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 following (most appropriate choice):

(a) mg/4 < F < mg/2	(i) Cube will move up.
(b) F > mg/2	(ii) Cube will not exhibit motion.
(c) F > mg	(iii) Cube will begin to rotate and slip at A.
(d) F = mg/4	(iv) Normal reaction effectively at a/3 from A, no motion.

Answer 1

(a) mg/4 < F < mg/2	(ii) Cube will not exhibit motion.
(b) F > mg/2	(iii) Cube will begin to rotate and slip at A.
(c) F > mg	(i) Cube will move up.
(d) F = mg/4	(iv) Normal reaction effectively at a/3 from A, no motion.

Explanation:

Consider the below diagram

Moment of the force F about point A, τ₁ = F × a .....(anti-clockwise)

Moment of weight mg of the cube about point A.

τ₂ = `mg xx a/2` .....(clockwise)

Cube will not exhibit motion, If τ₁ = τ₂  ......(∵ In this case, both the torque will cancel the effect of each other)

∴ F × a = `mg xx a/2`

⇒ F = `(mg)/2`

Cube will rotate only when, τ₁ > τ₂ 

⇒ F × a > `mg xx a/2`

⇒ `F > (mg)/2`

Let the normal reaction is acting at `a/3` from point A, then

`mg xx a/3 = F xx a` or `F = (mg)/3`  .......(For no motion)

When F = `(mg)/4` which is less than `(mg)/3`,   .....`(F < (mg)/3)`

There will be no motion.