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 match the following (most appropriate choice):
| a. | $\frac{\text{mg}}{4}<\text{F}<\frac{\text{mg}}{2}$ | i. | Cube will move up. |
| b. | $\text{F}>\frac{\text{mg}}{2}$ | ii. | Cube will not exhibit motion. |
| c. | $\text{F}>\text{mg}$ | iii. | Cube will begin to rotate and slip at A. |
| d. | $\text{F}=\frac{\text{mg}}{4}$ | iv. | Normal reaction effectively at $\frac{\text{a}}{3}$ |
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| a. | $\frac{\text{mg}}{4}<\text{F}<\frac{\text{mg}}{2}$ | i. | Cube will not exhibit motion. |
| b. | $\text{F}>\frac{\text{mg}}{2}$ | ii. | Cube will begin to rotate and slip at A. |
| c. | $\text{F}>\text{mg}$ | iii. | Cube will move up. |
| d. | $\text{F}=\frac{\text{mg}}{4}$ | iv. | Normal reaction effectively at $\frac{\text{a}}{3}$ |
Let us first consider the below diagram torque or moment of the force F about point A is given by I = aF
This is anticlockwise.
Torque of weight mg about A,
$=\frac{\text{q}}{2}$
This is clockwise.
N is acting at point A. So, torque due to normal reaction.
If $\tau_1=\tau_2,$ cube will not exhibit motion
$(\because$ In this case, both the torque will cancle the effect of each other$)$
$\therefore\ \text{F}\times\text{a}=\text{mg}\times\frac{\text{a}}{2}$
$\Rightarrow\text{F}=\frac{\text{mg}}{2}$
The cube will rotate only when $\tau_1>\tau_2$
$\Rightarrow\text{F}\times\text{a}>\text{mg}\times\frac{\text{a}}{2}$
$\Rightarrow\text{F}>\frac{\text{mg}}{2}$
Let normal rection is acting at $\frac{\text{a}}{3}$ from point A as, shown in diagram below.
$\text{N}\times\frac{\text{a}}{3}+\text{Fa}=\text{mg}\times\frac{\text{a}}{2}$ (Fot no rotation)
$\Rightarrow(\text{mg}-\text{F})\frac{\text{a}}{3}+\text{Fa}=\text{mg}\times\frac{\text{a}}{2}$ $\big[\because\ \text{N}+\text{F}=\text{mg}\big]$
$\Rightarrow\frac{2}{3}\text{F}=\frac{1}{6}\text{mg}$
$\Rightarrow\text{F}=\frac{\text{mg}}{4}$
about A will be zero.
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