A rigid mass less rod of length L is hinged at its one end and a weight W is hung at a distance l(<L) from this end. What a force P should be applied upwards at the other end so that the rod remains in equilibrium horizontally ?
Question
A rigid mass less rod of length L is hinged at its one end and a weight W is hung at a distance l(<L) from this end. What a force P should be applied upwards at the other end so that the rod remains in equilibrium horizontally ?
Solution
To solve this problem, we need to use the principle of moments. The principle of moments states that for a body to be in equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point.
Step 1: Identify the forces and their distances from the pivot point (hinge).
- The weight W is acting downwards at a distance l from the hinge.
- The force P is acting upwards at a distance L from the hinge.
Step 2: Set up the equation for the principle of moments.
- The clockwise moment due to the weight W is W*l.
- The anticlockwise moment due to the force P is P*L.
Step 3: According to the principle of moments, these two moments must be equal for the rod to be in equilibrium. So, we set them equal to each other and solve for P.
Wl = PL
Step 4: Solve for P. P = (W*l) / L
So, the force P that should be applied upwards at the other end of the rod to keep it in equilibrium horizontally is (W*l) / L.
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