Statics

CHAPTER 2

STATICS

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.

Force system Example

Coplanar like parallel force is straight. Weight of stationary train on rail off the track Coplanar concurrent 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 The weight of benches in class room Non- coplanar concurrent force A tripod carrying camera

Newton’s law 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 (m dv dt)

Third law: For every action, there is an equal and opposite reaction.

Principle of transmissibility of force: 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 force.

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

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

P1 P2 P3 P4 P2 P4 C P1 P3 E R B A R2 R1 D A B P P

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) = √

tan ( )

= 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.

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. P sin P sin P sin

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 body. P1 P2 P3  _P 2 P1 P3 a b c P2 B C D E P1 A

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.

Types of contacts/supports

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

 Flexible cable, belt, chain or rope

mg m

Free body diagram of just the block

̂ ̂ F3 F2 F1 V V _{V V} F M W=m g y x SMOOTH SMOO TH P 600N G W 600N R1 P R2

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

 Smooth surfaces

Contact force is compressive and is normal to the surface.

 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

N N N N F R N N θ θ

Weight of cable negligible

Weight of cable not negligible _θ θ T

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.

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.

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

A B R A M F V A Weld O r A R y Rx Rx Ry M

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

0 & M= M=0

R  F 

For collinear force system ∑ F ∑ F ∑ F For non-collinear force system ∑ ∑ ∑

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.

Wrench: When the direction of resultant couple ‘ ’ 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.

Friction

(i). Dry friction: Friction between the contact surface

(ii). Fluid Friction: Friction between the layers of fluid element.

(iii). 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 F<Fmax F= Friction force F= Fmax μN If F>Fmax F μKN= Friction force

μK= Kinetic friction co-efficient.

Virtual Work

Work: When a force acts on a body and moves it through some distance in its own direction, then work is said to be done. Thus, work may be defined as the product of the force and the distance moved in the direction of the force. Mathematically, we can write that

Work = Force × distance U =F × S

When the distance moved by the body is not in the direction of the force then to determine the work done, the component of the force in the direction of the distance moved may be multiplied with the distance moved For example if the force F is acting at an angle θ with the direction of the distance S moved, then work done is given by

U F cos θ × S

Virtual Displacement: It may be defined as the infinitesimally small imaginary (or hypothetical or virtual) displacement given to a body or to a system of bodies in equilibrium, consistent with the constraints. The virtual displacement may be either rectilinear or angular.

Virtual Work: The product of the force F and the virtual displacement δs in the direction of the force is called virtual work.

δU F δs

Principle of virtual Work

It states that if a system of forces acting on a body or a system of bodies is in equilibrium and if the system is supposed to undergo a small virtual displacement consistent with its geometrical constraints, the algebraic sum of the virtual work done by the system of forces is zero.

Trusses and Frames

Trusses are commonly used for construction of roofs of workshop factories and bridges. The trusses are subjected to mainly three types of loads, viz, dead load, live load and wind load. The dead load is self weight of truss live, load is the load which is applied to the truss e.g. the load acting on a bridge truss due to the passing of a train, load acting on a workshop truss due to an electric overhead travelling and the wind load due to the high velocities of wind blowing in a particular region.

When the number of members in a truss satisfies the condition, m = 2j – 3

where j is the number of joints, then the truss is known as a perfect truss, otherwise imperfect. The truss is called deficient or redundant, if m<(2j-3) or m> (2j – 3), respectively.

A pin jointed frame which has just sufficient number of members to resist the loads without undergoing deformation in space is called perfect frame. If number of member in frame is less than that that required for a perfect frame then it is called deficient frame. If number of members in frame is more than that required for perfect frame then it is called redundant frame. A redundant frame is indeterminate.

The following assumptions are made in solving trusses:

1. The members of truss are connected at the joints by friction less joints. 2. The members of truss lie in a common plane (plane truss).

3. The loads are applied only on the pins connecting the members and that the lines of action of the loads lie in the plane of the truss.

