Torque and Couple — Rotational Equilibrium

When a body is in equilibrium, two conditions must hold: net force is zero (translational equilibrium) and net torque is zero (rotational equilibrium). Most JEE/NEET problems on this topic test the second condition. Get torque calculations right and you walk away with full marks.

We will cover what torque really means, how to compute it for various forces, the special case of a couple, and how to apply rotational equilibrium to ladders, beams, and balanced systems. The chapter overlaps with rotational dynamics, so the formulas you learn here are reused later.

Key Terms & Definitions

Torque ( $\vec{\tau}$ ). The rotational analog of force. Torque is what makes objects spin or change their angular velocity.

Moment arm (lever arm). The perpendicular distance from the axis of rotation to the line of action of the force.

Couple. A pair of equal, opposite, and parallel forces with non-coincident lines of action. A couple produces pure rotation without translation.

Centre of gravity. The point where the entire weight of a body can be considered to act. For uniform bodies, it coincides with the geometric centre.

Rotational equilibrium. Net torque about any axis is zero, so the body has no angular acceleration.

Torque Formula

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

Magnitude: $\tau = r F \sin\theta$ , where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$ .

Equivalently: $\tau = F \cdot d$ , where $d = r\sin\theta$ is the moment arm.

The direction of $\vec{\tau}$ is given by the right-hand rule: curl the fingers from $\vec{r}$ to $\vec{F}$ , and the thumb points along $\vec{\tau}$ . For 2D problems, we use the convention: anticlockwise = positive, clockwise = negative.

How to Compute Torque — The 3-Step Method

The axis is wherever the body is hinged or pivoted. If there’s no fixed pivot, choose the axis cleverly — pick the point where the most unknowns “vanish” because their moment arm is zero.

The moment arm is the perpendicular distance from the axis to the line of action of the force. Extend the force vector backward or forward as a straight line — measure perpendicular distance from the axis to this line.

Anticlockwise torques are positive; clockwise are negative. Add all torques. For equilibrium, the sum must be zero.

What is a Couple?

A couple is a pair of equal, opposite, parallel forces $F$ separated by a perpendicular distance $d$ . The net force from a couple is zero, so no translation. But there is a net torque $\tau = F d$ that produces pure rotation.

Examples of couples in real life:

Turning a steering wheel (two hands push opposite tangentially).
Unscrewing a bottle cap.
The torsion of a key in a lock.

Important property: the torque of a couple is independent of the choice of axis. Compute torque about any point and you get the same answer $F d$ . This is unique to couples — no other system has axis-independent torque.

Rotational Equilibrium — The Two Conditions

For a body in static equilibrium:

$\sum \vec{F} = 0$ (forces balance — translational equilibrium)
$\sum \vec{\tau} = 0$ (torques balance about any axis — rotational equilibrium)

For 2D problems, condition 1 gives two scalar equations ( $\sum F_x = 0$ , $\sum F_y = 0$ ), and condition 2 gives one ( $\sum \tau = 0$ ). Three equations — sufficient to solve for three unknowns.

Choosing the axis cleverly. The axis can be any point — usually you pick the point through which the most unknown forces pass, since their torque is zero there. This eliminates them from the torque equation.

Solved Examples

Example 1 (Easy, CBSE)

A 4 m uniform plank weighing 50 N is supported on two saw-horses placed at its ends. A man weighing 700 N stands 1 m from the left support. Find the force on each support.

Take moments about the left support. Anticlockwise positive.

Forces:

Plank weight 50 N at the centre (2 m from left).
Man weight 700 N at 1 m from left.
Normal $N_R$ at right support (4 m from left), pushes up.

Torque equation: $N_R(4) - 50(2) - 700(1) = 0$ . So $N_R = (100 + 700)/4 = 200$ N.

Force balance: $N_L + N_R = 50 + 700 = 750$ . So $N_L = 550$ N.

Example 2 (Medium, JEE Main)

A ladder of length 5 m and weight 200 N rests against a smooth vertical wall. The base is on a rough floor, 3 m from the wall. Find the friction force at the floor needed to keep the ladder from slipping.

Set up: ladder from base (point B) to top (point A) at the wall. Height $h = \sqrt{5^2 - 3^2} = 4$ m.

Forces on ladder:

Weight 200 N at the centre (1.5 m from wall horizontally, 2 m up).
Normal from wall $N_w$ (horizontal, into the ladder at top).
Normal from floor $N_f$ (vertical, up at the base).
Friction $f$ (horizontal, at the base, pointing towards the wall).

Take moments about the base B. Wall is smooth, so wall friction is zero.

Anticlockwise: $N_w$ at top, lever arm = 4 m, torque = $4 N_w$ .

Clockwise: weight at centre, horizontal lever arm = 1.5 m, torque = $200 \times 1.5 = 300$ .

Balance: $4 N_w = 300 \Rightarrow N_w = 75$ N.

Force balance horizontally: $f = N_w = 75$ N.

Example 3 (Hard, JEE Advanced)

A uniform rod of mass $m$ and length $L$ is pivoted at one end and held horizontally. The rod is released. Find the angular acceleration just after release.

Torque about pivot: $\tau = mg(L/2)$ (weight acts at centre, lever arm $L/2$ ).

