# Finding the tension in a pulley system

Each of the weights A and B has a mass M and C has a mass of kM.

A) If the system is released form rest find the acceleration of the moveable pulley and the scale pan and show that the scale pan will ascend if k > 4.

B) When the system is moving freely find the tension in the string.

It is part B that is confusing me. In fact I don't need to solve part A to solve part B, but whether I use my answer from part A or not I get the same answer for part B which is different from the text book answer.

Here's my solution for part A:

I set the acceleration of the pan to a in a downwards direction.

Then the acceleration of C in the upward direction will be a/2, as the string to the left of C is not accelerating and the string to the right of C is accelerating at a.

For C:

$2T-kmg=ma/2$

For pan:

$2mg-T=2mga$

Now I do the algebra to eliminate T and arrive at:

$a=\frac{2g(4-k)}{8g+1}$

Great, if k > 4 then a is negative and so pan ascends - as was to be shown.

For part B I could substitute my value of a into one of the original two equations, or simply eliminate a from those original two equations.

Doing the latter:

$a=\frac{2mg-T}{2mg}$

Therefore

$8gt-4kmg^2=2mg-T$

$T=\frac{2mg(1+2kg)}{8g+1}$

$T=\frac{6kmg}{k+8}$

Which is not even close to mine.

• You might want to look at the units of your answer versus those of the book's answer. What units do you expect $T$ to have?
– John Hughes
Commented Apr 17, 2018 at 19:15

Since C has mass $km$ your equation of motion for C should be

$2T-kmg=kma/2$

$2T-kmg=ma/2$
$2mg-T=2ma$
$2mg-T=2mga$