2 sin 70 1.2 sin 36. 87o

Fy = 1.1594 kN

R = Fxi + Fyj

R = 1.644i + 1.1594j MA = [2 cos 70

o 0.15] + [2 sin 70o 2] + [1.2 cos 36.87 0.15] [1.2 sin 36.87 1.5] 0.5

MA = 2.72 0.5 MA = 2.22 Nm ccw

Example 12 A ladder rests at an angle, to the horizontal, with its ends resting on a smooth floor and against a smooth vertical wall, the lower end being attached by a string to the junction of the wall and the floor. (a) Find the tension in the string (b) Find also the tension in the string when a man whose weight is one-half that of the

ladder stands on the ladder at two-thirds of its length

A X

G D

S B

R

W W/2

Y

T C

y

x

500Nm

0.15m

0.15m

2kN 1.5m

A

3 4 1.2k

N

0.5

m 70o



Solution:

Let AB be the ladder resting against the wall BC. Let R and S be the reactions of the floor

and wall respectively and T be the tension in the string AC.

MB = 0 it gives

R AB cos = T (AB sin ) + W (AB

2cos )

R cos = T sin + W

2cos

Fy = 0 it gives

R = W

T sin = W cos W

2cos

T = W

2cot

When a man of weight W

2 stands at D where AD =

2

3 AB, then,

Fy = 0 it gives

R = 3

2 W

MB = 0 it gives

R AB cos = T AB sin + W ( AB

2cos ) + (

W

2

1

3ABco s )

Or R cos = T sin +W

2cos +

W

6cos

Or T sin = 3

2W cos

W

2cos

W

6cos

T = 5

6 W cot

Example 13

A jib crane is loaded as shown in figure. Determine the forces in the jib and the tie.

A

C

60 45

15

45

60 10 kN

tie T1

T2

jib T1

T2

B



Solution:

Using Lamis theorem at point A, we get 10

sin 15=

T1sin 45

= T2

sin 60

T1 = 10 sin 45

sin 15 = 10

1

2

1

0.25882 = 27.32 kN

T2 = 10 sin 60

sin 15 = 10

3

2

1

0.25882 = 33.46 kN

Otherwise, for the equilibrium of point A,

Fx= 0 gives

T1 cos 30 = T2 cos 45

T1 3

2= T2

1

2

T1 = T2 2

3

Fy= 0 gives

T1 cos 60 + 10 = T2 cos 45

T1 1

2+ 10 = T2

1

2

2

3

1

2T2 +

1

2T2 = 10

T2 (0.707 0.408) = 10

T2 = 10

0.299 = 33.45 kN

T1 = 2

3 33.45 = 27.31 kN

Example 14

Determine the reactions at the supports for the beam loaded as shown in figure.

Solution:

Fy= 0 gives

R1 + R2 = 500 + 1000 + 500 2 = 2500 N

MB = 0 gives

R1 6 = 500 4 + 1000 2 + 500 2 3

= 2000 + 2000 + 3000

R1 = 7000

6= 1166.67 N

R2 = 2500 1166.67 = 1333.33 N.

2 m 2 m 2 m

A B

500 N 1000 N

500 N/m

R1 R2



Example 15

Find the force, P that is capable of pulling the cylinder of figure. Over the block

Solution:

MC = 0 gives

W cos 30 (30)2 (20)2 = (P W sin 30) 20

100 3

2 22.36 = (P 100

1

2) 20

1936 = 20 P 1000

P = 936

20 = 46.8 N

Example 16

A uniform wheel 60 cm in diameter rests against a rigid rectangular block 15 cm thick in

figure. Find the least pull through the centre of the wheel to just turn the wheel over the

corner of the block. All surfaces are smooth. Find the reaction of the block. The wheel

weighs 10 kN.

Solution:

Let R be the reaction at A between the wheel and rectangular block and O be angle which

the pull P makes with R.

Now (AC)2 = (OA)2 (OC)2 = (30)2 (30 15)2 = 900 225 = 675

AC = 25.98 cm

C

P

B

30 cm

D

O

A

R

10 kN

15 cm

10 cm C

P

W = 100 N

30

30 cm



AD = AO sin = 30 sin

Taking moments about A, we get

10 CA = P DA

P = 10 25.98

30 sin =

8.66

sin

P will be least when sin is maximum, i. e. , = 90.

P = 8.66 kN

Now

cos AOC = OC

OA=

15

30= 0.5

AOC = 60

Resolving along R, we get; R = 10 cos AOC = 10 cos60 = 10

2= 5 kN

Example 17

A cylinder of diameter 1 m weighing 1kN and another block weighing 500 N are

supported by a beam of length 7 m weighing 250 N with the help of a cord as shown in

figure. If the surface of contact are frictionless, determine the tension in the cord.

