Rotational equilibrium problem

[ianb]

http://img95.imageshack.us/img95/9403/problempi4.png

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)

Final answer will be

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.

 

[OlderDan]

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

 

[ianb]

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

 

[Pyrrhus]

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

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.

 

[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.

 

[OlderDan]

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.

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.