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As shown in the Diagram, we have to find $a_1, a_2$ , Provided that, the surfaces are not friction-less, $\mu$ is the friction between the surfaces of plane and $M$ mass body, $\frac{\mu}{2}$ is the friction is the between the surfaces $M$ and $m$ bodies.

My approach was to make the free body diagram and indicating all the forces, according to me :
$(M+m)a_1=(M+m)gSin\theta - \mu(M+m)gCos\theta $
$ma_2=mgSin\theta-\frac{\mu}{2}mgCos\theta$.

But according to my teacher the answer is incorrect. He says, While considering only mass $M$ block, the equation of forces shopuld be, $Ma_1=(M+m)gSin\theta - \mu(M+m)gCos\theta - \frac{\mu}{2}mgcos\theta$

I considered the $M+m$ as a system, and so now I here omit the friction force between $M$ and $m$ because in this case it will be an internal force, and according to sir I cannot consider $M+m$ as a system because components are here in relative motion also by my equation he says the acceleration will be of centre of mass of both the blocks not the acceleration of the $M$. Please consider both the equations mine and my teacher's and explain who is correct and how. Now one more doubt is there was no relative motion between $m$ and $M$ then also, I think my equation will be same, am I correct in this also, or not ?

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closed as off-topic by user10851, user36790, John Rennie, Qmechanic Aug 31 '16 at 13:08

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If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Hi Adesh and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Aug 30 '16 at 15:19
  • $\begingroup$ You have assumed that the two masses have the same acceleration but the diagram indicates that they have different accelerations. So mass $m$ must be slipping relative to mass $M$. $\endgroup$ – Farcher Aug 30 '16 at 15:29
  • $\begingroup$ Yes, according to me the acceleration is due to both masses, am I wrong here. $\endgroup$ – Adesh Tamrakar Aug 31 '16 at 15:13
  • $\begingroup$ You can consider $M+m$ together, but the resulting acceleration will be that of the centre of mass. Your eqn for $a_1$ will be correct if the upper block does not move relative to lower block ($a_2=a_1$); otherwise wrong. New eqn not quite correct either : last term should contain $mg$ and should be + not - . $\endgroup$ – sammy gerbil Aug 31 '16 at 15:42
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You have not considered the frictional force acting on the big mass (M) due to the sliding of the small mass (m) on it. So your first equation should be;

(M+m)a1=(M+m)gSinθ−μ(M+m)gCosθ-(μ/2)mgCosθ

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  • $\begingroup$ I have edited the question. Please Check. $\endgroup$ – Adesh Tamrakar Aug 31 '16 at 15:12
  • $\begingroup$ The answer given by your teacher is incorrect (may be you have typed it wrongly). The frictional force created by small mass on big mass should be (μ/2)mgCosθ not (μ/2)cosθ. Teacher's answer is missing "mg". Check the equation in the answer that I have written. $\endgroup$ – Kosala Aug 31 '16 at 17:21
  • $\begingroup$ Ok, that was a mistake in writing, but consider the mass taken for acceleration I took $M+m$ and he took $M$ only, in this aspect who is correct, and my reason not to consider the frictional force applies by the $m$ on $M$ is that i'm considering the $M+m$ as a system and in that case this will be the internal force. $\endgroup$ – Adesh Tamrakar Sep 1 '16 at 8:43

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