Solving a Tricky Physics Pulley Problem: Finding M2 with Tension and Mass M1

[Arman777]

Thats correct.Leave m2 alone and you find the answer

 

[Arman777]

For correction we know that m2g-T=m2a and T=m1a
so
m2g-m1a=m2a
m2g=(m1+m2)a

As we expected.

 

[soolights]

For correction we know that m2g-T=m2a and T=m1a
so
m2g-m1a=m2a
m2g=(m1+m2)a

As we expected.

therefore since we're looking for m2 and only m2 we can simplify to m2=(m1+m2)a /g?

 

[Arman777]

oh, i missed that message, sorry
T=M1a , a = T/M1
M2g-T=m2a ,
M2g-T=M2(T/M1)

This is true.I just did that for to check that what we did was right or wrong.

The right equation is M2g-T=M2(T/M1).Leave M2 alone in this equation.Thats the answer.

sorry that I confused you

 

[soolights]

This is true.I just did that for to check that what we did was right or wrong.

The right equation is M2g-T=M2(T/M1).Leave M2 alone in this equation.Thats the answer.

sorry that I confused you

no problem, but shouldn't you be right if we are solving for M2? all the m2 should be on the LHS right?

 

[Arman777]

I didnt understand you

 

[Arman777]

no problem, but shouldn't you be right if we are solving for M2? all the m2 should be on the LHS right?

I diddnt understand you

 

[haruspex]

Leave M2 alone in this equation.

That might be causing some confusion. You mean, manipulate the equation so that M2 is alone on one side of the equation, and does not occur on the other side.
(Leaving it alone would mean not doing anything with it.)