In the diagram above, a rod 50 cm long of uniform cross-section is suspended horizontally on a fulcrum, F, by the action of two forces. What is the weight o...

In the diagram above, a rod 50 cm long of uniform cross-section is suspended horizontally on a fulcrum, F, by the action of two forces. What is the weight of the rod?

Answer Details

The weight of the rod can be determined by balancing the moments on either side of the fulcrum, F. The moment of a force is its magnitude multiplied by the perpendicular distance from the fulcrum to the line of action of the force. Let the weight of the rod be W, and the distance from the left end of the rod to the fulcrum be x. Then, the distance from the right end of the rod to the fulcrum is (50 - x). Since the rod is in equilibrium, the total clockwise moment about F must be equal to the total anticlockwise moment. The weight W acts downwards at the centre of gravity of the rod, which is at its midpoint, so its moment about F is W(50/2 - x) = 25W - Wx. The force acting to the left at the left end of the rod is 40N, and its moment about F is 40x. The force acting to the right at the right end of the rod is unknown, and its moment about F is P(50 - x), where P is the magnitude of the force. Setting the clockwise moment equal to the anticlockwise moment gives: 25W - Wx = 40x + P(50 - x) Simplifying, we get: 25W = 41x + 50P We need to find W, so we need to find P. We can use the fact that the rod is in equilibrium to say that the total upward force must be equal to the total downward force. Upward force = weight of rod + 40N Downward force = P Therefore: W + 40 = P Substituting this into the previous equation, we get: 25W = 41x + 50(W + 40) Simplifying, we get: 25W = 41x + 50W + 2000 Solving for W, we get: W = 80N Therefore, the weight of the rod is 80N.