The discussion revolves around a physics problem involving a pulley system with two masses, M1 on a frictionless table and M2 hanging off the side. Participants are trying to determine the relationship between M2, M1, and the tension T in the string connecting them.
The discussion is active, with participants exploring different interpretations of the problem. Some have offered guidance on drawing diagrams and setting up equations, while others are questioning the assumptions made about the system's motion. There is a recognition of the need to manipulate equations to isolate M2, but no consensus has been reached on the final form of the solution.
Participants note that the problem lacks specific information about acceleration and movement, which complicates the analysis. There is also a mention of homework constraints that may affect the approach taken by participants.
therefore since we're looking for m2 and only m2 we can simplify to m2=(m1+m2)a /g?Arman777 said:For correction we know that m2g-T=m2a and T=m1a
so
m2g-m1a=m2a
m2g=(m1+m2)a
As we expected.
This is true.I just did that for to check that what we did was right or wrong.soolights said:oh, i missed that message, sorry
T=M1a , a = T/M1
M2g-T=m2a ,
M2g-T=M2(T/M1)
no problem, but shouldn't you be right if we are solving for M2? all the m2 should be on the LHS right?Arman777 said: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
I diddnt understand yousoolights said:no problem, but shouldn't you be right if we are solving for M2? all the m2 should be on the LHS right?
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.Arman777 said:Leave M2 alone in this equation.