Question

A block of mass m = 5 kg is on the smooth inclined face of the wedge of mass M  = 10 kg , the inclination to the horizontal being θ = 37°.The wedge is resting on a smooth horizontal plane. Assuming the pulley P to be smooth and the string is light and inextensible. Find the acceleration of M, when M and m are always in contact.

[Given that $Sin {37}^o=\frac{3}{5}$ ]

• (A) 1.5 m/ sec²
• (B) 2.8 m/ sec²
• (C) 3.2 m/ sec²
• (D) None of these

Answer 1

The Correct Option is D

Explanation:
Given:Mass of the block on the inclined plane, m = 5 kgMass of the smooth inclined wedge, M = 10 kgAngle of inclined plane with horizontal, θ = 37°We have to find the acceleration of M, when M and m are always in contact.Given figure is reproduced below:

When the block slides down, the string between pulley and the wall shortens by the same amount as the distance moved by the block down the plane with respect to the wedge. Let, am = aM = awhere,am: acceleration of the block down the planeaM:  acceleration of the wedge horizontally towards the walla: acceleration of the block horizontally towards the wall.FBD of the smaller block is shown below:

Force equation of smaller block in the vertical direction,$mg-R\cos \theta -T\sin \theta =m\left(a\sin \theta \right)$.........(i)force equation in the horizontal direction,$T\cos \theta -R\sin \theta =m(a-a\cos \theta )$.......(ii)FBD of the wedge is shown below:

Force equation of the wedge in the horizontal direction$T+R\sin \theta -T\cos \theta =Ma$...... ..(iii)Multiplying eq. (i) by sinθ and eq. (ii) by cosθ and then subtracting the two equations, we get   $mg\sin \theta -T=ma(1-\cos \theta )$$\Rightarrow T=mg\sin \theta -ma(1-\cos \theta )$Adding eqs. (ii) and (iii), we get$T=Ma+ma(1-\cos \theta )$From eqs. (iv) and (v), we have$mg\sin \theta =2a(1-\cos \theta )+Ma$$\Rightarrow mg\sin \theta =a[M+2m(1-\cos \theta )]$$\Rightarrow a=\frac{mg\sin \theta }{M+2m(1-\cos \theta )}$Substituting the values, we get$a=\frac{5\times 10\times \sin {37}^{\circ }}{10+2\times 5(1-\cos {37}^{\circ })}$$\Rightarrow a=\frac{50\times (3/5)}{10+10(1-4/5)}$$\Rightarrow a=\frac{30}{12}=2.5\mathrm{m}/{\sec }^2$Hence, the correct option is (D).