# Rotational equilibrium problem

1. Nov 22, 2006

### ianb

I know this problem isn't hard, but I can't reach the final answer in the back.

OK, so if we project the forces along the x and y axis, we can easily conclude that:

F_(T,2) + F_(T,1) * sin50 = 10
F_(T,1) * cos50 = P

from here, though, I seem to be doing something wrong. The figure is in translational and rotational equilibrium, so net torque is zero. Let's take F_(T,2) as pivot. Then

F_(T,2)(0) + F_(T,1) * sin50(.30) = 10(.15)

F_(T,1) = 6.59

where as the book's answer is 11. Of course I can't continue from here and find the other forces, so I'll just leave it at that.

Heh. Thanks all.

Last edited: Nov 22, 2006
2. Nov 22, 2006

### OlderDan

What happened to the F applied at P in your torque equation?

3. Nov 22, 2006

### ianb

It's along the x-axis, so it isn't calculated (it goes through the pivot).

4. Nov 23, 2006

### Pyrrhus

No, OlderDan is right, you are taking moment (torque) about a point!, not about an axis (which is defined differently).

Unless you meant about the left down corner, which in that case you forgot one of the F(t,1) components torque. Maybe you should be more clear.

Last edited: Nov 23, 2006
5. Nov 23, 2006

### ianb

Wow, okay, then I guess we could say

F_(T,2)(0) + F_(T,1) * sin50(.30) + F_(T,1) * cos50(.30) = 10(.15)

but that will give F_(T,1) = 3.55, which is incorrect. Of course, I could have made something wrong there but there is a catch somewhere that I probably wasn't taught before.

Last edited: Nov 23, 2006
6. Nov 23, 2006

### OlderDan

If you use the lower left corner for the torque calculation, the two components of F1 produce torques in opposite directions and have different perpendicular distances.