Why Use Sin 53° Instead of Sin 37° for Torque Calculation?

[PeachBanana]

Homework Statement

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.

Homework Equations

clockwise torques = counterclockwise torques

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

 

[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.

 

[gneill]

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

 

[PeachBanana]

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

 

[asteriatic]

I would like to first address the importance of using the correct angles in calculations involving torque. Torque is a measure of the rotational force applied to an object, and it is dependent on both the magnitude of the force and the distance from the pivot point. In this case, the pivot point is the point where the pole AB is attached to the wall.

The tutor who suggested using the sine of 37° may have been trying to apply the parallel line theorem, which states that if two lines are parallel, then the corresponding angles are equal. However, this theorem only applies to lines that are parallel to each other, and in this scenario, the lines are not parallel. The force of gravity acting on the pole and the traffic light is not parallel to the pole itself, so the angle used in the calculation should be the angle between the force and the pole, which is 53°.

Furthermore, as a scientist, I would also like to point out the importance of using the correct units in calculations. In this case, the units for force are Newtons (N) and the units for distance are meters (m). It is important to use these units consistently throughout the calculation to ensure accuracy.

In conclusion, the correct approach to solving this problem is to use the sine of 53° in the calculation of the clockwise and counterclockwise torques. This will result in the correct value for the force, F, needed to keep the pole and traffic light in equilibrium. It is important to always carefully consider the physical situation and use the appropriate mathematical principles and units when solving problems in science.