Please explain this as well as how to solve: A uniform metre rule rests horizontally on a knife edge at the 60 cm mark when a mass of 10 g is suspended from one end. Draw a diagram of the arrangement.2.3.1. at which end must this mass be suspended? (2)2.3.2. what is the mass of the rule?

The metre rule and the knife edge form a lever and fulcrum system, like a "see-saw" or "teeter-totter". To balance the system, we will need the total moments or torques about the fulcrum to equal zero.

Begin by drawing a diagram of the arrangement. A metre rule is 100 cm long. The knife edge is located at 60 cm. The weight of the metre rule is located at its center, the 50 cm mark. To balance the weight of the metre rule, we must add a weight on the other side of the knife edge. Therefore, we should add the 10 g weight to the short end of the metre rule.

Next, use the free-body diagram to sum the moments or torques about the knife edge. Each torque is equal to the force times the distance from the knife edge. Make clockwise torques negative and counterclockwise torques positive. The weight of the metre rule is Mg, and it is 10 cm left of the knife edge. The weight of the 10 g mass is mg, and it is 40 cm right of the knife edge.

Summing the torques:

∑τ = Iα

(Mg) (10 cm) − (mg) (40 cm) = 0

(Mg) (10 cm) = (mg) (40 cm)

M = 4m

M = 4 (10 g)

M = 40 g

Therefore, the mass of the metre rule is 40 g.