Question

A block of mass 'm' is arranged on the wedge as shown in figure. The wedge angle is θ. If the masses of pulley and thread are negligible and friction is absent then find the acceleration of the wedge.

Note :- diagram is in attachment.​

Answer 1

The block would accelerate downward along the inclined plane wrt the wedge. Let this acceleration be $\small\text{ \bf{a_{bw}.} }$

Then the slanted portion of string increases and horizontal portion decreases. As a result the wedge would move rightwards with an acceleration, say $\small\bf{a_w.}$

Thus the net acceleration of the block is the resultant of $\small\text{ \bf{a_{bw}} }$ and $\small\bf{a_w,}$ i.e., $\small\text{ \bf{a_b=a_{bw}+a_w.} }$ The accelerations $\small\text{ \bf{a_{bw}} }$ and $\small\bf{a_w}$ make angle θ with each other.

If $\small\bf{a_b}$ makes an angle α with $\small\text{ \bf{a_{bw}} }$ then,

• component of $\small\bf{a_b}$ normal to inclined plane:- $\small\sf{a_b\sin\alpha=a_w\sin\theta\quad\dots(1)}$

• component of $\small\bf{a_b}$ along inclined plane:-  $\small\text{ \sf{a_b\cos\alpha=a_{bw}+a_w\cos\theta\quad\dots(2)} }$

[Recall that if R is the 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.

$\displaystyle\small\sum\bf{T\cdot x}\,\sf{=0}$

which on twice differentiation gives,

$\displaystyle\small\sum\bf{T\cdot a}\,\sf{=0}$

Fig. 1 shows the string in which net work done by tension at points A, B, C and D equals zero.

Second derivative of work done,

• at A = 0 because A is fixed point.

• at B $\small\sf{=Ta_w.}$

• at C $\small\sf{=Ta_w\cos\theta.}$

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

Now,

$\displaystyle\small\longrightarrow\sum\bf{T\cdot a}\,\sf{=0}$

$\displaystyle\small\longrightarrow\sf{Ta_w+Ta_w\cos\theta-Ta_b\cos\alpha=0}$

$\displaystyle\small\longrightarrow\sf{a_w+a_w\cos\theta=a_b\cos\alpha}$

From (2),

$\displaystyle\small\text{ \longrightarrow\sf{a_w+a_w\cos\theta=a_{bw}+a_w\cos\theta} }$

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

Consider FBD of the block (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 the forces normal to inclined plane,

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

From (1),

$\small\longrightarrow\sf{R_b=mg\cos\theta+ma_w\sin\theta\quad\dots(4)}$

On considering the forces along inclined plane,

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

From (2),

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

From (3),

$\small\longrightarrow\sf{T=mg\sin\theta-ma_w(1+\cos\theta)\quad\dots(5)}$

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

On considering horizontal forces,

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

Putting values of T and $\small\sf{R_b}$ from (4) and (5),

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

$\small\longrightarrow\sf{mg\sin\theta(1+\cos\theta)-ma_w(1+\cos\theta)^2-mg\sin\theta\cos\theta-ma_w\sin^2\theta=Ma_w}$

$\small\longrightarrow\sf{mg\sin\theta-ma_w[(1+\cos\theta)^2+\sin^2\theta]=Ma_w}$

$\small\longrightarrow\sf{mg\sin\theta-2ma_w(1+\cos\theta)=Ma_w}$

$\small\longrightarrow\sf{mg\sin\theta=a_w[M+2m(1+\cos\theta)]}$

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

Answer 2

Solution

It is obvious that when block m moves downward along the incline of wedge, the wedge moves to the right . As the length of thread is constant, the distance traversed by wedge along the incline is equal to distance traversed by wedge to the right . This implies that acceleration of wedge to the right is equal to downward acceleration of wedge .

Let a be the acceleration of wedge to the right .

Then the force acting on the block m are

(i) weight mg acting vertically downward

(ii) normal reaction `R_(1)`

(iii)tension T (up the incline )

(iv)Fictitious force ma to the left .

The free body diagram of mass m is shoen in figure .

For motion of block .m. on inclined plane

`mg sin theta +ma cos theta -T = ma ` ....(1)

As mass .m. does not breaks off the inclined plane , therefore for forces on .m. normal to inclined plane

`R_(1) +ma sin theta =mg cos theta ` ....(2)

The forces acting on the wedge are

(i) weight Mg downward

(ii) normal reactionof ground on wedge =`R_(2)`

(iii) normal reaction of block on wedge =`R_(1)`

(iv) tension (T,T) in string .

The free body diagram of wedge is shown in figure .

For motion of wedge in horizontal direction

` R_(1) sin theta +T-T cos theta = mg` ....(4)

From (1),

`T=mg sin theta + ma cos theta -ma` ....(5)

From (2), `R_(1) =mg cos theta - ma sin theta ` ....(6)

substituting these values in (3) , we get

`(mg cos theta - ma sin theta ) sin theta +(mg sin theta +ma cos theta - ma ) ( 1-cos theta )=Ma`

or `{M+m sin ^(2) theta +m(1-cos theta)^(2)}a`

`=mg cos theta sin theta +mg sin theta (1-cos theta )`

`a= (mg sin theta )/(M+m sin^(2) theta + m(1-2 cos theta+cos^(2) theta))=(mg sin theta)/ (M+m(sin^(2) theta +1 -2 cos theta+cos^(2) theta))`

`:. a=(mg sin theta)/(M+2m(1-cos theta))`