Question

Inside a lift moving vertically downwards with an acceleration $g$, two blocks of mass $m$ and $2m$ are placed as shown in the figure. The top block is given a sharp hit at an angle $\theta$ so as to move it. What is the value of $\theta$ for which both blocks move together without sliding on each other?

• A. $\theta <{53}^{\circ }$
• B. $\theta <{37}^{\circ }$
• C. $\theta >{53}^{\circ }$
• D. $\theta >{37}^{\circ }$

Answer 1

Answer: (C)

$\theta >{53}^{\circ }$

Lets calculate the effective $g$ first

$g_{eff}=$ downward acceleration of the lift - gravitational acceleration

$⟹g_{eff}=g-g=0$

Horizontal and vertical components of the applied force (F)

$F_H=Fcos\theta$

$F_V=Fsin\theta$

Lets assume that both blocks are moving together with velocity V.

Apply momentum conservation in the positive x direction on the system

$F_{H}dt=mV+\left(2m\right)V$

$F_{H}dt=3mV$. . . . .(1)

Now apply momentum conservation in the positive x direction 'm'

$F_{H}dt-fdt=mV$. . . . . (2); where $f$ is friction force on 'm'

By solving equation 1 and equation 2

$fdt=2mV$

Contact force between m and am $N=Fsin\theta$. . . . . . (3)

To move both the blocks together

$\mu Ndt>2mV$

$\mu Fsin\theta dt>2\left(\frac{Fcos\theta dt}{3}\right)$

$tan\theta >\frac{2}{3\mu }$

$⟹\theta >{53}^0$