Question

Consider hanging masses off the ends of a meter stick which is nailed to the wall, as shown the figure. Take the meter stick to be a 1.0 m long massless rod, the nail is d = 12 from the end as shown, and it is free to rotate about this point. If I put a mass m_1 = 5.0 kg onto the left of the meter stick, how much mass m_2 should I put onto the right end to ensure that it will not rotate? How much force, and in what direction, is the pivot point (the nail) acting on the meter stick with?

Answer 1

Part a

Moment of m1, = m1*d

Moment of m2, = m2*(1-d)

As per the condition for rotational equilibrium, the moment of forces must balance. So

m1*d = m2*(1-d)

5*0.12 = m2*(1-0.12)

m2= 0.68 kg

Part b

Weight of m1, w1 = m1*g downward

Weight of m2, w2 = m2*g downward

Since there is no net force hence force acting on stick = sum of weight = m1*g +m2*g = 55.664 N upward.