(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

2. Relevant equations

m1

[tex]\Sigma[/tex]F_{y}=N-m_{1}g = 0

[tex]\Sigma[/tex]F_{x}=T=m_{1}a

(Because theres no friction i see no opposing force to T)

m2

[tex]\Sigma[/tex]F_{y}=m_{2}g-T=m_{2}a

[tex]\Sigma[/tex]F_{x}=0

3. The attempt at a solution

m_{2}g-T=m_{2}a

T=m_{1}a

(m_{1}+m_{2})a=m_{2}g-T+T

I notice that the T's cancel when i add the equations together

so it becomes

(m_{1}+m_{2})a=m_{2}g

so...

a=(m_{2}g)/(m_{1}+m_{2})

so...

T=m_{1}a

T=(m_{1}) (m_{2}g)/(m_{1}+m_{2})

m_{1}=6.03kg

m_{2}=4.68kg

T=(6.03Kg)((4.68Kg)(9.8)/(6.03Kg+4.68Kg))

so...

T=25.8225N

T=m_{1}a

25.8225N = (6.03Kg)a

a=4.2823 m/s^{2}

did i do this right?

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

2. Relevant equations

3. The attempt at a solution

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

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