Question

In the arrangement shown, by what acceleration the boy must go up so that 100 kg block remains stationary on the wedge? The wedge is fixed and friction is absent everywhere $(g=10m{s}^{-2})(sin{53}^0=\frac{4}{5})$

Answer 1

Given,

mass of block $M_b=100$kg

Mass of mass $M_m=50$kg

$\theta ={53}^o$

FBD of mass on inclined plane.

There is no acceleration of mass

So $F_R=0$

$T-m_{b}g\sin \theta =0$

$T=m_{o}g\sin \theta$

$=100\times 10\times \sin 53$

$T=100\times 10\times \frac{4}{5}$

$T=800N$

Now, FBD of man, having acceleration a(Let)

$T=m_{m}g=m_{m}a$

$800-50\times 10=50\times a$

$300=50a$

$a=\frac{300}{50}=6$

$a=6m/s^2$

Acceleration of boy would be $6m/s^2$.