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