Rotational Statistics- Physics

Question

AHHH I've been at this question for literally a DAY and I still can't figure it out. PLEASE HELP!!

A 0.119 kg meterstick is supported at its 33 cm mark by a string attached to the ceiling. A 0.695 kg mass hangs vertically from the 4.08 cm mark. A mass is attached somewhere on the meterstick to keep it horizontal and in both rotational and translational equilibrium. The force applied by the string attaching the meter stick to the ceiling is 21.9 N.

a) Find the value of the unknown mass. The acceleration of gravity is 9.81 m/s^2. Answer in units of kg.

b) Find the point where the mass attaches to the stick. Answer in units of cm.

Answer 1

Since the yardstick is in equilibrium, the forces acting in the positive y direction have to equal the forces in the negative y direction. In other words:

∑F_y = 0

So that means the weights have to equal the upward force:

21.9 N = 0.695 kg (9.81) + 0.119 kg (9.81) + m kg (9.81)

21.9 N = 6.818 N + 1.167 N + 9.81 m

13.915 N = 9.81m

1.418 kg = m

So the mass has to equal 1.418 kg.

In order for it to be in rotational equilibrium, the clockwise torques have to equal the counter-clockwise torques

∑Τ_CW = ∑Τ_CCW

We'll pick the zero mark of the meter stick as the pivot point since all the distances are given from there.

The CW torques are created by the hanging weights:

0.695 kg * 9.81 * (.0408 m) = 0.2782 Nm

0.119 kg * 9.81 * 0.5 m = 0.5837 Nm

1.418 kg * 9.81 * x m = 13.91x Nm

The CCW torque is created by the string tension:

21.9 N * 0.33m = 7.227 Nm

Now we can solve

∑Τ_CW = ∑Τ_CCW

0.2782 Nm + 0.5837 Nm + 13.91x Nm = 7.227 Nm

13.91x Nm = 6.3651 Nm

X = 0.4576 m, or 45.76 cm from the end

Hope this helps