June 20, 2013

Posts by physics

Total # Posts: 331

m2a=m2g-m1a m2a+m1a=m2g a= m2g/(m1+m2) I got this m2a=m2g-m1a m2a + m1 a = m2g-m1a + m1 a m2 a + m1 a = m2g (m2 a + m1 a = m2g(m2 + m1)^-1 = a + a = (m2 + m1)^-1 m2g a + a = (m2 + m1)^-1 m2g 2a = (m2 + m1)^-1 m2g (2a = (m2 + m1)^-1 m2g)2^-1 = a = (2(m2 + m1))^-1 m2g a = (2(m2 ...

I was taking upwards in the y direction positve and downwards in the y dierction negetive my bad

a = m2^-1 (m1 a) -g solve for acceleartion it's the same post below they cancel out

y^2 + 6y -12 y -72 simplify

I think you mean the limits

ok how do I solve for a m1 a - m2 g = m2 a

I've asked a similar qustion earlier and now I think I have drawn the correct free body diagram and go about solving the problem. The only thing is I do not know how to solve it. a block (mass m1) on a smooth horizontal surface, connected by a thin cord that passes over a ...

Ok I guess the reason why I do not know how to do this problem is because I can not draw the proper free-body diagram... Three blocks on a frictionless horizontal surface are in contact with each other as shown in Fig. 4-5. A force F is applied to block 1 (mass m1). (a) Draw a...

is this right? ------Force F onto m1--->|m1| ---contact force onto m2--->|m2|-----contact force onto m3---->|m3|

Ok I guess I'm drawing the free body diagram wrong how do I draw it because this is not making much sense to me


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