Why Can't I Solve This Rotational Equilibrium Problem?

AI Thread Summary
The discussion revolves around a rotational equilibrium problem where the user struggles to find the correct force values. They initially derive equations based on force projections but arrive at an incorrect final answer of 6.59 instead of the book's answer of 11. Participants point out that the user is incorrectly calculating torque about a pivot point rather than an axis, leading to confusion in the calculations. They suggest reevaluating the torque contributions from both components of the force F_(T,1) and clarify that these components produce torques in opposite directions with varying distances. The user acknowledges the potential oversight but remains uncertain about the correct approach to resolve the problem.
ianb
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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.
 
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What happened to the F applied at P in your torque equation?
 
It's along the x-axis, so it isn't calculated (it goes through the pivot).
 
ianb said:
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.
 
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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.
 
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ianb said:
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.
 
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