Question

A uniform cube of mass M and side a is placed on a frictionless horizontal surface. A vertical force F is applied to the edge as shown in figure. Match the Column I with Column II.

	Column I		Column II
(A)	$$\frac{Mg}{4} < F < \frac{Mg}{2}$$	(p)	Cube will move up.
(B)	$$F > \frac{Mg}{2}$$	(q)	Cube will not exhibit motion.
(C)	F > Mg	(r)	Cube will begin to rotate and slip at A.
(D)	$$F = \frac{Mg}{4}$$	(s)	Normal reaction effectively at a/3 from A, no motion.

• A. A - p, B - q, C - s, D – r
• B. A - r, B - s, C - q, D – p
• C. A - q, B - r, C - p, D – s
• D. A - s, B - p, C - r, D – q

Answer 1

Answer: (C)

A - q, B - r, C - p, D – s

Answer 2

From figure, moment of force F about A,

$\tau_{1} = F \times a,$ anticlockwise.

Moment of weight Mg of cube about A.

$\tau_{2} = Mg \times \frac{a}{2},$ Clockwise.

The cube will not exhibit any motions,

If $\tau_{1} = \tau_{2}$

Or $F \times a = Mg \times \frac{a}{2}orF > \frac{Mg}{2}$

The cube will rotate only, when $\tau_{1} > \tau_{2}$

$F \times a > Mg\frac{a}{2}orF > \frac{Mg}{2}$

If we assume that normal reactions is effective at a/3 from A, then block would turn if

$Mg \times \frac{a}{3} = F \times aorf = \frac{Mg}{3}.$

When $F = \frac{Mg}{4} < \frac{Mg}{3},$ there will be no motion.

Hence, we conclude $A - q;B - r;C - p;D - s.$