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TS656577
Feb10-08, 04:26 PM
1. The problem statement, all variables and given/known data
Three blocks are attached by cords that loop over frictionless pulleys. Block B lies on a frictionless table; the masses are mA=6kg, mB=6kg, and mC=18kg. The accleration due to gravity is g.


2. Relevant equations
F=ma


3. The attempt at a solution

I thought that it would be wise to treat these as two separate problems. I thought I would solve one side by the following. Block B T=ma Block C F(g)-T=ma and thus mg-T=ma
I thought that to figure out the acceleration of block B (since we know C's is -9.8m/s^2) that i would use the equation a(B)=(m(C)g - (m(C)g))/m(B) But when I do that i would get 0/6. That cant be right because the block C would move because it is heavier than block A

mgb_phys
Feb10-08, 04:43 PM
A acts in the opposite direction to C so you could think of it as a negative weight on the same side as C. That will simplify it into a single falling block pulling on B.

TS656577
Feb10-08, 04:50 PM
Oops, i forgot to mention, i have to find the acceleration of block a. So would i still be on the right track? I feel like my signs are messed up in that equation i had

mgb_phys
Feb10-08, 04:53 PM
Since they are all on the same wire the accelaration of all the blocks is the same.

TS656577
Feb10-08, 04:55 PM
so the accleration of block A is +9.8 m/s^2?

TS656577
Feb10-08, 04:57 PM
or are you saying that once i figure out the acceleration of one block, then that would be the same for all of them