Question

a uniform metre rule of mass 150g is balanced on a knife edge of xm from a mass of 50g suspended at the 0.1m mark. find the position (x) of the knife edge

Answer 1

Given:

x - the mark on the ruler where the knife edge is

M - mass of the ruler

m - mass of the object attached to the ruler

a - the point where the mass m is attached

L - length of the ruler

Solution

The torque created by the hanging mass and the piece of ruler that is on the side where the mass is attached:

$$\tau_1=\frac{x}{L}M\frac{x}{2}+am.$$

The torque created by the ruler mass on the other side:

$$\tau_2=\frac{L-x}{L}m\frac{L-x}{2}.$$

Equilibrium means that

$$\tau_1=\tau_2,\\\space\\ \frac{x^2}{2L}M+am=\frac{(L-x)^2}{2L}M,\\\space\\ x=0.467\text{ m}.$$