Solution:

R

sin 90o=

S

sin 135o=

W

sin 135o

R = 10002 = 1414 N S = 1000 N From OAE

tan (22.5) =0.5

AE

AE = 0.5/ tan 22.5 AI = 1.207 Taking moment at A T(7) = R(AI) + 250 35 cos 45 + 500 7 cos 45o T = 442 . 114 N

A

O

90o

1000

N

E

R

T

50

0 N

D

250 N

45

S

m

C

45

22.5o

3.5 m

250 cos 45o

500 cos 45o 45

90o

135o

135o

1000 N

S

R



Example 18

Three cylinders weighing 100 N each and 16 cm in diameter are placed in a channel

rectangular in section as shown in figure. What is the pressure that the cylinder A is

exerting on the cylinder B at the point of contact? What is the pressure exerted by the

lower two cylinder on the channel base and walls at the contact points?

Solution:

Let the reactions at the points of contact be R1, R2, R3, R4 and P, Q be the pressure

between the cylinders A, C and A, B respectively.

cos =10

16= 0.625

= 51.32

Applying Lamis theorem at point O1, we get P

sin (90 51.32) =

Q

sin (90 51.32) =

100

sin (180 102.64)

P

0.62497=

Q

0.62497=

100

0.97576

P = Q = 100 0.62497

0.97576 = 64 N

Applying Lamis theorem at point O2, we get R1

sin (90 51.32) =

R1 100

sin 51.32 =

Q

sin 90

R1 = 64 0.62497 = 40 N

R2 100 = 64 0.78065

R2 = 150 N

By symmetry,

R1 = R4 = 40 N

R3 = R2 = 150 N

Example 19

A right circular cylinder is placed on a V block which is placed on a inclined plane as

shown in the figure along side. Find the value of when reaction at A is double of the

reaction at B.

R3

O3 R4 R1

R2

O1

O2

P

Q

B

10

cm

Q

P

H

G D

36 cm

A

100 N

C

H

100 N 100 N



Solution

FBD of the cylinder,

RA= 2RB

RBSM(135 + )

=RA

SM(135 )=

W

sin90o

sin(135 + )

sin(135 )=

RBRA

sin(135 + )

sin(135 )=

1

2

= 18.43o

RA RB

W

A B

45o



Assignment 1. For a particle in plane to be in

equilibrium

1. Sum of the forces along X-direction is

zero.

2. Sum of the forces along Y-direction is

zero.

3. Sum of the moments of all the forces

about any point is zero

Which of the following statements are

always correct?

(A) 1 & 2

(B) 2 & 3

(C) 1 & 3

(D) 1, 2 & 3

2. A system of concurrent forces P, Q & R

has the following magnitudes and

passing through origin and the

indicated points:

P = 280 N (12, 6, 4)

Q = 520 N (3, 4, 12)

R = 270 N (6, 3, 6)

The magnitude of the resultant of this

force system is

(A) 560 N

(B) 394 N

(C) 274 N

(D) None of these

3. At a point A in a plane there is a vertical

upward force of 2 N and a clockwise

couple of 4 Nm. These are equivalent to

a single vertical force of 2 N at a point B.

The distance of point B from point A is

(A) 1 m to the left

(B) 1 m to the right

(C) 2 m to the left

(D) 2 m to the right

4. A clockwise couple of 5 Nm acts at point

A on a plane and a counter clockwise

couple of 10 Nm acts at a point B (5m

right of A). These couple has to replaced

by an equivalent couple at a point C (1m

left of B) on the plane. The magnitude

and direction of the couple at C is

(A) 5 Nm clockwise

(B) 5 Nm counter clockwise

(C) 15 Nm clockwise

(D) 15 Nm counter clockwise

5. If the maximum and minimum resultant

forces of two forces acting on a particle

are 40 N and 10 N respectively, then

two forces are

(A) 25 N & 15 N (B) 20 N & 20 N

(C) 20 N & 10 N (D) 20 N & 5 N

6. A uniform beam AB pinned at A is held

by the cable BC in the position shown. If

the tension in the cable is 200 kgf, then

the weight of the boom and the reaction

of the pin at A on the boom are

respectively.

(A) 300 kgf, 100 3 kgf, 30o

(B) 400 kgf, 100 3 kgf, 60o

(C) 300 kgf, 200 3 kgf, 30o

(D) 400 kgf, 200 3 kgf, 60o

7. The coefficient of static friction between

block of 2kg and table shown is s = 0.2.

What should be maximum value of m so

that blocks do not move? Take

g = 10m/s2. Pulley and string are light

and smooth.

(A) 1 kg

(B) 0.4 kg

(C) 0.2 kg

(D) 2 kg

8. A uniform bar is supported with hinge

joint at point A and a smooth contact at

point B. The weight of bar is 300 N.

2kg

m

B

R

A

60o 60o

60o

C

60o

90o T

30o

W



Length of bar is 20 m. Point B is at

distance 15 m from A.

