Free Body Diagram And Equilibrium

Engineering Mechanics For

ME/CE By

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Contents

Contents #1.

#2.

#3.

#4.

Chapters Introduction

Page No. 1

 Introduction

1

Free Body Diagram and Equilibrium

2 – 28

 Introduction

2

 Equivalent Force System

2–3

 Newton’s 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

Trusses and Frames

29 – 42

 Trusses and Frames

29 – 31

 Solved Examples

31 – 38

 Assignment

39 – 40

 Answer Keys & Explanations

40 – 42

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

 Answer Keys & Explanations

50 – 51

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I

Contents

#5.

#6.

#7.

#8.

Principle of Virtual Work

52 – 59

 Principle of Virtual Work

52 – 54

 Solved Examples

54 – 59

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

 Answer Keys & Explanations

85 – 87

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

Kinematics and Dynamics of Rigid Body

95 – 107

 Center of mass and Center of Gravity

95

 Euler’s Equation of Motion

95 – 96

 Moment of Inertia

96 – 102

 Conservation of Angular Momentum

103 – 104

 Assignment

105 – 106

 Answer Keys & Explanations

106 – 107

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II

CHAPTER

12

"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, Newton’s 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.

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Free Body Diagram and Equilibrium

Force System Coplanar like parallel force is straight Coplanar concurrent force Coplanar non- concurrent force Non- coplanar parallel force Non- coplanar concurrent force Non- coplanar non-concurrent force

Example Weight of stationary train on rail off the track Forces on a rod resting against wall Forces on a ladder resting against a wall when a person stands on a rung which is not at its center of gravity The weight of benches in class room A tripod carrying camera Forces acting on moving bus

Newton’s 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. dv F ∝ (m ) 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. P

A P 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. ΣF = 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., ΣM = 0 ΣF × d = 0 This law is applicable only to coplanar, non-concurrent force systems. : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

3

Free Body Diagram and Equilibrium

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 P2

P3

P1

P3

R2

R P4

D

P4

E

R1 A

C P2

P1

B 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. P2 D E

θ

θ

P1

α

A

θ B

C

Resultant, (R) = √P12 + P22 + 2P1 P2 cosθ P2 sinθ tan α = ( ) P1 +P2 cosθ Where,

θ = Angle between two forces, α = 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. Lami’s 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.

P2

P1

γ α

β P3

α

c

P2

 b

P3 P1 a

β

P1 P2 P3 = = sinα sinβ sinγ

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4

Free Body Diagram and Equilibrium

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 600N W

600N

R1

G P

P R2 A Ladder Resting on Smooth Wall

F3

F2

V

F1

V

V

V

F

y

M

W=m g

x

A Cantilever Beam ĵ î mg

m

Free Body Diagram of Just the Block

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.

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5

Free Body Diagram and Equilibrium

Types of Contacts/Supports Following types of mechanical contacts can be found in various structures,  Flexible Cable, Belt, Chain or Rope θ

Weight of Cable Negligible

θ T

Weight of Cable not Negligible θ 

 

θ T

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

N



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

N



N

Collar or slider support force normal to guide only. There is no tangential force as surfaces are considered to be smooth. Pin Connection M Rx Ry R A freely hinged pin supports a force in any direction in the plane normal to the axis; usually y shown as two components Rx and Ry. A pin not free to turn also supports a couple M. Built in or Fixed End A A M A O F Weld r V Rx



A built-in or fixed end supports an axial force F, a transverse force V, and a bending moment M. : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

6

Free Body Diagram and Equilibrium

Coplanar Non-Concurrent Forces Varignon’s 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. B

d2 R d

P2 d1

A

P1

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.