4. The weight of members is negligible as compared to the applied loads.

5. The truss is rigid and that it does not deform or change its shape upon the application of the loads.

The member of a truss may be in tension or compression. A member in tension is called a tie and a member in compression a strut.

Methods of Solution: Two methods are generally used for determining the forces in various members of a truss. These methods are

1. Analytical methods

(a) Method of joints (concurrent force system). (b) Method of sections (non-concurrent force system). 2. Graphical method

 Large truss in which only few forces are required

 Situation where method of joints fail.

While determining the reactions at the supports, the following points should be remembered (i) At simply supported (i.e., pinned or roller support) support there can be only a vertical

reaction.

(ii) At fixed support, the reaction can take an arbitrary direction.

A frame in which all the member lies in a single plane is called plane frame. While a frame in which all the members do not lie in a single plane is called space frame.

For perfect frame, m = (2j -3)

For deficient frame, m < (2j -3)

If there is only one force acting at joint, then for the equilibrium, this force should be equal to zero. 3 1 2 3 1 2 4 5 4

If there are two forces acting at a joint then, for the equilibrium, forces should act along the same straight line. The two forces should be equal and opposite. If (only) two forces acting at a joint are not along the same straight line, then for the equilibrium of the joint each force should be equal to zero.

If three forces act at a joint and two of them are along the same straight line then, for the equilibrium of the joint, the third force should be equal to zero.

Solved examples

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

Solution

Let , and be the tensions in the parts AB, BC and CD respectively, as shown in Fig.

For the equilibrium of point B, we have

= = = 5 = × = 16.73 N = 5 = × = 13.66 N For the equilibrium of point C, we have

= = = = 13.66 × × = 23.66 N = × = 27.32 N Example 2

A fine string ABCDE whose extremity A is fixed has weights and 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 and respectively with the vertical, find

(A) Tensions in the portions AB, BC, CD and DE (B) The value of the weights, and , and (C) The pressure on the pulley axis.

A B C D 5 N W 45 120 T 1 _T 3

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 and respectively, as shown in Fig.

For the equilibrium of point B, we have

= =

and for the equilibrium of point C,

= = Hence = 20 × = 20 × = 17.32 N = 20 × = 20 × = 10 N Thus = × = × = 11.55 N = × = 10 × × = 5.77 N

Pressure on the pulley

F = √ × × cos = 20 √ × = 20 √ = 38.6 N A D E W2 W1 B C 20 N 60 3 20 N 20 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.

Solution From pulley D 2T W

T W

Taking moments about A W× Wl x Wl P A P P A B C D W

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 to each other. The pole is supported by another wire attached to the middle point of the pole and inclined at to the horizontal. Show that the tension in this wire is 4 times that of the telephone wire.

Solution

Let the tension in the two portions of the telephone wire be each and the tension in another wire be , as shown in Fig.

Then T = 2 cos Let AC = BC = l

Taking moments about B, we get T × cos ×

= 2 × × 2 = 4 = 4 .

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 Fig. The complete cylinder weighs W N. The plane of contact of both of its halves is vertical. Determine the minimum value of P for which both halve of the cylinder will be in equilibrium on a horizontal plane.

W/2 r W P P P G (a) (b) A A B C D A B C D

Solution

Given the problem as shown below. We draw the Free Body Diagram as follows. Note the following salient points

In the FBD.

 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 acting on Center of Gravity of each semi cylindrical half. To calculate the CG location, we know that

∫

∫

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

Figure 1: Question Description

∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ = (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.

( ) (

)

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,

Again taking moment about point C, we get (

) (

) ( )

clearly, as number of turns

The minimum force to hold

the halves together

Figure 4: FBD for the general n rope turn case

Example 6

Given the pulley rotating at constant angular velocity 𝜔as shown below, which is using a rope of thickness d meters to lift a mass M. What is the tension in the rope?

