Question: Find the mass M of the hanging block in the following figure which will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulleys are light.


What I did: Considering that M-mass block is moving downwards, I assumed that the string would transmit force $Mg$ to the inclined plane. Thus the inclined plane would move forward with acceleration $\frac{Mg}{M^-}$ towards the right, while the m-mass block would move backward, in a horizontal direction, with the same acceleration. Calculating components, the acceleration up the incline would be $\frac{Mg}{M^-}\cos \alpha$. The acceleration of the block on its own would be $g\sin \alpha$. Equating the two, I got $M = M^-\tan \alpha$

New approach (considering horizontal component of normal force exerted by small block) Considering the fact that the normal force exerted by the wedge on the block to be $mg\cos \alpha$, the block exerts an equal force on the wedge. The horizontal component of this normal force will be $mg\cos \alpha\sin \alpha$. Thus the acceleration of the wedge will be $$\frac{Mg - m\cos \alpha\sin \alpha}{M^-+m}$$ This should be equal to the downward acceleration of the small block: $g\sin \alpha$. Equation the two and solving for $M$, I got: $$M = \sin \alpha(M^- + m) + m\cos \alpha\sin \alpha$$ Still not the correct answer: $\frac{M^- + m}{\cot \alpha -1}$


closed as off-topic by John Rennie, Ali, Jim, Kyle Kanos, Kyle Oman Jul 11 '14 at 14:27

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  • 2
    $\begingroup$ You've been asked to do this by Newtonian methods? You poor thing. Free-body diagram all three mobile parts; note the 3rd-law pairs (including the tension); set up the system of equations; wail away at the math; collapse in a sobbing heap; ???; profit! $\endgroup$ – dmckee Jul 4 '14 at 20:32
  • $\begingroup$ That said, you should have edited this into you previous question. $\endgroup$ – dmckee Jul 4 '14 at 20:43
  • $\begingroup$ @dmckee Agreed. Well, if not by Newtonian methods, would you mind telling me another method? :D $\endgroup$ – Gummy bears Jul 5 '14 at 5:58
  • $\begingroup$ It might be easier by Lagrangian methods, but if you have to ask you don't have that tool yet. It is usually introduced in a upper-division mechanics course. $\endgroup$ – dmckee Jul 5 '14 at 14:00
  • $\begingroup$ Actually, I notice that you have been asked for a very particular thing rather than a full solution. That is manageable with Newtonian methods, but you have got to have the right frame of mind to do it quickly. Start by imagining the string is attached to a infinitely powerful moter and ask how fast it needs to accelerate the system to achieve the same goal then generalize. $\endgroup$ – dmckee Jul 5 '14 at 14:07

For hanging block $$Mg-T=Ma$$ For block on platform, consider the two$(M^-,m)$ as a system $$T=(M^-+m)a\\\implies a=\frac{Mg}{M^-+M+m}$$ Now consider the block $M^-$ at rest(change your reference frame), So now a pseudo-force $ma$, should be acting on $m$ leftwards horizontally.
Taking components parallel to plane: $$ma\cos\alpha=mg\sin\alpha\implies a=g\tan\alpha\\ \implies \frac{Mg}{M^-+M+m}=g\tan\alpha\implies M=\frac{M^-+m}{\cot\alpha-1}$$


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