A meterstick hangs from a string at position 0.2 m. A 0.3 kg mass hangs from the

Question

A meterstick hangs from a string at position 0.2 m. A 0.3 kg mass hangs from the meterstick at position 0 m, and the meterstick balances. What is the mass of the meterstick?

The answer choices are:

a. 0.1 kg

b. 0.2 kg

c. 0.3 kg

d. 0.4 kg

e. 0.5 kg

I know the answer has to be either 0.1 or 0.2 kg because the 0.3 kg must be heavier than the meterstick in order to balance the much greater distance (0.8 m) on the other side. I also know that torques and forces must balance, but there is hardly any information here to derive an answer. All I can find is the force for the .3 kg, which is 3 N. I also know it has no torque because its distance is 0 m.

Explanation / Answer

if the system is in equilibrium the net torque is zero.

take the pivot point at 0.2 m mark, and solve for mass of the stick.

(0.2 m)Mg = (0.5 m - 0.2 m)mg

Hence, the mass of the stick is,

m = (0.2 m)(M)/(0.3 m) = 0.2*0.3 / 0.3 = 0.2 kg

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