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7

Free Body Diagram and Equilibrium

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, μ Fmax = μN FFmax F = μk N= Friction Force μk = Kinetic Friction Co-efficient

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8

Free Body Diagram and Equilibrium

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. A 45°

D

T1 1

30°

120°

T3 3

B

T2

T2 C

5N W

For the equilibrium of point B, we have T1 T2 5 (From Lami’s theorem) = = ° ° sin 60 sin 165° sin 135 5 × 0.86602 sin 60° = = 16.73 N T1 = 5 ° 0.25882 sin 165 sin 135° 5 × 0.70710 T2 = 5 = = 13.66 N sin 165° 0.25882 For the equilibrium of point C, we have T2 T3 W (From Lami’s theorem) = = ° ° sin 150 sin 120 sin 90° sin 120° √3 2 T3 = T2 ( ) = 13.66 × × = 23.66 N ° sin 150 2 1 T2 sin 90° 13.66 × 1 W= = = 27.32 N ° sin 150 0.5 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 : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

9

Free Body Diagram and Equilibrium

T1 T2 W1 = = ° ° sin 90 sin 120 sin 150° And for the equilibrium of point C, T2 20 W = = ° ° sin 150 sin 90 sin 120° 20 N

A

D

30° 60° T1

E

20 N

B

T2

20 N

C

T2

W2

W1

sin 120o √3 = 20 × = 17.32 N o sin 90 2 1 sin 150o = 20 × = 10 N T2 = 20 × o 2 sin 90 sin 90o 10 × 2 Thus, T1 = T2 × = = 11.55 N sin 120o √3 sin 150° 1 2 W1 = T2 × = 10 × × = 5.77 N o sin 120 2 √3 Pressure on the pulley Hence, W2 = 20 ×

F = √(20)2 + (20)2 + 2 × 20 × 20 cos 30o = 20 √2 + 2 ×

√3 = 20 √2 + √3 = 38.6 N 2

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.

C

T1 l

T1

T1

A B

x

D

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10

Free Body Diagram and Equilibrium

Solution: From pulley D 2T1 = W W T1 = 2 T1 = A

l

W 2

x P

Taking moments about A W MA = 0 = × l − p. x = 0 2 Wl = px 2 Wl x= 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 4√3 times that of the telephone wire. Solution: Let the tension in the two portions of the telephone wire be T1 each and the tension in another wire be T2 , as shown in figure. T1 T

A

60° T1 C

T2 B

60°

D

Then T = 2 T1 cos 30o = √3T1 Let AC = BC = l Taking moments about B, we get T × 2l = T2 cos 60° × l T2 = 2T1 × √3 × 2 = 4 √3T1 T2 = 4√3 T1 : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

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Free Body Diagram and Equilibrium

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. P r G P

4r 3π A

P

W/2 (b)

W (a)

Solution: Given the problem as shown in below figure. We draw the free body diagram as follows. Note the following salient points P 1 W 2 W

1 W 2

P

P

P Nx N

P C

P

Nx Ny

Ny

𝐅𝐫𝐞𝐞 𝐁𝐨𝐝𝐲 𝐃𝐢𝐚𝐠𝐫𝐚𝐦

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. 1

 The left two force, are gravity of each half which is 2 acting on Center of Gravity of each semi cylindrical half. To calculate the CG location, we know that ∫ xdA CG∞ = A ∫A dA Now, employing polar coordinates and taking a infinitesimal element as shown in figure, we get as follows. : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

12

Free Body Diagram and Equilibrium

2 3 π R R ∫A yda ∫0 ∫0 r cos θ (rdr)dθ 4R 3 CG∞ = = = = & CGy = = π R π R 1 2 3π dA ∫A dA ∫ ∫0 ∫0 (rdr)dθ ∫0 ∫0 (rdr)dθ A R π 2 (which is clear from symmetry of the body about X axis) Y x = r sinθ ∫A xdA

π

R

∫0 ∫0 r sin θ rdrdθ

y = r cosθ

θ

dA = (r dθ)dr

dθ dr

r

X

Semi-Cylinder COG Calculation Diagram 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 4R 2W ( ) ( ) + (P)(R) − (P)(2R) = 0 ⇒ P = 2 3π 3π 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? Let’s 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, (n + 1)P

