# Homework Help: Torques and a Traffic Light

1. Apr 2, 2012

### PeachBanana

1. The problem statement, all variables and given/known data

A traffic light hangs from a pole as shown in the figure . The uniform aluminum pole AB is 7.50 m long and has a mass of 12.0 kg. The mass of the traffic light is 21.5 kg.

2. Relevant equations

clockwise torques = counterclockwise torques

3. The attempt at a solution

This may seem silly but my professor said to use the sin of 53° but one of the tutors told me to use the sin of 37 ° (which ended up being incorrect) because of the parallel line theorem. Could someone explain to me why the tutor was mistaken?

(12.0 kg)(3.75m)(9.8 m/s^2)(sin 53°) + (21.5 kg)(9.8 m/s^2)(7.50 m)(sin 53°) = F * 3.80 m

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2. Apr 2, 2012

### PhanthomJay

The tutor may have just erred in the geometric or trigonometric relationships. The moment, M, of a force, F, about a point, is the cross product of the position vector, r, times the force , that is, M = r X F = (r)(F)(sin theta), where theta is the angle in between the force and position vectors.

3. Apr 2, 2012

### Staff: Mentor

Is it possible that the tutor suggested the use of cos(37°) rather than sin(37°)?

4. Apr 2, 2012

### PeachBanana

Thanks you two! No, he definitely said (sin 37°).