Solution

This is a deceptively simple question. The important point to be noted here is the fact that the rope is of finite non-zero thickness. So, as time proceeds, more and more rope winds in the pulley and hence, radius of pulley increases and hence by the relation, velocity of mass V = no,

mass starts accelerating by

𝜔

Now for every πr rad. rotation of pulley, radius changes by rope thickness d for every 𝜔 rad/s rotation, radius changes by

Now, on drawing the FBD for the mass we get as follows. From the FBD, we see that by Newton’s second law,

(

𝜔

)

Hence, solved.

Figure 6: Free Body Diagram

Example 7

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 Fig. Find the reactions at thesurface of contact if there is no friction and the weight of the cylinder is 150 N.

Solution

Let and be the reaction of the 10 and 30 planes respectively. Using Lami’s theorem we get

= = = W W × = 0.778 W × = 116.6 N = W 150 × = 40.52 N Example 8

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 Fig. 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.

C E A B P W W D r b-2r r b O P r r A O W 30 30 10 10 140 40 170 B

Solution

Let and 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

The forces acting at the point A are ( - W), and P. Using Lami’s theorem we get

= =

P =

= ( cot

The forces acting at the point B are W, and P Again using Lami’s theorem

= = cot P = For r = 25 cm and b = 90cm cos × × 100 cot = 133.3 N P = = 166.66 N = P sin + W = 166.66 sin + 100 = 200 N = (200 – 100) cot = 133.33 N. Example 9

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 Fig. 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, = reaction on the wheel at A

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

Using Lami’s theorem we have

= P = W tan tan = AE = = √ = = 0.245 m tan = 0.8165 = 39.23 P = 1.5 × 0.8165 = 1.225 kN Example 10

Two rollers of weights and are connected by a flexible string AB. The rollers rest on two mutually perpendicular planes DE and EF, as shown in Fig. Find the tension in the string and the angle that it makes with the horizontal when the system is in equilibrium. Take = 60 N, = 120 N and

Solution

Let and be the reaction on the planes at A and B respectively and T the tension in the string AB. These forces are shown in Fig.

B A D E F 90 - 𝛉 0.2m 0.2m 1.5 kN 0.25 m A B C B C D D P P (b) (a ) W E A

Roller A[Fig (b)]

Applying Lami’s theorem at A we have

= = … Roller B [Fig.(c)] = = …. (2) sin = cos = tan = . cot = × cot = 40.89 = 40.89 - 30 = 10.89 T = = = 79.38 N. T T 90 - 90 - - 90 + 90 + + (b) (c)

A

B

T

T

90

(a)

Example 11

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.

sin T sin T sin T × sin N T × sin N T N T N T N T = 294.3 N A 450 300 B C 150 30k g D

3

4

36. Example 12

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 × cos cos sin sin

[ cos × ] [ sin × ] [ cos × ] [ sin × ]

N-m CCW Example 13

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 T C X G D S B R W W/2 Y y x 500N. m 0.15m 0.15m 2kN 1.5m A 3 4 1.2k N 0.5 m

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 the tension in the string AC.

= 0 gives × × × R = T = 0 gives R = W T T =

When a man of weight stands at D where AD = AB, then = 0 gives R = W = 0 gives R × AB = T × × + × AB or R = T or T = W cos = W T = W Example 14

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

A C 60 15 45 60 10 kN Tie jib B

Solution

Using Lami’s theorem at point A we get

sin sin sin = 10 = 10 × × = 27.32 kN = 10 = 10 × × = 33.46 kN

Otherwise, for the equilibrium of point A, = 0 gives cos × cos × × = × √ = 0 gives cos cos × × - √ × + = 10 (0.707 – 0.408) = 10 = = 33.45 kN = √ × 33.45 = 27.31 kN Example 15

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

Solution. = 0 gives + = 50 + 1000 + 500 × 2 = 2500 N = 0 gives × × × × × = 2000 + 2000 + 3000 N 2 m 2 m 2 m A B 500 N 1000 N 500 N/m

