Question

A 0.102 kg meterstick is supported at its 39.8 cm mark by a string attached to the ceiling. A 0.687 kg mass hangs vertically from the 4.67 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 21.6 N. Find the value of the unknown mass. The acceleration of gravity is 9.81 m/s 2 . Answer in units of kg.

Answer 1

1.41 kg

Step-by-step explanation:

Draw a free body diagram. There are four forces on the meterstick:

Weight force Mg pulling down,

Weight of the first mass m₁g pulling down,

Weight of the second mass m₂g pulling down,

and tension force T pulling up.

Sum of forces in the y direction:

∑F = ma

T − Mg − m₁g − m₂g = 0

Solve for m₂:

m₂ = (T − Mg − m₁g) / g

Plug in values:

m₂ = (21.6 − 0.102×9.81 − 0.687×9.81) / 9.81

m₂ = 1.41 kg