M1 Pulley Question

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?

Pulley Question.jpg503.9KB

PulleyQuestionAttempt1a.jpg497.6KB

PulleyQuestionAttempt1b.jpg499.3KB

Reply 1

mqb2766

For the B part it is
2ma = ...
Youve just got ma.

Reply 2

JonnyK81

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)

Reply 3

mqb2766

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.

Reply 4

JonnyK81

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?

Reply 5

mqb2766

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.

Reply 6

JonnyK81

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!