= 2500 – 1166.67 = 1333.33 N. Example 16

Find the force P that is capable of pulling the cylinder of Fig. over the block

Solution = 0 gives W cos × √ sin × 100 × × ( × ) × 1936 = 20 P – 1000 P = = 46.8 N Example 17

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

C P 15 cm B 30 cm D O A R 10 cm C P W = 100 N 30 30 cm

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 = -

= – = 900 – 225 = 675 AC = 25.98 cm

AD = AO sin sin Taking moments about A, we get 10 × ×

P = 10 ×

P will be least when sin is maximum, i.e., . P = 8.66 kN

Now cos AOC =

Resolving along R, we get

R = 10 cos AOC = 10 cos 5 kN Example 18

Two identical prismatic bars PQ and RS, each weighing 150 N are welded together to form a T and are suspended in a vertical plane as shown in Fig. 2.20. Calculate the value of the angle θ that the bar PQ will make with the vertical when a vertical load of 200 N is applied at S.

Solution

Let PQ = RS = 2l

= 0 gives

× sin × sin cos sin = 0

or 150 sin sin cos sin = 0 850 sin cos tan . R Q S 200 N 150 N 150 N P

Example 19

A cylinder of diameter 1 m weighing 1 kN 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 Fig. If the surface of contact are frictionless, determine the tension in the cord.

Solution

Let R and S be the reactions at D and E respectively where the cylinder is touching the wall and the beam For the point O applying Lami’s theorem we get

= =

R = 1000 = 1414 N S = W = 1000 N

For the equilibrium of the beam, = 0 gives

T × × × cos × cos T× × tan ×

× = 4800.4

T = 685.8 N. Example 20

Three cylinders weighing 100 N each and 16 cm in diameter are placed in a channel rectangular in section as shown in Fig. 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?

S R O 50 cm 1000 N E B T G R T A 500 N D 45 45 250 N 7 m C

Solution

Let the reactions at the points of contact be and P, Q be the pressure between the cylinders A, C and A, B respectively.

cos

= 51.32

Applying Lami’s theorem at point , we get

= = = = P = Q = 100 × = 64 N

Applying Lami’s theorem at point , we get

= = × = 40 N - 100 = 64 × = 150 N By symmetry 40 N = 150 N.

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’

P Q B 100 N 100 N 10 cm Q P H G D 36 cm A 100 N C H

Ans:- F.B.D of the cylinder, RA= 2RB       =18.430 Example 21

A couple M is applied to the shaft fixed to the lever OA, as shown in figure. Determine the force P on the smooth slider block required to keep OA in the horizontal position. The length of the link AB is b. h M x M P B b A O RA RB W A B 450 

Solution

The active forces acting on the system include the force P and the couple M. A small rotation of the arm OA is accompanied by a small horizontal movement of the slider block at B.

Now x h b

or

The rotation δθ of the arm OA clockwise with the value δθ δh a

Work done by P during virtual displacement is,

and the work done by couple M is Thus, = - P = Example 22

A rigid, weightless bar rests against two frictionless walls as shown in figure. It is held in equilibrium by two forces F and F parallel to the walls. What is the relationship between F and F in the position shown?

Solution

The displacement of the bar compatible with the constraints is shown in figure by dotted lines. From the geometry of the system, we have

cos sin

sin cos

Virtual work done by force F = F sin Virtual work done by force F = F cos Hence = - sin cos

cot

Finding Kinematic relations in Rope-Pulley Systems

Often in GATE, one faces problems where there is a rope-pulley system and one needs to find the accelerations/energies/Work Done etc. But the problem is that the relation between the motion of each block/pulley/rigid body might not always be known simply by looking.

This section suggests a Generic Method for solving this question. The basic idea behind this method is

Rope Length Remains Constant throughout the motion. So, the whole idea boils down to following step.

1. Step 1: Label the critical/junction/bend points on the pulley system with convenient symbols.

2. Step 2: Displace the system from the given position incrementally by some distance x & draw the new displaced configuration of the system.

