Statics and friction equations for a rope system

2 replies · 2K views · last reply
chrisyuen
Messages
55
Reaction score
0

Homework Statement



Textbook diagram of two rough rings A and B on a horizontal rod, joined by a string through ring C carrying weight W, with tensions T, normal reactions R1 and R2, friction forces F1 and F2, and weights 2W and W labelled.


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 [tex]\theta[/tex]. Am I right?

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

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

Thank you very much!
 
chrisyuen said:

Homework Statement



Textbook diagram of two rough rings A and B on a horizontal rod, joined by a string through a smooth ring C carrying weight W, with tensions T, reactions R1 and R2, and friction forces F1 and F2 labelled.


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 [tex]\theta[/tex]. Am I right?

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

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

Thank you very much!

I tried some methods and found the followings finally:

tan [tex]\theta[/tex] = 3u = 3u / 1

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

we can get cos [tex]\theta[/tex] = (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 [tex]\theta[/tex] = 3u = 3u / 1

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

we can get cos [tex]\theta[/tex] = (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.