Consider a setup similar to that used in the last part of the lab. A mass m1 hangs d1 cm to the right of the center of the bar. A string attached to the bar d2 cm to the right of the center passes over a pulley. A mass m2 is attached to its other end. We see that the student has made a mistake, and the string is at an angle θ from the vertical. (a) Assume that the string is vertical (it is perpendicular to the bar). Use the equilibrium equation for the torques to find m2. (Your answer should be in terms of m1, d1, and d2.) m2 = (b) Now consider the actual setup with the string at an angle θ to the vertical. Again use the equilibrium equation for the torques to find m2. (Your answer should be in terms of m1, d1, d2, and θ.)

A) The value of m2 using equilibrium of torques when the string is vertical is; m2 = (m1 × d1)/d2

B) The value of m2 using equilibrium of torques the actual setup with the string at an angle θ to the vertical is; m2 = (m1 × d1)/(d2 cos θ)

We are given;

First mass; m1

Distance of first mass to the right of the center; d1 cm

Second mass; m2

Distance of bar from the right of the center that passes through a pulley; d2 cm

A) Now, we are told that the string is perpendicular to the bar. Thus, taking equilibrium of torques, we have;

m1 * g * d1 = m2 * g * d2

g will cancel out to get;

m1 * d1 = m2 * d2

We want to find m2. Thus;

m2 = (m1 × d1)/d2

B) Now we are told that the string is set up at an angle θ to the vertical. Using equilibrium of torques, we now have

m1 * g * d1 = m2 * g * d2 * cos θ

g will cancel out to get;

m1 * d1 = m2 * d2 * cos θ

Making m2 the subject gives;

m2 = (m1 × d1)/(d2 cos θ)

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Explanation:

part A.

m1g×d1 - m2×g×d2 = 0

m1×d1 - m2×d2 = 0

m1×d1 = m2×d2

(m1×d1)/d2 = m2'

part B

same thing just d2× cos(?) so,  

(m1×d1)/(d2 ×cos(?)) = m2