3. Step 3: Label the displaced points in the new configuration with some modified symbols. Example: Suffixing a dash symbol ie A point labelled in step 2 becomes A’ etc 4. Step 4: Write the total string length before and after the incremental motion & equate

both as string length remains constant.

5. Step 5: Cancel out common terms on both the sides and look for further simplifications from system constraints. Example: Displacements of rigidly attached bodies will be same etc. Simplify the resulting expression.

That’s it!! You will have a nice kinematic relation between displacements in the system. On differentiating it, you can get the velocity and acceleration relation for the given system and proceed with solving We’ll do -3 examples by which this will become clear.

Example 23

Find the force (F) needed to maintain equilibrium. The red, blue, and green ropes are three separate ropes. Block W has a weight of 500 Kg, pulley A has a weight of 10 Kg, and pulley B has a weight of 20 Kg. basically, this is a pulley

System which has a series arrangement on 2 copies of the same pulley system shown below and hence, we can analyze this by breaking it down studying one copy. Displace the system from initial configuration ABCDC to displaced configuration

A’BC’DC’ Equating rope lengths before and after the motion

AB + BC + CD + DC = A'A + AB + BC’ + C’D + DC’ = A'A + AB + (BC — C’C + (CD — CC’ +

(DC — C’C

A'A = 3C’C Displacement at W xw = C’C A'A = 9xw From Conservation of Energy, as work

done remainssame F x A'A = W x xw F =

A

B

W F

Figure 3: Example 2 system labeled in initial and displaced configuration

2.

rope lengths as follows. AB + BC’ + CC’ + C’D + DB' + B'B = A'A + AB + 2BB' + BC’ +

CC’ + C’D + DB’ AA' = -BB' ie the pulley is going to move the same distance as rope and in same direction as assumed direction of BB’ is upward and answer is negative So basically the pulley starts falling down ie the mechanism cannot be constructed. This is the famous Fool’s Tackle. If you draw the force diagram, we can see that pulley B has unbalanced excess resultant force downward which causes it to accelerate downward.

Figure 4: Force Balance diagram showing why the Fool’s Tackle can’t be constructed

Example 24

A weight W is to be lifted by means of a force P applied to the system of two equal pulleys shown in the figure. Find the relationship between P and W for equilibrium conditions.

Solution

Let P move downward by a small virtual displacement then W will move up by an amount . Hence

× × Or .

Example 25

A horizontal force P is applied to the end of one of three identical hinged links each of which has a weight W and length . Determine the equilibrium configuration.

Solution

This is a problem having three degree of freedom. Thus three virtual displacement will be required to solve the problem. These are the three angular coordinates to completely specific the position of the bars.

With reference to figure

h cos θ 𝛉 𝛉 𝛉 4cm 4 cm P /2

h cos θ cos θ

h cos θ cos θ cos θ sin θ sinθ sinθ

If θ is allowed to vary first and θ and θ are held constant, the principle of virtual work requires [ ] . Thus

θ δθ W sinθ δθ

tanθ

Next θ is allowed to vary and θ and θ are held constant, so that [ ] Thus

θ δθ W sinθ δθ W sin θ δθ

tanθ

Lastly, θ is allowed to vary while θ and θ are held constant. Thus [ ] gives

cosθ δθ W sinθ W sinθ δθ θ δθ tan θ P

Example 26

The truss shown in Fig. is loaded and supported shown. Find the stresses in all members.

Solution kN ∑ = 0 gives × × × × kN 12kN 8kN C B D A 2m 2m E 20kN

kN kN [ ] F sin F × kN F cos F F kN Joint B [Fig.(b)] sin sin (C) C 12 D 21 (D) A 19 (A) (B) B 8

× × × × ×

Thus, we find that the initially assumed direction of force in member EB is to be reversed. Joint C [Fig.(c)] × From (a) and (b) 2 =

Hence the assumed direction of force in EC has to be reversed. Joint D[Fig. (d)] cos ×

