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 and mark the correct choice from the given codes.
Column-I Column-II
(i) $\frac{Mg}{4} < F < \frac{Mg}{2}$ (p) Cube will move up
(ii) $F > \frac{Mg}{2}$ (q) Cube will not exhibit motion
(iii) F > Mg r) Cube will begin to rotate and slip at,4.
(iv) $F = \frac{Mg}{4}$ (s) Normal reaction effectively at $\frac{a}{3}$ from A, no motion
Codes

Moderate

Solution

From figure, Moment of force F about A,
$τ_{1} = F \times a$ , anticlockwise.
Moment of weight Mg of cube about A,
$τ_{2} = Mg \times \frac{a}{2}$ , clockwise
The cube will not exhibit any motion, if $τ_{1} = τ_{2}$
or F x a = $Mg \times \frac{a}{2} or F > \frac{Mg}{2}$
The cube will rotate only, when $τ_{1} > τ_{2}$
$F \times a > Mg \frac{a}{2} or F > \frac{Mg}{2}$
If we assume that normal reaction is effective at $\frac{a}{3}$ from A, then block would turn if
$Mg \times \frac{a}{3} = F \times a or F = \frac{Mg}{3}$
When F = $\frac{Mg}{4} < \frac{Mg}{3},$ there will be no motion.
Hence, we conclude (i) - q: (ii) - r; (iii) -p; (iv) - s.

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