Did I Solve the Equations for Statics and Friction Correctly?

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The discussion centers on solving equations related to statics and friction, specifically focusing on the equilibrium conditions at points A and B. The user correctly identifies the equations F1 = T sin θ and F2 = T sin θ for equilibrium. Additionally, they derive cos θ = (9u² + 1)⁻⁰.⁵ from the tangent relationship tan θ = 3u, suggesting a potential typo where "a" should be "9". The community confirms the user's reasoning and calculations.

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Homework Statement



DSCF0001.jpg


Homework Equations



F = uR

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!
 
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chrisyuen said:

Homework Statement



DSCF0001.jpg


Homework Equations



F = uR

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?
 
chrisyuen said:
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.
 

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