Concept of Equilibrium

Concept of Equilibrium

• Before starting the further discussion it is necessary to understand some concepts of equilibrium and free body diagram (F.B.D). It will be discussed in detail in upcoming chapters.

• A body is said to be in equilibrium provided it remains at rest nitially at rest, or has constant velocity if initially it was moving.

• However, the concept of 'equilibrium' is mostly used for 'static equilibrium' of bodies, i.e. for bodies at rest.

• The condition for equilibrium is that, the resultant must be zero (R = 0). Hence the conditions for equilibrium of a concurrent force system are,

• If there are three forces in equilibrium then Lami's theorem can be used.

1. Forces in Strings / Cables / Ropes

• Strings/ropes/cables oppose increase in length. When the forces acting on a string tend to elongate the string, internal opposing forces are developed inside the string which are known as tensile forces.

• For example, when two rollers A and B, connected by a string are kept on two inclined planes as shown in Fig. 2.11.1 (a), tensile forces develop in the string which tend to pull A towards right and B towards left as they tend to move apart.

• On the other hand, no tensile force develops inside the string if the rollers are kept on inclined planes as shown in Fig. 2.11.1 (b) as A and B tend to move towards each other which is not opposed by the string.

• The direction of tensile force is always away from the object under consideration inside the string. 

• When cables are passed over pulleys, tension in the cable is same on both sides of the pulley provided there is no friction between the cable and the pulley. 

• In many practical applications, masses of the pulleys are negligible. Hence using equilibrium of the pulleys, tension in other cables can be obtained. 

• For example, in Fig. 2.11.2, if tension in AB is taken as T, then tension in BC, DE and EF will also be T. Tension in BG can be obtained using equilibrium of the pulley 1 as 27. Similarly tension in EH is 2T and tension in IJ is 3T.

2. Forces in Rods

a) Coaxial forces :

• If external forces on a straight rod are acting only at the end points and the rod is in equilibrium, the forces inside the rod are coaxial.

• Coaxial forces can be either tensile or compressive as rods oppose elongation as well as compression.

• Tensile forces are directed away from and compressive forces towards, the point or object under consideration along the length of the rod.

• Rods having coaxial forces are called two force members.

b) Non-coaxial forces :

• If external forces acting on the rod are not all acting at the end and atleast one of them is not along the axis, the force in the rod is not coaxial. 

• Minimum number of forces required for equilibrium of rod is three. Such rods are known as multiforce members.

3. Force in Spring

• When a spring is deformed through length x, the restoring force set up in the spring is given by F = kx, where,

K = Spring constant or stiffness constant

x = Deformation of spring

• Negative sign indicates that the force is directed opposite to deformation. The force is always along the length of the spring. Refer Fig. 2.11.5.

4. Free Body Diagram (F.B.D.)

• It is the isolated diagram of an object/system of objects/any point in the system in which all forces and couple moments acting on it are shown including support reactions.

• When objects are supported by some supporting mechanisms, the supporting mechanism exerts a force or couple moment on the object which tries to restrict the motion of the object.

• These forces or couple moments which oppose or restrict the motion by objects are called support reactions.

• The different types of supports and their corresponding reaction required in this chapter (Friction) are as follows (The other supports and their reactions will be discussed in upcoming chapters).

i) Block rest on horizontal surface

• Consider a block of weight W which is supported on a horizontal surface as shown in Fig. 2.11.6. It is also subjected to horzontal force P.

• For this case, weight is acting vertically downwards at its centre of gravity and one normal reaction (perpendicular to surface). Refer Fig. 2.11.6.

ii) Block rest on inclined surface

• Consider a block of weight W which is supported on an inclined surface at an angle θ and it is also subjected to force P as shown in Fig. 2.11.7

• For this case, weight W is acting vertically downwards at its C.G. and normal reaction is perpendicular to surface at an angle (90 - θ) with horizontal or with vertical.

• Generally, in such cases parallel to surface is considered as x-axis and perpendicular to that as y-axis.

iii) Body is connected to a cable or wire

• Consider a body connected to a cable as shown in Fig. 2.11.8.

• In this case, tensile force or tension (T) is acting on the body which is generally away from the body.

• In addition to tension T, normal reaction (N) is also acting on the body.

iv) Two rollers or two blocks attached to each other

• If two rollers are attached to each other as shown in Fig. 2.11.9. then their F.B.D. includes reaction along the line joining two centres. Also, includes normal reaction at all contacting surfaces.

• Similarly if two blocks are attached then also F.B.D. is same. Refer Fig. 2.11.10

v) Rod or ladder supported on two surfaces

• For rod or ladder supporting on surface, there is reaction perpendicular to surface. Refer Fig. 2.11.11.