Moment of inertia of uniform rod about end: $I = mL^2/3$ .

Newton’s second law for rotation: $\tau = I\alpha \Rightarrow \alpha = \tau/I = \dfrac{mg(L/2)}{mL^2/3} = \dfrac{3g}{2L}$ .

Exam-Specific Tips

JEE Main weightage. Torque and equilibrium together: ~1 question per shift. Mostly pure rotational equilibrium (ladder, beam, plank).

JEE Advanced. Often combines torque with rotational dynamics ( $\tau = I\alpha$ ). Tougher numerical setup.

NEET. Centre of gravity and simple beam balance — 1 question typically. Direct application.

CBSE class 11. A 3- or 5-marker on torque, couple, or equilibrium. Use the proper FBD and you score full marks.

Common Mistakes to Avoid

Mistake 1: Forgetting the perpendicular distance. Torque is $\tau = r F\sin\theta$ , not $rF$ . If you forget $\sin\theta$ , you overcount.

Mistake 2: Wrong sign convention. Mixing anticlockwise and clockwise sign conventions in the same equation is the #1 mistake. Pick one, stick with it.

Mistake 3: Picking a bad axis. Choosing an axis where many unknowns appear in the torque equation makes the algebra painful. Pick an axis through one or more unknown force points.

Mistake 4: Treating a couple as needing an axis. A couple has the same torque about any axis. Don’t try to “choose” an axis for it.

Mistake 5: Forgetting that the centre of gravity is where the weight acts. Treat weight as a single force at the CG of the body. For uniform bodies, that’s the geometric centre.

Practice Questions

Q1. A force $\vec{F} = 3\hat{i} + 4\hat{j}$ N is applied at position $\vec{r} = 2\hat{i} + \hat{j}$ m. Find the torque about the origin.

$\vec{\tau} = \vec{r}\times\vec{F} = (2 \times 4 - 1 \times 3)\hat{k} = 5\hat{k}$ Nm.

Q2. A uniform beam 6 m long, weight 100 N, is supported at both ends. A 200 N weight is placed 2 m from the left end. Find the support forces.

Take moments about left support: $N_R(6) - 100(3) - 200(2) = 0 \Rightarrow N_R = 700/6 \approx 116.7$ N. Force balance: $N_L = 300 - 116.7 = 183.3$ N.

Q3. Two equal forces of 10 N act at opposite ends of a 0.5 m rod, perpendicular to the rod. Find the torque of the couple.

$\tau = F d = 10 \times 0.5 = 5$ Nm.

Q4. A door of width 1 m is pushed at its outer edge with a force of 20 N perpendicular to the door. Find the torque about the hinges.

$\tau = 20 \times 1 = 20$ Nm.

Q5. Why is it easier to open a door by pushing at the edge than at the middle?

Torque = force × moment arm. At the edge, moment arm is largest, so the same force produces more torque, making the door easier to open.

Q6. A wrench of length 25 cm is used to tighten a bolt. If the maximum force the user can apply is 80 N, find the maximum torque.

$\tau = 80 \times 0.25 = 20$ Nm.

Q7. A see-saw is balanced when a 30 kg child sits 2 m from the pivot and a 40 kg child sits at distance $d$ on the other side. Find $d$ .

$30 \times 2 = 40 \times d \Rightarrow d = 1.5$ m.

Q8. A uniform metre stick is balanced on a knife edge at the 50 cm mark. A 100 g mass is hung at the 20 cm mark. To rebalance, where must a 60 g mass be hung on the other side?

Take moments about pivot. Left: $100 \times 30 = 3000$ . Right: $60 \times d = 3000 \Rightarrow d = 50$ cm. So at the 100 cm mark — the very end.

FAQs

Q: Is torque a vector or a scalar?

Torque is a vector. Its direction is given by the right-hand rule applied to $\vec{r}\times\vec{F}$ . In 2D problems we treat it as a scalar with sign convention.

Q: What’s the difference between torque and moment?

In Indian physics textbooks, “moment of a force” is the same as torque. Engineering uses “moment” more often, physics uses “torque.” Same concept.

Q: Can torque be zero even if force is non-zero?

Yes — when the force passes through the axis of rotation (so $r = 0$ ) or when the force is parallel to $\vec{r}$ (so $\sin\theta = 0$ ).

Q: Why does a couple cause only rotation, not translation?

The two forces of a couple are equal and opposite, so they cancel as a translational pair. But they don’t lie on the same line, so they create a non-zero torque, leading to pure rotation.

Q: How do I apply this to a body in 3D?

Use the cross product $\vec{\tau} = \vec{r}\times\vec{F}$ with full 3D coordinates. The result is a vector that points along the rotation axis.

Q: What’s the SI unit of torque?

Newton-metre (Nm). It has the same dimensions as energy (joule), but represents a different physical quantity, so we keep them distinct.

Q: Can the axis of rotation be outside the body?

Yes. For a body pivoted by a string from the ceiling, the axis is at the ceiling, not inside the body. The choice of axis is a matter of convenience.

Q: How is torque related to angular momentum?

$\vec{\tau} = d\vec{L}/dt$ . Net torque equals the rate of change of angular momentum — the rotational version of Newton’s second law.