(n + 1)P

1 w 2

1 w 2 P

P Nx Ny

nP

Nx nP

Ny

FBD for the General n Rope Turn Case

W 4R Again taking moment about point C, we get ( ) ( ) + (P)(R) − ((n + 1)P)(2R) = 0 2 3π 1 2W ⇒P= 2n + 1 3π 1 Clearly, as number of turns n ↑⇒ ↓⇒ The minimum force to hold the halves together P ↓ 2n + 1 : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

13

Free Body Diagram and Equilibrium

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. R1

R2 40° 140°

170° O 30° 10° 30°

W

B

10°

A

Solution: Let R 1 and R 2 be the reaction of the 10° and 30° planes respectively. Using Lami’s theorem, we get W R1 R2 = = sin 40° sin 150° sin 170° sin 150° 0.5 R1 = W = W × = 0.778 W sin 40° 0.64278 = 0.778 × 150 = 116.6 N sin 170° 0.17365 R2 = W = 150 × = 40.52 N sin 40° 0.64278 Example 7 Two smooth spheres of weight, W and radius, r each are in equilibrium in a horizontal channel of width (b<4r) and vertical sides, as shown in figure. Find the three reactions from the sides of the channel which are all smooth. Also find the force exerted by each sphere on the other. Calculate these values if r = 25 cm, b = 90 cm and W = 100 N.

R2

R3 P

A

C

r

θ

B

E r

O P R1 W

W D r

b−2r b

r

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14

Free Body Diagram and Equilibrium

Solution: Let R 1 R 2 and R 3 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, b − 2r cos θ = 2r The forces acting at the point A are (R 3 − W), R 1 and P. Using Lami’s theorem, we get R3 − W R1 P = = sin θ sin(90° − θ) sin 90° R3 − W ∴ P = sin θ R 1 = (R 3 − W) cot θ The forces acting at the point B are W, R 2 and P. Again using Lami’s theorem. R2 P W W = = ; R 2 = W cot θ ; P = sin(90° − θ) sin 90° sin θ sin θ For r = 25 cm and b = 90 cm 90 − (2 × 25) 40 cos θ = = = 0.8 2 × 25 50 ∴ θ = 36.87° ∴ R 2 = 100 cot 36.87° = 133.3 N 100 P = = 166.66 N sin 36.87° R 3 = P sin θ + W = 166.66 sin 36.87° + 100 = 200 N R 1 = (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, R A = reaction on the wheel at A The three forces P, W and R A are in equilibrium. Since P and W meet at D, therefore R A must pass through D. Using Lami’s theorem, we have P W = sin(90° + θ) sin(180° − θ) D

RA

P

D

0.25 m

θ

C

1.5 kN

P

C A B (a)

0.2m

A

E W B

0.2m

(b)

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15

Free Body Diagram and Equilibrium

P = W cos θ sin θ P = W tan θ AE tanθ = DE AE = √(AC)2 − (CE)2 = √(0.25)2 − (0.25 − 0.2)2 ⇒ √0.0625 − 0.0025 = √0.06 = 0.245 m 0.245 = 0.8165 tan θ = 0.3 θ = 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, θ that it makes with the horizontal when the system is in equilibrium. Take, W1 = 60 N, W2 = 120 N and α = 30° E B

A

θ

(90° − α) D

α

W2

F

W1

Solution: Let R A and R B 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. RB RA

B

T

T θ

A W1 (90° − α)

(a)

W2

α

Roller A[Fig (b)] Applying Lami’s theorem at A, we have T W1 = sin(90° + α) sin{180° − (α + θ)} T W1 = … . (1) cos α sin(α + θ) Roller B [Fig. (c)] T W2 = sin(180° − α) sin{90° + (α + θ)} : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

16

Free Body Diagram and Equilibrium

T W2 = … . (2) sin α cos ( θ + α) W1 cos α W2 sin α sin(θ + α) = ; cos (θ + α) = T T RA

