# Homework Help: Dynamic Pulley System

1. Oct 3, 2009

### Roderic Day

1. The problem statement, all variables and given/known data
http://img53.imageshack.us/img53/6141/41622697.png [Broken]

2. Relevant equations
F = ma, all that basic stuff.

3. The attempt at a solution
Alright
g = 32.2
Ma = 150/g = 4.66lb
Mb = 75/g = 2.33lb

so Box A
__
|....}==== 2T
|__|

box B

3T
-----|....|______ 60lb
===={__|

Sum of Fx = ma

so for A
2T = 4.66aA
for B
60 - 3T = 2.33aB

then for the lengths of the wire I did
_________
O________
........n__O
|-sa-||sb|

so L = 2sa + 3sb
0 = 2aA + 3aB
but clearly aA and aB must have opposite signs so
2aA = 3aB

using simultaneous equations with the following you get
60 = 2.33aB +3T
0 = 4.66(3/2)aB - 2T
which makes aB = 7.22ft s-2 (ANS)
and T = 14.4 lb

then i figured there were two ways to go about work done

KEi(A+B) + Wd = KEf(A+B)
or Wd = 2T*sA + (60-3T)*sB

Which is great cause you get to doublecheck your result

the problem is that for method 1 I get 334.1J
and for method 2 I get 216.6J

So, yeah, any help is appreciated.

EDIT1: Formatting Errors
EDIT2: I found my mistake. My simultaneous eqs were wrong, it was 6.99, not 4.99.
Now the answers are aA = 7.02 ft s-2 and W = 140.4J
I'd still appreciate some double-checking on my method since I'm not really sure if I'm doing this right.

Last edited by a moderator: May 4, 2017