Ask your own question, for FREE!
Physics 42 Online
OpenStudy (anonymous):

What horizontal force F must be constantly applied to M so that M1 and M2 do not move relative to M? neglet friction http://imageshack.us/photo/my-images/823/immaginelq.jpg/

OpenStudy (anonymous):

Since M2 is not moving, the tension of the string is T=M2∗g. Plugging this into the euation T=M1∗a we find that the acceleration of M1 is a=(M2/M1)∗g Since M1 is at rest with respect to M, M have the same accelleration a So, F=(M+M2)∗a. or F=(M2/M1)∗(M+M2)∗g that's what i found. But the problem solution states that F=(M2/M1)∗(M+M1+M2)∗g which i don't see why it should be like that. Intuitively if i push on the block M, i push both M and M2, that's why i found (M+M2), but i don't push on M1, since there is no friction. It's movements are due to the force of M2 pulling down.

OpenStudy (anonymous):

but the question states that the force should be applied on M, so Newton's law should be applied on M, as far as I know, Thus, \[F=(M.M _{2}.g)/M _{1}\]

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Latest Questions
REYSHAWN23134: Hello guys. How are you all
1 hour ago 0 Replies 0 Medals
Gucchi: ````````````` english help
1 hour ago 5 Replies 0 Medals
Gucchi: english help `
2 hours ago 11 Replies 1 Medal
Gucchi: english help
3 hours ago 16 Replies 3 Medals
luhivqqcherry: guys so what will happen if I don't get my iron up fast enough ? .
3 hours ago 15 Replies 2 Medals
luhivqqcherry: how do I fix low iron ?
3 hours ago 64 Replies 0 Medals
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!