# Forces due to Tension

1. Feb 19, 2012

### sb13

1. The problem statement, all variables and given/known data

Suppose that in the figure shown in the URL below, the masses of the blocks are M = 2.6 kg and m = 6.6 kg. What are (a) the magnitude of the acceleration and (b) the tension in the cord?

2. Relevant equations

F= ma

3. The attempt at a solution

I found the Fnet, y and Fnet, x for the hanging block (H) and for the sliding block (S).

For S Fnet, y = May which becomes FN = FgS, and Fnet becomes T= Ma

For H Fnet, y becomes T-FgH = may, which becomes T-mg= -ma

I combined the two equations to cancel out T and solve for acceleration.

T-mg+ma = T- Ma
-mg+ma = - Ma
-mg+ma+Ma = 0
ma+Ma = mg
a(m+M) = mg
a = mg/(m+M)

That gave me the correct answer for (a) but when I went to try to cancel out a and solve for tension

T = Ma, T-mg = -ma
a = T/M a= (T-mg)/(-m)

a-T/M = a-(T-mg)/(-m)
-T/M = -(T-mg)/(-m)
(T-mg)/(-m) - (T/M) = 0
M(T-mg)/(-m) - (-m)(T)/M = 0
MT-Mmg/(m+M) -mT/(m+M) = 0
T((M-mg)-m)/(m+M) = 0
T= -M+mg +m/(m+M)

That gives a negative number for tension. What did I do wrong?

Last edited: Feb 19, 2012
2. Feb 19, 2012