Reaction at support B at equilibrium

(A) 200 N

(B) 173.2 N

(C) 100 N

(D) 300 N

9. Three cables are joined at junction ring

C. Determine tension in cable AC caused

by weight of 30 kg cylinder (take g = 10

m/s2)

(A) 155.3N

(B) 268.97 N

(C) 15.53 N

(D) 26.897 N

10. Ratio of lift force L to drag force D for a

simple airfoil is L/D = 10. If the lift

force on a short section of the airfoil is

200 N, compute the angle which

resultant makes with horizontal,

(A) 84.3o

(B) 10o

(C) 42.15o

(D) None of these

11. A cubical block of ice of mass, m and

edge, L is placed in large tray of mass M.

If the ice melts, how for does the centre

of mass of the system ice plus tray

comes down.

(A) 2(M+m)L

m

(B) L

2

(C) mL

M+m

(D) mL

2(M+m)

12. Compute the resistive moment at

support, O. Forces lie in one plane

(A) 5.98

(B) 2.5

(C) 2.73

(D) 1.4

13. A mass 35 kg is suspended from a

weightless bar AB which is supported by

a cable CB and a pin at A as shown in. The

pin reactions at A on the bar AB are

(A) Rx = 343.4 N, Ry = 755.4 N

(B) Rx = 343.4 N, Ry = 0

(C) Rx = 755.4 N, Ry = 343.4 N

(D) Rx = 755.4 N, Ry = 0

14. The 30 N force, P is applied

perpendicular to the portion BC of the

bent bar as shown in the figure.

Determine moment of P about point A

(in Nm)

m

T

B

275mm

A

12

5 m

m

y

x

0

1.8m 0.6m 0.6m

0.6m

4kNm 1.4kN

3kN

30o

B

A

D

L

A

45

30

B

30

D

C

60o

Smooth Contact

B



(A) 96 Nm

(B) 57.94 Nm

(C) 81.94 Nm

(D) 24 Nm

15. If two equal forces of magnitude, P act

an angle , their resultant will be

(A) 2P sin

2

(B) 2P cos

2

(C) 2P tan

2

(D) P sin

2

16. A 50 kg block rests on a 20o inclined

plane, as shown in figure. s = 0.40.

Maximum horizontal force, P that can

be applied to the block without causing

it to slide.

(A) 439 N (B) 78.52 N

(C) 495 N (D) 139.2 N

Answer Keys & Explanations

Assignment 1

1. [Ans. D]

By definition of equilibrium

2. [Ans. B]

Resultant force can be calculated by

adding the three force vectors.

3. [Ans. D]

4. [Ans. B]

5. [Ans. A]

6. [Ans .D] W

sin 90o=

T

sin(90o + 60o)=

R

sin(90o + 30o)

OrW

sin 90o=

200

cos 60o=

R

cos 30o

From which, W = 400 kgf and R = 200

3 kgf, and the angle which R makes

with horizontal is 60o

7. [Ans. B]

FBD of 2kg mass: FBD of small mass

hanging

T = sN

= s(Mg) (1)

T = mg (2)

By 1 and 2,

mg = s(Mg)

m = sM = 0.2 2 = 0.4 kg

8. [Ans. C]

FBD

Taking moment about A MA = 0

W cos 60 10 B 15 = 0

B = 300 0.5 10

15= 100 N

10 m

15 m

w cos 60 w 60

Ay

Ax

B

Mg

T

N sN Mg

T

P 50kg

20o

B

A

P

1.6m

45o

1.6m

C



9. [Ans. A]

TCD

sin 105o=

TABsin 150o

TAB =sin 150o

sin 105o TCB = 155.3 N

10. [Ans. A]

11. [Ans. D]

Xcm = Mx1 + mx2

m + m (1)

When ice melts, centre of mass of water is on surface of tray (since tray is large) x2

` = x2 L/2

xnew = Mx1 + m(x2 L/2)

M + m (2)

(1) (2)

xnew = xcm = mL

2(M + m)

() means comes down.

12. [Ans. A]

1.8 3 + 1.4 1.2 4 + 2.4 1.4 sin 60

= 5.98 kNm. So change any option with 5.98

13. [Ans. D]

Tsin = mg

tan =15

275

= 24.45o

T=829.5 N

Rx = Tcos24.45 = 755.4 N

Ry = 0

14. [Ans. C]

m = 0,

P cos 45o (1.6 + 1.6 cos 45o) + P sin 45o

(1.6 sin 45o)

value is 81.94 N-m

15. [Ans. B]

R = P2 + R2 + 2P(cos )

= 2P2[1 + cos ]

= 2P2 [1 + 2 cos2

2 1]

= 2Pcos

2

16. [Ans. A]

(+)Fx = 0; P cos 20o sN 50g sin 20

o

= 0

(+)Fy = 0, N P sin 20o 50g cos 20o

= 0

By (1) and (2),

P cos 20o 50g sin 20o = s (P sin 20o +

50g cos 20o)

P = s(50g cos 20

o) + 50g sin 20o

cos 20o ssin 20o

= 438.58N

P

20o 20o

N

f = sN

x y

50g

20o

P

45o

B

A

C

1.6m

1.6m

mg

Rx

B

X1 X2

A D

B

105o

150o

105o

1.Engineering Mechanics2.Coverpage3.Contents_EMChapter 2

free body diagram and equilibrium

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