The discussion focuses on solving a physics problem involving two masses, M1 and M2, connected by a frictionless pulley. Participants utilize Newton's Laws to derive the relationship between the tension (T), mass M1, and mass M2. The final equations established are M2g - T = M2(T/M1) and M2 = (M1 + M2)a/g, emphasizing the importance of isolating M2 to find its value in terms of T and M1.
PREREQUISITESStudents in introductory physics courses, educators teaching mechanics, and anyone interested in understanding the dynamics of pulley systems and forces in motion.
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.