JonnyK81
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Hello all,

I'm stuck with a pulley question... Please see attached.

My attempt is also attached. Not sure where I'm going wrong... any ideas?
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mqb2766
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For the B part it is
2ma = ...
Youve just got ma.
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JonnyK81
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(Original post by mqb2766)
For the B part it is
2ma = ...
Youve just got ma.
I'm sorry, which line?

Are you saying that 14/9 mg = 2 ( 4/9 mg + mew mg)
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mqb2766
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(Original post by JonnyK81)
I'm sorry, which line?

Are you saying that 14/9 mg = 2 ( 4/9 mg + mew mg)
When you find the resultant force / Newton 2 for B its
2mg - T = 2ma = 8/9 mg
not 4/9mg on the right hand side.
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JonnyK81
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(Original post by mqb2766)
When you find the resultant force / Newton 2 for B its
2mg - T = 2ma = 8/9 mg
not 4/9mg on the right hand side.
Ok,

so 2mg - T = 8/9 mg. I can see this now by resolving forces at B.

Now what? Sub in T = 14/9 mg?
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mqb2766
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(Original post by JonnyK81)
Ok,

so 2mg - T = 8/9 mg. I can see this now by resolving forces at B.

Now what? Sub in T = 14/9 mg?
Err, I think that value used the previous (wrong) equation.
You had about the right approach. Just recalculate T and solve for mu.
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JonnyK81
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(Original post by mqb2766)
Err, I think that value used the previous (wrong) equation.
You had about the right approach. Just recalculate T and solve for mu.
Yes, I see now.

When I resolved B, the equation should have read: 2mg - T = 2 x m x 4/9 g
So: - T = 8/9 mg - 2mg
and: T = 2mg - 8/9 mg
thus: T = 10/9 mg

Then for the second part, equating would give: 10/9 mg = 4/9 mg + mew mg
minus 4/9 mg both sides: 6/9 mg = mew mg
cancelling mg: 6/9 = mew = 2/3

Again... not enough attention paid when writing the equations... I'm getting old!
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JonnyK81
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QED. Thank you!

For anyone interested... full solution is now attached.
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Last edited by JonnyK81; 4 weeks ago
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