Member Magnitude of force, kN Nature of force

AB 38/ - BC 30 - CD 42 - DE ₂₁ + EC 18 + EB 22 + AE 19 +

Solution ∑ 0 gives cos kN ∑ F gives sin kN ∑ = 0 gives × × × cos × sin × = 60 + 40 + 5 kN kN

Joint A [Fig. (a)]

F sin F × kN F cos F × F F kN Joint B. [Fig. (b)] F kNF kN Joint C [Fig.(c)] 30 + 20 sin F sin F cos cos 3@ 2m = 6m D F G 30kN E _10kN C B 20kN A

× ×

Joint D [Fig. (d)]

cos cos cos cos

× ×

cos cos cos cos

× ×

or … a

… b

Subtracting (b) from (a) gives

Joint E [Fig. (e)]

cos cos sin sin F andF kN 33.09 (e) 11.54 20 B 69.76 (b) (a) A 20 _(c) C 70.41 69.76 (d) D 54.64 G 22.44 (f) 38.86

F sin

F ×

Member Magnitude of Force, kN Nature AB AC BC BD CD CE DE DF EF EG FG 69.76 70.41 20.00 69.76 54.64 33.09 11.54 44.88 0 38.86 44.88 - - - - + + + - 0 + - Example 27

Determine the forces in all the members of the truss loaded as shown in Fig.

Solution ∑ F gives sin θ = 40 × = 22.19 kN* cos θ sin θ + ∑ F gives cos θ × ∑ gives × × × D E F 2m C 3m A 𝛉 O 𝛉 3m 20kN 10kN 10kN B 𝛉

kN kN Joint A [Fig. (a)]

21.26 = 10 cos θ F sin θ 21.26=10× F × F × sin θ F cos θ F F × × kN Joint B [Fig. (b)]

23.33 cos θ sin θ F cos θ F cos θ

23.33 ×

×

F F … a

23.33sin θ cos θ F sin θ F sin θ

23.33×

× ×

F F

From (a) and (b)

F kNF kN

Joint D [Fig. (d)]

15 cos θ sin θ F cos θ

15 × × = F F A 10 22.18 21.26 θ (a) 20 _θ θ θ B 23.33 (b) F 12.02 (e) 15.25 θ 12.02 21.67 (c) C 2.77 θ θ 10 (d) D 15 θ θ

sin θ F F sin θ cos θ 15 × F × × F Joint C [Fig. (c)] 21.67sin θ F sin θ 21.67 × F F cos θ F F cos θ 21.67 × E F kN Joint F [Fig. (f)] E sin θ F ×

Member Force, kN Nature AB AC BC BD CD CE CF DE EF 23.33 2.77 21.67 15.00 12.02 0 15.25 21.67 21.67 - + - - - - + - + Example 28

Find the stresses in the members BD, BE, EG, FG and HI of the truss as shown in Fig

(a) (b) (c) (c) (d) L 5m (d) (a) (b) C E G I K A B D F H J 6@ 3m = 18m

Solution

kN

× × × × kN kN

kN

The members for which the stresses are wanted must be cut by sections such as aa, bb, cc and dd. The section to the left of aa is taken for determining the stress in BD and BE. Using the equation∑ , where E is the intersection of BE and CE, we get

F × × F × ∑ F gives F cos θ F × × To find the force in EG, consider section bb, ∑ gives

F × ×

F kN

To find the force in FG, consider section cc and ∑ F gives

F kN

To find the force in HI, consider the right position of the section dd. ∑ F gives

F

F kN

Example 29

Determine the stresses in BD, BE and CE of the truss shown in Fig.

A @ F H 80 kN 3m G D B 4@ 3m=12m 60 kN 3m C 40 kN 4m E θ @

Solution kN ∑ gives × × × × kN cos θ sin

To determine the forces in BD, BE and CE, the truss is cut by section aa. ∑ gives F cos θ × F sin θ × × × F × F = ∑ gives F × × F kN

To find the force in BE, ∑ F gives F cos F cos θ

F ×

× F = 14.14 kN (+)