5. General Procedure for Solving Problems

1. Draw F.B.D. of a point in the system/an object in the system/complete system of objects in equilibrium. Consider all forces acting on the system including support reactions.

2. If forces are concurrent, use

3. For three concurrent forces in equilibrium, use Lami's theorem,

4. For an object hanging freely from a cable, the tension in the cable will be equal to the weight of the object.

5. For equilibrium of a system of more than one objects, draw free body diagrams of all the objects. Start solving with that F.B.D. which has minimum number of unknowns.

Solved Examples for Understanding

Example 2.11.1 

Determine the magnitude and angle θ of F so that particle P, shown in Fig. 2.11.12 is in equilibrium.

Solution :

Example 2.11.2 

Determine the magnitude and direction of the force 'P' which keeps the concurrent system in equilibrium (Fig. 2.11.13).

Solution :

Example 2.11.3 

Fig. 2.11.14 shows a 10 kg lamp supported by two cables AB and AC. Find the tension in each cable.

Solution: 

The F.B.D. of point A is shown in Fig. 2.11.14 (a). Using Lami's theorem,

Example 2.11.4 

Find the forces in cables AB and CB shown in Fig. 2.11.15. The remaining two cables pass Over frictionless pulleys E and F and support masses 1200 kg and 1000 respectively.

Solution: 

All the forces are concurrent at B. The FBD of B is shown in Fig. 2.11.15 (a).

Example 2.11.5 

Fig. 2.11.16 shows a system of cables in equilibrium condition under two vertical loads of 300 N and 500 N. Determine the forces developed in the different segments.

Solution: 

Free body diagrams of points B and C are shown in Fig. 2.11.16 (a).

Using Lami's theorem for C,

Example 2.11.6 

Draw the freebody diagram of sphere shown in Fig. 2.11.17 and determine the reactions at the points of contact using Lami's theorem.

Solution: 

The F.B.D. of sphere is shown in Fig. 2.11.17 (a). 

Using Lami's theorem,

Example 2.11.7 

Determine the tension in the string and the reaction at the contact surface for the cylinder of weight 1000 N placed as shown in Fig. 2.11.18.

Solution: 

The F.B.D. of cylinder is shown in Fig. 2.11.18 (a).

Using Lami's theorem,

Example 2.11.8 

Two identical rollers, each of weight 500 N, are supported by an inclined plane making an angle of 30° to the horizontal and a vertical wall as shown in the Fig. 2.11.19. 

i) Sketch the free body diagrams of the two rollers.

ii) Assuming smooth surfaces, find the reactions at the support points.

Solution: 

The F.B.D. of each roller is shown in Fig. 2.11.19 (a). The reaction between the two rollers makes angle 30° with horizontal as the line joining the centres is parallel to the inclined plane.

For B, using Lami's theorem,

Example 2.11.9 

Two cylinder P and Q rest in channel as shown in the Fig. 2.11.20. The cylinder P has diameter of 100 mm and weighs 200 N where as the cylinder Q has diameter of 180 mm and weighs 500 N. If the bottom width $ 500 N.  If the bottom width of the box is 180 mm, with one side vertical and other inclined at 60°, determine the reactions at all the four points of contact.

Solution : 

To find the angle made by the reaction between P and Q with horizontal, use constructions as shown in Fig. 2.11.20 (a).

The free body diagram of the two cylinders are shown in Fig. 2.11.20 (b).

Using Lami's theorem for P,

Example 2.11.10 

Three identical spheres P, Q, R of weight 'W' are arranged on smooth inclined surfaces as shown in Fig. 2.11.21. Determine the angle 'α' which will prevent the arrangement from collapsing.

Solution: 

The free body diagrams of the three spheres are as shown in Fig. 2.11.21 (a).

When the arrangement is about to collapse, P and Q tend to move away from each other. As a result, reaction between P and Q tends to zero.

From FBD of R:

From FBD of P :

Example 2.11.11

Determine the reactions at contact points for the system shown in Fig. 2.11.22.

Solution: 

The reaction between the two rollers at Q is along the line joining their centres. The angle for this reaction with horizontal can be obtained as shown in Fig. 2.11.22 (a).

From Fig. 2.11.22 (a),

cos α = 40/50

⸫ α = 36.87°

The free body diagrams of the two rollers are

shown in Fig. 2.11.22 (b).

Using Lami's theorem for F.B.D. of B,

Example 2.11.12 

A 3 kN crate is to be supported by the rope and pulley arrangement shown in Fig. 2.11.23. Determine the magnitude and direction of the force F, which should be exerted at the free end of the rope.

Solution: 

The F.B.D of movable pulley is shown in Fig. 2.11.23 (a).

From given Fig. 2.11.23,

Example 2.11.13 

Two rollers weighing 5 kN and 3 kN are connected by a string and are supported on mutually perpendicular smooth inclined planes as shown in Fig. 2.11.24 Find the angle of the string with the horizontal, when the system is in equilibrium.

