Answer to Question #119186 in Physics for henok

Question

A uniform meter stick of mass 0.5kg is pivoted at a distance of 40cm from one end. If a 2kg
mass is suspended from this end, what mass must be suspended from the other end in order for
the meter stick to be in equilibrium?

Answer 1

According to the equilibrium condition, the torque acting on the one side of the stick should be equal to the one acting on another side:

$$M_1 = M_2$$

By definition, the torque is:

$$M = mgL$$

where $m$ is the mass which is suspended at distance $L$ from the pivot point.

As far as the stick is uniform, it's whole mass can be replaced with a point suspended in it's center of the mass: $0.5 m$ from the end.

Thus, on the one side from pivot point the torque will be:

$$M_1 = 2kg\cdot g\cdot 0.4m = 0.8g$$

On the one side from pivot point the torque will consist of two terms: the torque due to stick itself and the torque due to the mass to be found. Thus

$$M_2 = m_{stick}\cdot g(0.5-L_1) + m\cdot g\cdot 1m = \\ =0.5kg\cdot g\cdot (0.5m-0.4m) +m\cdot g\cdot 1m= 0.05g + mg$$

where $m$ is the mass to be found.

According to $M_1 = M_2$ :

$$0.8g = 0.05g + mg\Rightarrow m = 0.75 kg$$

Answer. m = 0.75 kg.