# Homework Help: Two masses connected by a pulley with a frictionless table

1. Sep 24, 2010

### AHinkle

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

2. Relevant equations
m1
$$\Sigma$$Fy=N-m1g = 0
$$\Sigma$$Fx=T=m1a
(Because theres no friction i see no opposing force to T)

m2
$$\Sigma$$Fy=m2g-T=m2a
$$\Sigma$$Fx=0

3. The attempt at a solution

m2g-T=m2a
T=m1a

(m1+m2)a=m2g-T+T
I notice that the T's cancel when i add the equations together
so it becomes

(m1+m2)a=m2g
so...
a=(m2g)/(m1+m2)

so...
T=m1a
T=(m1) (m2g)/(m1+m2)

m1=6.03kg
m2=4.68kg

T=(6.03Kg)((4.68Kg)(9.8)/(6.03Kg+4.68Kg))
so...
T=25.8225N

T=m1a
25.8225N = (6.03Kg)a

a=4.2823 m/s2
did i do this right?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 24, 2010

### collinsmark

You know, you could have just plugged the numbers in here and saved yourself some work.
'Looks right to me.