JEE Class main Answered

Let us consider the system consists of triangular wedge of mass M and sliding block of mass m .

In absence of external force , accleration of center of mass of system consists of wedge and block is zero .

As shown in figure , block moves with acceleration ( g sinθ ) parallel to the inclide surface of wedge.

Vertical resolved component of block is  -( g sinθ )sinθ = - g sin²θ 

When block of mass sliding, Let a be the acceleration of wedge in the vertical direction .

Since acceleration of center of mass is zero , we get

M a - m g sin² θ = 0

Hence acceleration of wedge is 

a = ( m / M ) g sin² θ ..........................(1)

Free body diagram of wedge is shown below

Mg is weight of wedge and mg is weight of block . N is the normal force .

If wedge has acceleration a in vertical direction , by Newton's second law , net force equals the product of mass

and acceleration .

N - (M+m)g = M a

N = (M+m)g + M a

If we substitute acceleration from eqn.(1) , then above expression is written as

N = ( M+m)g + [ M  × ( m / M ) g sin² θ ] =  [ M + m ( 1 + sin²θ ) ] g

Balance measures the normal force as weight of wedge .

Hence , balance reading when block of mass m sliding on wedge is [ M + m ( 1 + sin²θ ) ] g