**Branch :**First Year-Engineering Syllabus

**Subject :**Elements of Mechanical Engineering

## Equilibrium of bodies II

**Equilibrium of bodies II:**** **we can now represent a force applied on a body pivoted at a point as the sum of the same force on it at the pivot and a couple acting on it. This is shown in figure 1. Thus if the bar shown in figure 1 is in equilibrium, the pivot must be applying a force - and a counter couple moment on it.

To see the equivalence, let us take the example above and add a zero force to the system at the pivot point. This does not really change the force applied on the system. However, the trick is to take this zero force to be made up of forces and - as shown in figure 2. Now the original force and - at the pivot are separated by distance *d * and therefore form a couple moment of magnitude *Fd *. In addition there is a force on the body at the pivot point. The combination is therefore a force at the pivot point P and a couple moment . Notice that I am not saying about the pivot. This is because a couple is a free vector and its effect is the same no matter at which point it is specified.

**Example: ** You must have seen the gear shift handle in old buses. It is of Zigzag shape. Let it be of the shape shown in figure 3: 60cm at an angle of 45° from the x-axis, 30cm parallel to x-axis and then 30cm again at 45° from the x-axis, all in the x-y plane shown in figure 3. To change gear a driver applies a force of on the head of the handle. We want to know what is the equivalent force and moment at the bottom i.e., at the origin of the handle.

For this again we can apply a zero force i.e., ( and -) at the bottom so that original force and - give a couple moment

Thus equivalent force system is a force at the bottom and a couple equal to .

Having obtained equivalent force systems, next we wish to discuss what kind of forces and moments do different elements used in engineering mechanics apply on other elements.