Question

A 0.113 kg meterstick is supported at its 39.5 cm mark by a string attached to the ceiling. A 0.607 kg mass hangs vertically from the 4 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 19.7 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

1.29 kg

Step-by-step explanation:

mass of stick (Ms) = 0.113 kg

first mass (Ma) = 0.607 kg

second mass (Mb) = ?

force on the string (Fs) = 19.7 N

acceleration due to gravity (g) = 9.81 m/s^{2}

Find the value of the second mass (Mb).

since the meter stick is in rotational and translational equilibrium, the sum of upward forces is equal to the sum of downward forces irrespective of where the forces act on the meter stick.

therefore

• force on string (Fs) = force of first mass (FA) + force of second mass (FB) + force of meter stick's weight (FW)
• Fs = FA + FB + FW
• force = mg
• Fs = Ma.g + Mb.g + Ms.g

• Fs = (Ma + Mb + Ms) x g

• 19.7 = (0.607 + Mb + 0.113) x 9.81
• 19.7/9.81 = 0.607 + 0.113 + Mb
• 2.01 = 0.720 + Mb
• Mb = 2.01 - 0.720
• Mb = 1.29 Kg

the value of the second mass is 1.29 kg