Statics Problem - finding angle

[btbam91]

Hey guys I'm having trouble with this simple statics problem.

[PLAIN]http://img443.imageshack.us/img443/7218/37statics.png

[PLAIN]http://img841.imageshack.us/img841/6734/372.png

My set up.

[PLAIN]http://img543.imageshack.us/img543/9349/ya1x.jpg

I'm confused on where I'm supposed to take it from here. The solutions are: 6.12, 33.8 degrees.

Thanks!

 

[btbam91]

Can a moderator delete this thread please? I'm going to post it in the physics section.

 

[PhanthomJay]

I'm not a moderator, but if you post to the Physics section, they'll probably kick it back to this section, as it is apparently a homework problem. Your equation appears correct, but I think you need a trig identity to solve it, and you get 2 solutions. You might want to try the alternate definition for torque: Torque = r X F = rFsintheta, and see what you get.

Edit: To clarify, I mean to say that T = r X F is the vector cross product, where r is the position vector from the pivot to the point of application of the force, and F is the applied force. The magnitude of the torque (moment) about the pivot is T=(r)*(F)*(sin theta), where theta is the interior angle between the force and position vectors. You need a bit of geometry/trig to find theta as a function of alpha. You will get 2 solutions for alpha.

 

[berkeman]

Yes, this post belongs here and not in the general technical forums. Thanks for the help, PhantomJay.

 

[btbam91]

Thanks for the responses!

I'm confused as to why taking the inverse sin of something results in two angles.

Here's a solution to an extremely similar kind of problem.

[PLAIN]http://img178.imageshack.us/img178/959/statiscs11.png

I don't understand how taking the inverse sin of .99658 results in 85 degrees and 95 degrees, two angles that are only about 10 degrees apart. Is there a trig identity that I'm missing?

Thanks!

Thanks!

 

[PhanthomJay]

Thanks for the responses!

I'm confused as to why taking the inverse sin of something results in two angles.

Here's a solution to an extremely similar kind of problem.

I don't understand how taking the inverse sin of .99658 results in 85 degrees and 95 degrees, two angles that are only about 10 degrees apart. Is there a trig identity that I'm missing?

Thanks!

Thanks!

The sin function is positive in both the first and 2nd quadrants. When you plug the inverse into your calculator, it only gives you the first quadrant result. For example, in your first posted problem, you find, approximately, that sin (70 + alpha) = .97 , therefore, 70 + alpha = 76, and alpha =6 degreees. But arc sin 0.97 is also 103 degrees, therefore 70 + alpha = 103, and alpha = 33 degrees. Draw a sketch to confirm this.