Question

Three identical spheres, each of mass m, are kept in contact inside a box as shown in figure. If box is moving vertically upward with an acceleration g/4, then (neglect friction) (Sphere A is not touching the box)

• A. Normal force applied by the spheres on the bottom of the box is $\frac{9}{4}$ mg.
• B. Normal force applied by the spheres on the bottom of the box is $\frac{15}{4}$ mg.
• C. Normal force between spheres A and B is $2\sqrt{3}$ mg
• D. Normal force between spheres A and B is $\frac{5mg}{4\sqrt{3}}$.

Answer 1

Answer: (B), (D)

The correct options are
B Normal force applied by the spheres on the bottom of the box is $\frac{15}{4}$ mg.
D Normal force between spheres A and B is $\frac{5mg}{4\sqrt{3}}$.

Given,

Each sphere is of mass, $mkg$

Upward acceleration, $a=\frac{g}{4}$

Total mass of box contains three spheres, $3mkg$

(i) Net normal force on base of box, $N=3mg+3ma=3mg+3m\frac{g}{4}=\frac{15mg}{4}$

(ii) Horizontal component of normal reaction at center of A,

$N_1\sin {30}^o=N_2\sin {30}^o$

$N_1=N_2=N$

Vertical component of normal reaction at center of A,

$N_1\cos {30}^o+N_2\cos {30}^o=m(g+a)$

$2N\cos {30}^o=m(g+\frac{g}{4})$

$N=\frac{5mg}{4\sqrt{3}}$

Hence, net normal reaction at base of box is $\frac{15mg}{4}$, and normal reaction on A is $\frac{5mg}{4\sqrt{3}}$ .