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 regarding equilibrium at points A and B. The user questions whether F1 and F2 should equal T sin θ for both points, seeking confirmation of their understanding. They also inquire about deriving the equation cos θ = (au² + 1)⁻⁵, suspecting a typo where "a" should be "9." Responses indicate agreement with the user's reasoning and confirm the suspicion about the typo. The conversation emphasizes the importance of correctly applying trigonometric relationships in static equilibrium problems.
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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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