Statics and Friction 01

[chrisyuen]

1. The problem statement, all variables and given/known data

http://i689.photobucket.com/albums/vv256/physicsforums/DSCF0001.jpg

2. Relevant equations

F = uR

3. The attempt at a solution

I don't understand the followings:

For equilibrium at A, the equation should be F1 = T sin \theta. Am I right?

While for the one at B, the equation should be F2 = T sin \theta. Am I right?

In addition, how can I get cos \theta = (au2 + 1)-0.5 at the last answer?

Thank you very much!

[chrisyuen]

1. The problem statement, all variables and given/known data

http://i689.photobucket.com/albums/vv256/physicsforums/DSCF0001.jpg

2. Relevant equations

F = uR

3. The attempt at a solution

I don't understand the followings:

For equilibrium at A, the equation should be F1 = T sin \theta. Am I right?

While for the one at B, the equation should be F2 = T sin \theta. Am I right?

In addition, how can I get cos \theta = (au2 + 1)-0.5 at the last answer?

Thank you very much!

I tried some methods and found the followings finally:

tan \theta = 3u = 3u / 1

Put the above result to a right-angled triangle with adjacent side = 1 and opposite side = 3u,

we can get cos \theta = (9u2 + 1)-0.5.

I guessed that the "a" is actually a "9" which is a typo instead. Am I right?

[Shooting Star]

I tried some methods and found the followings finally:

tan \theta = 3u = 3u / 1

Put the above result to a right-angled triangle with adjacent side = 1 and opposite side = 3u,

we can get cos \theta = (9u2 + 1)-0.5.

I guessed that the "a" is actually a "9" which is a typo instead. Am I right?

You seem to be quite right.