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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.
 one year ago
 one year ago
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.
 one year ago
 one year ago

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maheshmeghwal9Best ResponseYou've already chosen the best response.0
Answer is \[\LARGE{\color{blue}{a=\frac{mg \sin \alpha }{M+2m(1\cos \alpha)}.}}\]\[\color{red}{\text{But how to get this answer?}}\]
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.0
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 xcomponent, second line ycomponent) 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.
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.0
* altitude above ground is actually x sinα + H where H is height of the pulley. H disappears in the derivation.
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.0
I also have a solution based on forces and N's laws, although I prefer the more global energy approach.
 one year ago

yrelhan4Best ResponseYou've already chosen the best response.0
also sir, what would be the displacement of the wedge? this is a common question, but i always end up getting stuck. @VincentLyon.Fr
 one year ago
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