T

RB (90°+ θ + α)

θ (90°+ α) α

α

T

α (90° −) (90° − α)

(α) (90° − α)

W2 (c)

W1 (b) 60 W1 cot α = × cot 30° = 0.866 120 W2 θ + α = 40.89°; θ = 40.89° − 30° = 10.89° 60co s 30° W1 co s α = = 79.38 N ∴ T = sin (α + θ) si n 40.89°

tan(θ + α) =

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. D A 45o

C

15o

30o 30kg

B Solution:

Let T1 and T2 be the tension in the string AC and BC respectively. 45o

T1

T = 294.3 N 45o 60o

30o

15o 120o

135o

T2

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17

Free Body Diagram and Equilibrium

294.3 T1 T2 = = o o sin 105 sin 135 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. 1.5m

A

2kN 70o

0.5 m

0.15m 0.15m

y 500Nm 3

1.2k N

4

x

Solution:

Fx = 2 × cos 70 + 1.2 cos 36. 87o Fx = 1.644 kN Fy = 2 sin 70 − 1.2 sin 36. 87o 3 Fy = 1.1594 kN 36.87o 4 R = Fx i + Fy j R = 1.644i + 1.1594j MA = −[2 cos 70o × 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, α 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 S G R A

Y B

D

W/2 α W T

C

X

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18

Free Body Diagram and Equilibrium

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. ΣMB = 0 it gives AB R × AB cos α = T × (AB sin α) + W × ( cos α) 2 W R cos α = T sin α + cos α 2 ΣFy = 0 it gives R = W W ∴ T sin α = W cos α − cos α 2 W T = cot α 2 2 W stands at D where AD = AB, then, When a man of weight 3 2 ΣFy = 0 it gives 3 R = W 2 ΣMB = 0 it gives W 1 AB cos α) + ( × ABco s α) R × AB cos α = T × AB sin α + W × ( 2 3 2 W W Or R cos α = T sin α + cos α + cos α 2 6 3 W W Or T sin α = W cos α − cos α − cos α 2 2 6 5 T = W cot α 6 Example 13 A jib crane is loaded as shown in figure. Determine the forces in the jib and the tie. A T1

tie

45°

60°

15°

60° T1 B

T2

10 kN

jib T2

45°

C

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19

Free Body Diagram and Equilibrium

Solution: Using Lami’s theorem at point A, we get 10 T1 T2 = = sin 45° sin 60° sin 15° sin 45° 1 1 T1 = 10 = 10 × × = 27.32 kN sin 15° 0.25882 √2 sin 60° 1 √3 T2 = 10 = 10 × × = 33.46 kN sin 15° 2 0.25882 Otherwise, for the equilibrium of point A, ΣFx = 0 gives T1 cos 30° = T2 × cos 45° 1 √3 T1 × = T2 × 2 √2 2 T1 = T2 × √ 3 ΣFy = 0 gives T1 cos 60° + 10 = T2 cos 45° 1 1 T1 × + 10 = T2 × 2 √2 2

∴ − √3 ×

1

T 2 2

1

+2 T2 = 10

T2 (0.707 – 0.408) = 10 10 T2 = = 33.45 kN 0.299 2 T1 = √ × 33.45 = 27.31 kN 3 Example 14 Determine the reactions at the supports for the beam loaded as shown in figure. 500 N/m 500 N

1000 N

A R1

B 2m

2m

2m

R2

Solution: ΣFy = 0 gives R 1 + R 2 = 500 + 1000 + 500 × 2 = 2500 N ΣMB = 0 gives R 1 × 6 = 500 × 4 + 1000 × 2 + 500 × 2 × 3 = 2000 + 2000 + 3000 7000 R1 = = 1166.67 N 6 R 2 = 2500 − 1166.67 = 1333.33 N. : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

20

Free Body Diagram and Equilibrium

Example 15 Find the force, P that is capable of pulling the cylinder of figure. Over the block P

