In the arrangement shown in Fig. 1.24 the masses m of the bar and M of the wedge, as well as the wedge angle alpha , are known.The masses of the pulley and the thread are negligible. The friction is absent. Find the acceleration of the wedge M.

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Answer is \[\LARGE{\color{blue}{a=\frac{mg \sin \alpha }{M+2m(1-\cos \alpha)}.}}\]\[\color{red}{\text{But how to get this answer?}}\]

- Vincent-Lyon.Fr

Let x, v and a be displacement, velocity and acceleration of wedge.
All these values are positive (to the right).
Velocity of m wrt wedge is (first line x-component, second line y-component)
-v cosα
-v sinα
Velocity of wedge wrt ground is:
v
0
Adding them up gives velocity of m wrt ground:
v (1 – cosα)
-v sinα
KE of wedge is KE1 = ½ M v²
KE of mass is KE2 = ½ m v² [(1 - cosα)² + sin²α] = m v² (1- cosα)
Altitude of m above ground is -x sinα , as it goes x down along the inclined when wedge moves of x.
PE of mass is PE = -m g x sinα
The system is conservative.
So total energy is conserved.
Adding up all derivatives of above energies must give zero:
0 = Mva +2mva(1 - cosα) - mg v sinα
Simplifying by v and solving for a leads to the solution.

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## More answers

- Vincent-Lyon.Fr

* altitude above ground is actually -x sinα + H
where H is height of the pulley.
H disappears in the derivation.

- Vincent-Lyon.Fr

I also have a solution based on forces and N's laws, although I prefer the more global energy approach.

- yrelhan4

also sir, what would be the displacement of the wedge? this is a common question, but i always end up getting stuck. @Vincent-Lyon.Fr

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