Ncert Exercises & Solutions

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 the figure. Match the following (most appropriate choice).

	List- I		List- II
(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.

1.	a - (i), b - (iv), c - (ii), d - (iii)
2.	a - (ii), b - (iii), c - (i), d - (iv)
3.	a - (iii), b - (i), c - (ii), d - (iv)
4.	a - (i), b - (ii), c - (iv), d - (iii)

Hint: Recall the concept of toppling.

Step 1: Find torque due to

F

$F$ and

m g

$mg$ about point

A .

$A.$

Consider the given diagram, the moment of the force $F$ about point $A,$ $τ _1 = r\times F$ (anti-clockwise)
The moment of weight $mg$ of the cube about point $A.$

$τ _2 = mg × \frac{a}{ 2}$ ( clockwise )

Step 2: Equate the torques for no motion and find the value of $F.$

Cube will not exhibit motion if

τ_{1} = τ_{2}

$\tau_1=\tau_2$

( $∵$ $∵$ In this case, both the torque will cancel the effect of each other)
$F × a = mg × \frac{a}{ 2} ⇒ F = \frac{mg}{ 2}$

Step 3: Find the value of $F$ for which the cube will rotate.

Cube will rotate only when, $𝜏 _1 > 𝜏 _2$
$⇒ F × a > mg × \frac{a} {2} ⇒ F > \frac{mg} {2}$

Step 4: If the normal reaction is acting at $a/3$ from point $A$ and for no motion find $F$ and interpret.
Let the normal reaction is acting at $a/3$ from point $A,$ then
$mg × \frac{a}{ 3} = F × a~\text{ or }~ F = \frac{mg}{ 3}$ (For no motion)

When $F = \frac{mg}{ 4},$ which is less than $mg/3.$ $\left( F < \frac{mg}{ 3}\right )$
there will be no motion.

Hence, option (2) is the correct answer.

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