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Uniform cube of mass m and side a is placed in a frictionless horizontal surface. A vertical force F is applied to the edge as shown in the figure.

Match the following

a) $\frac{mg}{4}<F<\frac{mg}{2}$ i) cube will move up

b) $F>\frac{mg}{2}$ ii) cube will not exhibit motion

c) $F>mg$ iii) cube will begin to rotate and slip at A

d) $F=\frac{mg}{4}$ iv)normal reaction effectively at $\frac{a}{3}$ from A, no motion

Answers (1)

(a) – (ii), (b) – (iii), (c) – (i), (d) – (iv)

Let $\tau _{z}$ be the moment of force due to F at A and $\tau '_{z}$ be the moment of force due to mg at A

$\tau _{z}$ is anti-clockwise whereas is $\tau '_{z}$ clockwise

$\tau '_{z}=\frac{mga}{2}$ and $\tau _{z}=ax\; \vec{F}$

$a\times\; \vec{F}=\frac{mga}{2}$

$\vec{F}=\frac{mg}{2}$

Cube will rotate only if $\tau _{z}>\tau '_{z}$

$\frac{Fa}{2}>\frac{mg}{2}$

$F>\frac{mg}{2}$

If normal reaction force acts effectively at $\frac{a}{3}$ from A

$\tau '_{z}=mg \; \frac{a}{3}$

$\tau _{z}=F.a$

$\tau _{z}=\tau '_{z}$

implies

$F=\frac{mg}{3}>\frac{mg}{4}$

So due to $F = \frac{mg}{4}$ block will not move

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infoexpert23

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