It is found that a uniform wooden latch 100 cm long and mass 95 g balances on a knife-edge when a 5 g mass is hung 10 cm from one end. How far is the knife edge from the centre of the latch?
Let x be the distance from the knife-edge to the center of the latch.
The torque produced by the 5 g mass hanging 10 cm from one end is (5 g)(10 cm) = 50 g·cm.
The torque produced by the uniform wooden latch is (95 g)(x cm).
Since the latch balances on the knife-edge, the torques on both sides of the latch are equal, so 50 g·cm = 95 g·cm.
Therefore, (95 g)(x cm) = 50 g·cm.
Dividing both sides of the equation by 95 g gives x cm = (50 g·cm)/(95 g) = 0.5263 cm.
Therefore, the knife edge is approximately 0.5263 cm from the center of the latch.