Physics Help Forum friction force

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 Jun 11th 2017, 07:55 AM #1 Junior Member   Join Date: May 2017 Location: Sri Lanka Posts: 19 friction force The block m is on block M. An external force apply to M. If we think m=3Kg and M=6Kg ,between M and floor there is no friction and between m and M μ=0.4. What is the maximum force with which the lower block can be pulled horizontally so that the two blocks move together without slipping? The friction between m and M: μR=0.4*30=12N(F1) When we apply Fmax external force for M then m forms 12N of friction force on M for the opposite direction of Fmax force. Then M forms 12N on m as the reaction of the friction. Then m takes the necessary acceleration from that force. F1=ma 12=3a a=4ms-2 Then the M acceleration also should be 4ms-2 Fmax-F1=Ma Fmax-12=6*4 Fmax=36N According to that 36N is the maximum force with which the lower block can be pulled horizontally so that the two blocks move together without slipping. then if we form a force less that 36N the blocks moves together without slipping. But now see this. If we add 30N for M as an external force, the M tries to move, as a result m add the friction force for the opposite. The friction between m and M: μR=0.4*30=12N(F1) Then the acceleration of M: F=Ma 30-12=6a 18=6a a=3ms-2 As the 12 N for M, M forms the reaction for m,12N >this direction Acceleration of m:12=ma 12=3a a=4ms-2 I am confuse that why there is different acceleration for each block. Can you show where am I wrong.
Jun 11th 2017, 11:41 AM   #2
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Join Date: Aug 2008
Posts: 113
 The block m is on block M. An external force apply to M. If we think m=3Kg and M=6Kg ,between M and floor there is no friction and between m and M μ=0.4. What is the maximum force with which the lower block can be pulled horizontally so that the two blocks move together without slipping?
refer to the attached diagram ...

$f \le \mu \cdot mg$

proceeding by treating it as an unknown value ...

$\color{blue}{f} = ma_1 \implies a_1 = \dfrac{\color{blue}{f}}{m}$

$F-\color{red}{f} = Ma_2 \implies \dfrac{F-\color{red}{f}}{M} = a_2$

no slippage, so $a_2 = a_1$

$\dfrac{F-\color{red}{f}}{M} = \dfrac{\color{blue}{f}}{m}$

Newton's 3rd Law says $\color{blue}{f} = \color{red}{f}$ ...

solving for $f$

$f = \dfrac{mF}{M+m} < \mu \cdot mg \implies F < \mu g(M+m) = 36 \, N$
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 Tags force, friction, newton

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