Question

For the given system, pulley and string are ideal, and friction is absent.
a) Relate the speed of the wedge and the block, if the wedge moves with speed v towards the left
b) Find the acceleration of the wedge​

Answer 1

The block displaces downward along the inclined plane wrt the wedge. Let this displacement be $\small\text{ \bf{x_{bw}.} }$

Then the slanted portion of string increases and horizontal portion decreases. As a result the wedge moves leftward with a displacement, say $\small\bf{x_w.}$

Thus net displacement of the block is the resultant of $\small\text{ \bf{x_{bw}} }$ and $\small\bf{x_w,}$ i.e., $\small\text{ \bf{x_b=x_{bw}+x_w.} }$ By differentiating wrt time we can obtain $\small\text{ \bf{v_b=v_{bw}+v_w} }$ and $\small\text{ \bf{a_b=a_{bw}+a_w.} }$

Here $\small\bf{x_w}$ makes angle (180° - θ) with $\small\text{ \bf{x_{bw}.} }$ If $\small\bf{x_b}$ makes angle α with $\small\text{ \bf{x_{bw}} }$ then,

• component of $\small\bf{x_b}$ normal to inclined plane:- $\small\sf{x_b\sin\alpha=x_w\sin(180^o-\theta)=x_w\sin\theta\quad\dots(1)}$

• component of $\small\bf{x_b}$ along inclined plane:-  $\small\text{ \sf{x_b\cos\alpha=x_{bw}+x_w\cos(180^o-\theta)=x_{bw}-x_w\cos\theta\quad\dots(2)} }$

[Recall that if R is resultant of two vectors P and Q, both of which make an angle θ with each other, and R makes an angle α with P then R·sinα = Q·sinθ and R·cosα = P + Q·cosθ (Fig. 4)]

We know the net work done by the tension equals zero.

Fig. 1 shows the string in which net work done by tension at points A, B, C and D equals zero. A is a fixed point (no displacement), B and C are points on pulley which move along wedge, and D is point of contact of block and string.

Work done,

• at A = 0.

• at B $\small\sf{=Tx_w.}$

• at C $\small\sf{=Tx_w\cos(180^o-\theta)=-Tx_w\cos\theta.}$

• at D $\small\sf{=Tx_b\cos(180^o-\alpha)=-Tx_b\cos\alpha.}$

Now,

$\displaystyle\small\longrightarrow\sf{Tx_w-Tx_w\cos\theta-Tx_b\cos\alpha=0}$

$\displaystyle\small\longrightarrow\sf{x_w-x_w\cos\theta=x_b\cos\alpha}$

From (2),

$\displaystyle\small\text{ \longrightarrow\sf{x_w-x_w\cos\theta=x_{bw}-x_w\cos\theta} }$

$\displaystyle\small\text{ \longrightarrow\sf{x_{bw}=x_w\quad\dots(3)} }$

Now magnitude of $\small\bf{x_w}$ and $\small\text{ \bf{x_{bw}} }$ are same and angle between them is (180° - θ), so magnitude of $\small\text{ \bf{x_b=x_{bw}+x_w} }$ is,

$\small\text{ \longrightarrow\sf{x_b=\sqrt{(x_{bw})^2+(x_w)^2+2x_{bw}x_w\cos(180^o-\theta)}} }$

$\small\text{ \longrightarrow\sf{x_b=\sqrt{(x_w)^2+(x_w)^2-2(x_w)^2\cos\theta}} }$

$\small\text{ \longrightarrow\sf{x_b=\sqrt{2(x_w)^2(1-\cos\theta)}} }$

$\small\text{ \longrightarrow\sf{x_b=\sqrt{2(x_w)^2\cdot2\sin^2\left(\dfrac{\theta}{2}\right)}} }$

$\small\text{ \longrightarrow\sf{x_b=2x_w\sin\left(\dfrac{\theta}{2}\right)} }$

Differentiating wrt time,

$\small\text{ \longrightarrow\sf{v_b=2v_w\sin\left(\dfrac{\theta}{2}\right)} }$

Here $\small\sf{v_w=v.}$

$\small\text{ \longrightarrow\sf{\underline{\underline{v_b=2v\sin\left(\dfrac{\theta}{2}\right)}}} }$

Consider FBD of block of mass m (Fig. 2). It is acted upon tension in the string T (along inclined plane), reaction with wedge $\small\bf{R_b}$ (normal to inclined plane) and weight mg downwards which is resolved in directions along and normal to inclined plane.

On considering forces normal to inclined plane,

$\small\longrightarrow\sf{mg\cos\theta-R_b=ma_b\sin\alpha}$

From (1) we get $\small\sf{a_b\sin\alpha=a_w\sin\theta.}$

$\small\longrightarrow\sf{R_b=mg\cos\theta-ma_w\sin\theta}$

On considering forces along inclined plane,

$\small\longrightarrow\sf{mg\sin\theta-T=ma_b\cos\alpha}$

From (2) we get $\small\text{ \sf{a_b\cos\alpha=a_{bw}-a_w\cos\theta.} }$

$\small\text{ \longrightarrow\sf{T=mg\sin\theta-m(a_{bw}-a_w\cos\theta)} }$

From (3) we get $\displaystyle\small\text{ \sf{a_{bw}=a_w.} }$

$\small\longrightarrow\sf{T=mg\sin\theta-ma_w(1-\cos\theta)}$

Consider FBD of wedge of mass M (Fig. 3). It is acted upon two tensions (horizontal and along inclined plane), reaction with block $\small\bf{R_b}$ (normal to inclined plane), weight Mg downward and reaction with floor $\small\bf{R_f}$ upward.

Considering horizontal forces,

$\small\longrightarrow\sf{T(1-\cos\theta)+R_b\sin\theta=Ma_w}$

Putting values of T and $\small\sf{R_b,}$

$\small\longrightarrow\sf{[mg\sin\theta-ma_w(1-\cos\theta)](1-\cos\theta)+(mg\cos\theta-ma_w\sin\theta)\sin\theta=Ma_w}$

Further solving for $\small\sf{a_w,}$ we get,

$\small\text{ \longrightarrow\sf{\underline{\underline{a_w=\dfrac{mg\sin\theta}{M+2m(1-\cos\theta)}}}} }$