A uniform cube of mass $m$ and side $a$ rests on a frictionless horizontal surface. Its bottom edge on one side is labelled A and serves as the possible pivot for tipping. A vertical, upward force $F$ is applied at the top edge on the far side of the cube — that is, at a horizontal distance $a$ from the pivot edge A. Match each range or value of $F$ in Column I with the correct resulting behaviour in Column II. Column I: (a) $mg/4 mg/2$; (c) $F > mg$; (d) $F = mg/4$. Column II: (i) The cube will move up (lift off). (ii) The cube will not exhibit any motion. (iii) The cube will begin to rotate and slip at A. (iv) The normal reaction acts effectively at $a/3$ from A, and there is no motion.

A uniform cube of mass m and side a rests on a frictionless horizontal surface. Its bottom edge on one side is

NCERT Exemplar · Q17

m

(a)

mg/4 < F < mg/2

;

(b)

F > mg/2

;

(c)

F > mg

;

(d)

F = mg/4

.
Column II:

(i) The cube will move up (lift off).

(ii) The cube will not exhibit any motion.

(iii) The cube will begin to rotate and slip at A.

(iv) The normal reaction acts effectively at

a/3

from A, and there is no motion.

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Concept understanding — Torque

What is Torque? The Intuition

Think about opening a door. You push near the handle, not near the hinges. Why? Because the same push is far more effective when applied farther from the hinge. That "turning effectiveness" of a force is exactly what torque measures.

If you push a door with 10 N of force right at the hinge, the door barely moves. Push with the same 10 N at the handle, and it swings open easily. The force is identical — what changed is the distance from the axis of rotation. Torque combines force and distance into one quantity that tells you how much a force will make something spin.

Note

Torque is to rotation what force is to translation. Force makes things move in a straight line; torque makes things rotate.

The Precise Definition

Torque $\vec{\tau}$ about a point (or axis) is defined as the cross product of the position vector $\vec{r}$ (from the axis to the point where force acts) and the force vector $\vec{F}$ :

\vec{\tau} = \vec{r} \times \vec{F}

The magnitude is:

\tau = rF\sin\theta

where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$ .

The direction of torque is given by the right-hand rule: curl your fingers from $\vec{r}$ toward $\vec{F}$ , and your thumb points along $\vec{\tau}$ . This direction is perpendicular to both $\vec{r}$ and $\vec{F}$ .

Tip

For maximum torque, push perpendicular to the lever arm ( $\theta = 90^\circ$ , so $\sin\theta = 1$ ). Pushing directly toward or away from the axis ( $\theta = 0^\circ$ or $180^\circ$ ) produces zero torque — that's why you can't open a door by pushing straight into it.

Why Torque Equals Rate of Change of Angular Momentum

This is the rotational analogue of Newton's second law: $\vec{F} = d\vec{p}/dt$ .

For a particle of mass $m$ at position $\vec{r}$ with velocity $\vec{v}$ , its linear momentum is $\vec{p} = m\vec{v}$ . Angular momentum about the same point is:

\vec{L} = \vec{r} \times \vec{p}

Now differentiate with respect to time:

\frac{d\vec{L}}{dt} = \frac{d\vec{r}}{dt} \times \vec{p} + \vec{r} \times \frac{d\vec{p}}{dt}

Since $d\vec{r}/dt = \vec{v}$ and $\vec{p} = m\vec{v}$ , the first term is $\vec{v} \times m\vec{v} = 0$ (cross product of parallel vectors). The second term uses $d\vec{p}/dt = \vec{F}$ , giving:

$\frac{d\vec{L}}{dt} = \vec{r} \times \vec{F} = \vec{\tau}$ …

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Only vertical forces act: the upward pull $F$ at the far edge, the weight $mg$ at the centre, and the normal reaction $N$ . Vertical equilibrium fixes $N=mg-F$ , and taking torques about the pivot edge A locates $N$ at $x=a(mg/2-F)/(mg-F)$ . As $F$ grows, $N$ moves toward A, then leaves the base (tipping), then vanishes (lift-off).

Set-up

Take torques about the bottom edge A. Distances from A: weight $mg$ acts at the centre, $a/2$ from A; the applied upward force $F$ acts at the far top edge, $a$ from A; the normal reaction $N$ acts somewhere in the base, a distance $x$ from A.

Equations

Vertical equilibrium: $N + F = mg \ \Rightarrow\ N = mg - F.$
Torque balance about A (upward forces to the right of A tend to lift the far side; the weight resists):

N\,x + F\,a - mg\,\frac{a}{2} = 0 \ \Rightarrow\ x = \frac{a\left(\tfrac{mg}{2} - F\right)}{mg - F}.

The four cases

(d) $F = mg/4$ : $x = \dfrac{a(mg/2 - mg/4)}{mg - mg/4} = \dfrac{a(mg/4)}{3mg/4} = \dfrac{a}{3}.$ The reaction sits at $a/3$ from A, inside the base → no motion → (iv). …

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Set-up

Equations

The four cases

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