30 cm

10 cm

C W = 100 N 30° Solution: ΣMC = 0 gives W cos 30° × √(30)2 − (20)2 = (P − W sin 30°) × 20 1 √3 × 22.36 = (P − 100 × ) × 20 2 2 1936 = 20 P – 1000 936 P = = 46.8 N 20

100 ×

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. P 30 cm R

D θ A

O

C

15 cm B

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 : 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

21

Free Body Diagram and Equilibrium

AD = AO sin θ = 30 sin θ Taking moments about A, we get 10 × CA = P × DA 25.98 8.66 P = 10 × = 30 sin θ sin θ P will be least when sin θ is maximum, i. e. , θ = 90°. P = 8.66 kN Now OC 15 cos AOC = = = 0.5 OA 30 ∠AOC = 60° Resolving along R, we get; R = 10 cos AOC = 10 cos60° =

10 = 5 kN 2

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. C

45°

T 90o O

D

S m

45°

R 1000 N

E

45°

500 cos 45o

50 0N

22.5o

o 250 N 250 cos 45 3.5 m

A

Solution: R S W = = o o sin 90 sin 135 sin 135o R = 1000√2 = 1414 N S = 1000 N From ∆ OAE 0.5 tan (22.5) = 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

R 135o 135o

S 90o

1000 N

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22

Free Body Diagram and Equilibrium

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? 36 cm A O1 Q R2 Q D

O2

P 100 N

H P R 3

R1

B 10 100 N cm

R4

O3

G

C H 100 N

Solution: Let the reactions at the points of contact be R 1 , R 2 , R 3 , R 4 and P, Q be the pressure between the cylinders A, C and A, B respectively. 10 cos θ = = 0.625 16 ∴ θ = 51.32° Applying Lami’s theorem at point O1 , we get P Q 100 = = sin (90° − 51.32°) sin (90° − 51.32°) sin (180° − 102.64°) Q 100 P = = 0.62497 0.62497 0.97576 0.62497 P = Q = 100 × = 64 N 0.97576 Applying Lami’s theorem at point O2 , we get R1 R 1 − 100 Q = = sin (90° − 51.32°) sin 51.32° sin 90° R 1 = 64 × 0.62497 = 40 N R 2 − 100 = 64 × 0.78065 R 2 = 150 N By symmetry, R 1 = R 4 = 40 N R 3 = R 2 = 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’.

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23

Free Body Diagram and Equilibrium B

A 45o 

Solution FBD of the cylinder, RB

RA

W

RA= 2RB RB RA W = = SM(135 + ) SM(135 − ) sin90o sin(135 + ) R B = sin(135 − ) R A sin(135 + ) 1 = sin(135 − ) 2 ∴  = 18.43o

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24

Free Body Diagram and Equilibrium

Assignment 1.

2.

3.

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 (C) 1 & 3 (B) 2 & 3 (D) 1, 2 & 3

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 (C) 20 N & 10 N (B) 20 N & 20 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. C T 90o

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 (C) 274 N (B) 394 N (D) None of these 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 (C) 2 m to the left (B) 1 m to the right (D) 2 m to the right

R 60oo 60o 60

o

60

B

30o A

(A) (B) (C) (D) 7.

W

300 kgf, 100 √3 kgf, 30o 400 kgf, 100 √3 kgf, 60o 300 kgf, 200 √3 kgf, 30o 400 kgf, 200 √3 kgf, 60o

The coefficient of static friction between block of 2kg and table shown is μs = 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. 2kg

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

m (A) 1 kg (B) 0.4 kg 8.

(C) 0.2 kg (D) 2 kg

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.

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25

Free Body Diagram and Equilibrium

Length of bar is 20 m. Point B is at distance 15 m from A.

11.

B Smooth Contact

60o

Reaction at support B at equilibrium (A) 200 N (C) 100 N (B) 173.2 N (D) 300 N 9.

