Issue with angles in torque problem?

[fightboy]

Ok so for this example you have to calculate the net torque about an axis that passes through the hinges in each of the following cases in the image:

a) and c) were fairly straightforward to calculate but i had issues with b) and c). The solution listed the angle for b) as sin(115°) and for c) as sin(160°). I'm confused on where the 115° and 160° came from when the picture showed the angles as 25° and 70° respectively. Am I missing something? Thanks!

 

[berkeman]

Ok so for this example you have to calculate the net torque about an axis that passes through the hinges in each of the following cases in the image:
View attachment 73087

a) and c) were fairly straightforward to calculate but i had issues with b) and c). The solution listed the angle for b) as sin(115°) and for c) as sin(160°). I'm confused on where the 115° and 160° came from when the picture showed the angles as 25° and 70° respectively. Am I missing something? Thanks!

There's a small typo in your text above. I believe you mean that a) and d) are easy, and b) and c) are confusing you.

It is indeed a bit strange to use sin() to calculate the torque and show those angles, but it can be done of course. It's much more normal to use the cos() function, given the angles listed in the diagrams. Are you familar with the dot product?

Also, are you familiar with using the vector cross product to calculate the vector torque (it has magnitude and direction)? If you haven't gotten there yet, don't worry about it for now. If you have seen it, then you know that the sin() is used in it, but for the complementary angle to the one shown in your problem's diagrams.

Does that make sense?

 

[HallsofIvy]

All four look equally easy to me as log as you keep in mind that it only the force perpendicular to the arm that is relevant. If the given force makes angle \theta with the vertical, then the component that is perpendicular to the arm is F cos(\theta).

 

[restolano]

Oh, I think you must post the entire problem not just the diagram.