Force, Work, Power and Energy

Second Law of motion

When a force acts on a rigid body, two kinds of motion are exhibited.

Linear or Translational Motion: When a rigid body, which is free to move, starts moving in a straight path along the direction of applied force, then such motion is called linear or translation motion. Rotational Motion: When a rigid body, pivoted at a point, starts rotating about the axis passing through the pivoted point when a force acts on it, then such motion is called rotational motion.Torque (Moment of Force)

The given figure shows a wrench and a nut. When a force, F, is applied to the handle of the wrench, the nut turns in a direction as shown in the given diagram.

It is interesting to note that the greater the distance between the nut and the point of application of force (denoted by D in the figures), the easier it will be to turn the nut.

Therefore, the turning of the nut depends on two factors:

i. The greater the applied force F, the more easily the nut can be turned.

ii. The greater the distance d, the more easily the nut can be turned.

It is clear from these points that the turning effect can be increased either by increasing F or by increasing d (distance between the nut and the point of application of force F)

On account of these points, we define a quantity called torque.

Torque (τ) = Force (F) × Perpendicular distance (d)

Torque represents the turning force acting on an object. It can either be clockwise or anticlockwise, depending upon how the force is applied. The given figures show clockwise as well as anticlockwise torque.

A clockwise torque tends to turn an object in the clockwise direction. Similarly, an anticlockwise torque tends to turn an object in the anticlockwise direction.

Torque is also known as moment of force.

Unit of Torque

Torque (τ) = Force (F) × Perpendicular distance (d)

Since the unit of force is N and the unit of distance is m, the unit of torque is Nm (Newton-metre).

Translational Equilibrium

An object is said to be in translational equilibrium if the net force acting on the object is zero. A translational equilibrium corresponds to the state of rest or to a straight-line motion at a constant speed. It means that if an object is in translational equilibrium, the object remains at rest or continues its motion in a straight line at constant speed. The essential condition for translational equilibrium can be given in the form of an equation as:

Σ F = 0

The given figure shows a block on which two forces are acting.

The net force on this block is 120 N − 120 N = 0 N.

Therefore, the given block is said to be in translational equilibrium.

It means that if the block is already at rest, then it will continue to be at rest after the two forces start acting simultaneously.

Also, it means that if the block was already in motion, it will continue the same type of motion after the two forces start acting simultaneously.

Rotational Equilibrium

An object is said to be in rotational equilibrium if the net torque acting on the object is zero.

Consider a drum fitted such that it can rotate around its axis. The given figures show two such drums. Arun and Pankaj tie up ropes, as shown in the given figures and pull the ropes with equal forces.

In which of the given situations will the drum not rotate?

Yes, that is right! The drum will not rotate in situation II. Therefore, in situation II, the drum is in rotational equilibrium. The torque exerted by Arun is equal but opposite to the torque exerted by Pankaj.

Clockwise and Anticlockwise Moment

Torque is also known as moment of force.

The given figure shows a metre scale that can rotate about the fixed point O.

The moment of force F1 is given by:

τ1 = F1 × d1

This tends to turn the metre scale in the clockwise direction.

Hence, it is a clockwise moment.

Similarly, the moment of force F2 is given by:

τ2 = F2 × d2

It is an anti-clockwise moment. It tends to turn the metre scale in the anticlockwise direction.

For an object to remain in remain in rotational equilibrium:

Clockwise moment = Anticlockwise moment

Example

Let us consider a light rod AB of negligible mass with centre at C. Two parallel forces, each of magnitude F, are applied at the ends such that the forces are perpendicular to the rod, as shown in the figure below.

Here, total force on the rod = F − F = 0. This is because the forces act in opposite directions.

The net force on the rod is zero. Therefore, the rod is in translational equilibrium.

The moments of both forces about C are equal (= aF) but they are not opposite. They act in the same direction and cause the anticlockwise rotation of the rod. Thus, the rod is not in rotational equilibrium.

Couple A pair of parallel forces, which are equal and opposite and are not acting along the same line, form a couple. These two equal and opposite forces always act at two different points. Such couple is always needed to produced a rotation. Examples: Turning of a tap water, tightening the cap of a bottle, turning a steering wheel are some examples where couple is used. Moment of Couple It is the product of either force and perpendicular distance between the two forces or couple arm. Moment of force F at the end A = OA × F (anticlockwise) Moment of force F at the end B = OB × F (anticlockwise) Total moment of couple = OA × F + OB × F = OB × F = F × d (anticlockwise) Principle of MomentsAccording to the principle of moments, a body will be in rotational equilibrium if algebraic sum of the moments of all forces acting on the body about a fixed point is zero.

Example − Take an ideal lever comprising of a rod AB of negligible mass pivoted at a point O.

Here, F1 and F2 are two parallel forces. For translational equilibrium, net force should be equal to 0.

∴ R − F1 − F2 = 0

R = F1 + F2 … (i)

For rotational equilibrium, the algebraic sum of moments of forces about O must be zero. If AO = d1 and OB = d2, then

F1 × d1 − F2 × d2 = 0 … (ii)

[Anticlockwise moments are taken as positive and clockwise as negative]

From equation (ii),

F1d1 = F2d2 … (iii)

In case of the lever, force F1 is usually some weight to be lifted (called load) and its distance from the fulcrum (AO = d1) is called the load arm.

Force F2 is the effort applied to lift the load and its distance from the fulcrum (OB = d2) is called the effort arm.

From equation (iii), we have

Load × Load arm = Effort × Effort arm

The ratio F1/F2 is called mechanical advantage (M.A.) of the lever i.e.,

Usually, M.A. > 1 i.e., F1 > F2 i.e., a small effort is applied to lift a heavy load.

Centre of Gravity

The centre of gravity of a body is the point at which the body’s whole weight is said to act from. The concept of centre of gravity makes it easy to do calculations for objects having extended size. Generally, it turns out that the centre of gravity is also the centre of mass. The centre of mass of an object is the point where the whole mass of the object is said to be concentrated. The Centre of gravity of a body of irregular shape can be determined by the following method.Suspend the body from some point such as A.

Draw the vertical line AA1. This line AA1 passes through Centre of Gravity, which lies somewhere on this line.

Similarly, draw vertical lines BB1 and CC1 by suspending the body from some other points B and C, etc. The point of intersection of these verticals gives us the position of Centre of Gravity (G) of the irregular body. Centre of Gravity of Some Regular Bbjects

Object Position of Centre of Gravity 1. Rod Mid point of rod 2. Circular Disc Geometric centre 3. Solid or hollow Sphere Geometric centre of sphere 4. Solid or hollow Cylinder Mid point on the axis of cylinder 5. Solid Cone At a height h/4 from the base, on its axis 6. Hollow Cone At height h/3 from the base, on its axis 7. Circular Ring Centre of ring 8. Triangular Lamina The point of intersection of medians 9. Parallelogram, rectangular lamina The point of intersection of the diagonalsExample

Consider the motion of a body that is rotating as well as translating, as shown in the given figure.

Let the mass of this irregular object be “M” and the net force acting on the object be “F”. Each of the particles of the object follow complex paths as the object is rotating as well as translating. It is, therefore, difficult to apply the equation “f = ma” on each particle. But it is possible to use this equation if the whole mass “M” of the object is considered to be concentrated at the centre of gravity of gravity (G) of the object. During the journey of the shown object, it is the point G that does not revolve. All other particles of the object revolve around the point G.

Even though it is not possible to easily track the motion of individual particles, it is possible to track the motion of point G using the

To view the complete topic, please