45°

2(M+m)L

(B)

L

(C)

m

(D)

2

mL M+m mL 2(M+m)

Compute the resistive moment at support, O. Forces lie in one plane 1.4kN 4kNm 3kN 0 B A 0.6 0.6m 1.8m 30o 0.6m m (A) 5.98 (C) 2.73 (B) 2.5 (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

D C

(A)

12.

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

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.

30°

30° B

10.

(C) 15.53 N (D) 26.897 N

125 mm

(A) 155.3N (B) 268.97 N

A

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 θ which resultant makes with horizontal,

275mm

x

(A) (B) (C) (D)

D

14. (C) 42.15o (D) None of these

B

y

L

(A) 84.3o (B) 10o

T

m

Rx = 343.4 N, Ry = 755.4 N Rx = 343.4 N, Ry = 0 Rx = 755.4 N, Ry = 343.4 N Rx = 755.4 N, Ry = 0

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)

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26

Free Body Diagram and Equilibrium

16.

P C 1.6m 45o

B

1.6m

A 50 kg block rests on a 20o inclined plane, as shown in figure. μs = 0.40. Maximum horizontal force, P that can be applied to the block without causing it to slide. 50kg

P

A

(A) (B) 15.

96 Nm 57.94 Nm

(C) (D)

81.94 Nm 24 Nm

20o

(A) 439 N (B) 78.52 N

If two equal forces of magnitude, P act an angle θ, their resultant will be (A) 2P sin

θ

(C) 2P tan

2 θ

θ

(B) 2P cos 2

(C) 495 N (D) 139.2 N

θ 2

(D) P sin 2

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]

FBD of 2kg mass: FBD of small mass hanging T = μsN = μs(Mg) …(1) T = mg …(2) By 1 and 2, mg = μs(Mg) m = μsM = 0.2 × 2 = 0.4 kg 8.

[Ans. C] FBD B

6.

7.

[Ans .D] W T R = = o o o o sin 90 sin(90 + 60 ) sin(90 + 30o ) W 200 R Or = = o o sin 90 cos 60 cos 30o From which, W = 400 kgf and R = 200 √3 kgf, and the angle which R makes with horizontal is 60o [Ans. B] Mg

T T

μsN

N

15 m 10 m 60°

w

w cos 60

Ax Ay

Taking moment about A ∑MA = 0 W cos 60 × 10 − B × 15 = 0 10 B = 300 × 0.5 × = 100 N 15

Mg

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27

Free Body Diagram and Equilibrium

9.

[Ans. A]

14.

P 45o

C

D

105o

A

[Ans. C]

𝐂

1.6m

105o 150o

B 1.6m

B

TCD TAB = o sin 105 sin 150o sin 150o TAB = × TCB = 155.3 N sin 105o 10.

[Ans. A]

11.

[Ans. D]

A

Σm = 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 θ)

X2

= √2P2 [1 + cos θ]

X1

Mx1 + mx2 … (1) m+m When ice melts, centre of mass of water is on surface of tray (since tray is large) x2` = x2 − L/2 Mx1 + m(x2 − L/2) xnew = … (2) M+m (1) – (2)  mL xnew = −xcm = − 2(M + m) (−) means comes down. Xcm =

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] Tsinθ = mg 15 tan θ = 275 ⇒ θ = 24.45o T=829.5 N R x = Tcos24.45 = 755.4 N Ry = 0 θ

= √2P2 [1 + 2 cos2 = 2Pcos 16.

θ − 1] 2

θ 2

[Ans. A]

50g P

20o

20o

y x f = μs N

20o

N (+↗)ΣFx = 0; P cos 20o − μs N – 50g sin 20o =0 (+↖)ΣFy = 0, N – P sin 20o – 50g cos 20o =0 By (1) and (2), P cos 20o – 50g sin 20o = μs (P sin 20o + 50g cos 20o ) μs (50g cos 20o ) + 50g sin 20o P = cos 20o − μs sin 20o = 438.58N

B Rx

mg

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28