Solution: 

The FBDs of the rollers are shown in Fig. 2.11.24 (a). 

The forces are drawn away from the centre as shown in Fig. 2.11.24 (b) for using Lami's theorem.

For 5 kN roller,

Example 2.11.14 

Three cylinders A, B and C weighing 150 N, 400 N and 200 N respectively are piled in a channel as shown. Determine the reactions offered by the walls and floor. Radit of A, B and C are 40 cm, 60 cm and 50 cm respectively. Refer Fig. 2.11.25.

Solution: 

The reactions between the cylinders are along the line joining their centres. The angles α1 and α2 made by these reactions with horizontal can be obtained from Fig. 2.11.25 (a).

The FBD for the three cylinders are shown in Fig. 2.11.25 (b).

For C:

Example 2.11.15 

Two cylinders of masses m1 = 100 kg and my = 50 kg are connected by a rigid bar of negligible weight hinged at each cylinder. The cylinders are resting on smooth inclined planes and are in equilibrium in the position shown under a force P. Determine the magnitude of force P.

Refer Fig. 2.11.26.

Solution: 

The FBD of each cylinder is shown in Fig. 2.11.26 (a). The rigid bar is subjected to forces only at the ends and hence the force in it will be axial. As the tends length of bar to R1 130° decrease, there will be compressive force Fc in it which is directed towards the cylinders as shown.

For cylinder 1 by Lami's theorem,

Example 2.11.16 

A lamp weighting 5 N is suspended from the ceiling by a chain. It is pulled aside by horizontal cord until the chain makes an angle of 60° with the ceiling as shown in Fig. 2.11.27 cord is horizontal as shown in Fig. 2.11.27. Find the tensions in the chain and the cord by applying Lami's theorem.

Solution: 

The F.B.D. of point where force is applied, is shown in Fig. 2.11.27 (a).

By Lami's theorem,

Example 2.11.17 

If the resultant of the three concurrent forces acting on the bolt shown in Fig. 2.11.28 is zero, determine the orientation of 0 and the required magnitude of F3.

Solution :

Dividing equation (1) by (2),

Example 2.11.18 

Find the length of cord PR in Fig. 2.11.29 so that the 10 kg bell is suspended in the position shown. The underformed length of the spring PQ is 0.5 m and the spring stiffness is 300 N/m.

Solution: 

Using Lami's theorem for the F.B.D. of point P shown in Fig. 2.11.29 (a).

Example 2.11.19 

Determine the tensions in various segments of the connected flexible cables as shown in the Fig. 2.11.30.

Solution: 

The free body diagrams of Q and S are shown in Fig. 2.11.30 (a)

Using Lami's theorem for S,

Example 2.11.20 

Two cylinders, having weight WA = 2000 N and WB = 1000 N are resting on smooth inclined planes having inclination 60° and 45° with the horizontal respectively as shown in Fig. 2.11.31. They are connected by a weightless bar AB with hinge connections. The bar AB makes 15° angle with the horizontal. Find the magnitude of the force P required to hold the system in equilibrium. 

Solution: 

Let Fc be the compressive force in rod. The free body diagrams of cylinders A and B are shown in Fig. 2.11.31 (a)

From equations (1) and (2),

RA = 2732.05 N

Fc = 2449.5 N

Example 2.11.21 

The 200 kg crate in Fig. 2.11.32 is suspended using the ropes AB and AC. Each rope can withstand a maximum of 10 kN before it breaks. If AB always remains horizontal, determine the smallest angle to which the crate can be suspended before one of the ropes breaks.

Solution: 

The FBD of point A is shown in Fig. 2.11.32 (a).

Example 2.11.22 

Two cylinders E, F of diameter 60 mm and 30 mm. Weighing 160 N and 40 N respectively are placed as shown in Fig. 2.11.33. Assuming all the contact surfaces to be smooth, find the reactions at A, B and C.

Solution: From Fig. 2.11.33 (b)

Example 2.11.23 

A roller of radius r and weight W = 1000 N is to be pulled over a curb of height h = r/2 as shown in Fig. 2.11.34 by applying a horizontal force P at the end of a string wound around the circumference of the roller. Find the magnitude of force P required to start the roller over the curb.

Solution: 

The F.B.D. of roller is shown in Fig. 2.11.34 (a). To pull the roller over the curb, the reaction at the bottom surface tends to zero.

As there are three forces acting on the roller, the direction of reaction at A must pass through the point of intersection of the lines of action of P and 1000 N. As RA has line of action from A to B, its angle a with horizontal will be given by,

As the forces are concurrent at B as shown in Fig. 2.11.34 (b). Using Lami's theorem,