Concept Card: Moment of a Force & Torque
1. Physical Framework & Core Definitions:
Torque ($\vec{\tau}$), also called the moment of a force, quantifies the turning or rotational tendency of a force about a specified pivot point or rotation axis. It is the rotational analogue of force in linear dynamics.
- Vector Cross-Product Definition:
$$\mathbf{\vec{\tau} = \vec{r} \times \vec{F}}$$
where:- $\vec{r}$ is the position vector of the point of application of the force measured from the reference origin or pivot point $O$.
- $\vec{F}$ is the applied force vector.
- Vector Classification: An axial vector (pseudo-vector) perpendicular to both $\vec{r}$ and $\vec{F}$, pointing along the axis defined by the Right-Hand Rule.
- SI Unit: Newton-meter ($\text{N}\cdot\text{m}$).
Crucial Distinction: Although $\text{N}\cdot\text{m}$ shares the same base dimensions as the Joule ($[M L^2 T^{-2}]$), torque is NEVER expressed in Joules. Joules are reserved exclusively for scalar energy and work, whereas torque is a directed rotational vector.
2. Geometric Methods for Computing Torque Magnitude
- Angle Formulation:
$$\tau = r F \sin\theta$$
where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$ when placed tail-to-tail ($0^\circ \le \theta \le 180^\circ$). - Lever Arm (Moment Arm) Method ($r_{\perp}$):
$$\mathbf{\tau = F \cdot r_{\perp}}$$
where $r_{\perp} = r\sin\theta$ is the perpendicular distance from the pivot $O$ to the line of action of the force. - Perpendicular Force Component Method ($F_{\perp}$):
$$\mathbf{\tau = r \cdot F_{\perp}}$$
where $F_{\perp} = F\sin\theta$ is the component of force perpendicular to the radial vector $\vec{r}$. The radial component $F_{\parallel} = F\cos\theta$ passes through the pivot and produces zero torque.
3. Cartesian Determinant Form
Given $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and $\vec{F} = F_x\hat{i} + F_y\hat{j} + F_z\hat{k}$:
$$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ F_x & F_y & F_z \end{vmatrix} = (y F_z – z F_y)\hat{i} + (z F_x – x F_z)\hat{j} + (x F_y – y F_x)\hat{k}$$
4. Torque about a Fixed Axis & Theory of Couples
- Torque about an Axis: For an axis $L$ along unit vector $\hat{u}_L$, $\tau_L = (\vec{r} \times \vec{F}) \cdot \hat{u}_L$.
- Zero Torque Condition: A force produces zero torque about an axis if its line of action is parallel to the axis or intersects the axis ($r_{\perp} = 0$).
- Couple: A pair of equal and opposite forces ($\vec{F}$ and $-\vec{F}$) with parallel lines of action separated by distance $d$:
- Net force is zero ($\vec{F}_{\text{net}} = \vec{0}$), causing zero linear acceleration of the centre of mass.
- Moment of a couple: $\vec{\tau} = \vec{r}_{12} \times \vec{F}$, with magnitude $\tau = F \cdot d$.
- Fundamental Theorem: The moment of a couple is completely independent of the choice of reference point or origin.
- Complete Mechanical Equilibrium:
$$\Sigma \vec{F}_{\text{ext}} = \vec{0} \quad \text{(Translational Equilibrium)}$$
$$\Sigma \vec{\tau}_{\text{ext}} = \vec{0} \quad \text{(Rotational Equilibrium)}$$
If $\Sigma \vec{F} = \vec{0}$, then if net torque is zero about one point, it is zero about ALL points in space!
5. Common JEE Pitfalls & Traps
- Trap 1 (Reference Point Dependency): Torque is not an absolute quantity; it exists only relative to a chosen reference point. Always specify the pivot when writing torque equations.
- Trap 2 (Forces Passing Through the Pivot): Forces whose lines of action pass through the chosen pivot exert zero torque ($r_{\perp} = 0$). Selecting unknown hinge or normal reaction points as the pivot eliminates them from the torque balance.
- Trap 3 (Unit Confusion): Never write torque in Joules ($\text{J}$). Use Newton-meters ($\text{N}\cdot\text{m}$).
Solved Examples
Example 1 (Direct Conceptual Application – Vector Determinant & Pivot Shift):
A force $\vec{F} = (3.0\hat{i} – 2.0\hat{j} + 4.0\hat{k})\text{ N}$ acts at point $P$ with position vector $\vec{r}_P = (2.0\hat{i} + 3.0\hat{j} – \hat{k})\text{ m}$ relative to origin $O$.
(a) Determine the torque vector $\vec{\tau}_O$ about the origin $O$.
(b) Determine the torque vector $\vec{\tau}_A$ about point $A$ with position vector $\vec{r}_A = (\hat{i} – \hat{j} + 2.0\hat{k})\text{ m}$.
(c) Find the angle $\theta$ between $\vec{r}_P$ and $\vec{F}$, and verify the torque magnitude using $\tau = r F \sin\theta$.
Solution:
(a) Torque about origin $O$:
$$\vec{\tau}_O = \vec{r}_P \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2.0 & 3.0 & -1.0 \\ 3.0 & -2.0 & 4.0 \end{vmatrix}$$
$$\vec{\tau}_O = [(3)(4) – (-1)(-2)]\hat{i} + [(-1)(3) – (2)(4)]\hat{j} + [(2)(-2) – (3)(3)]\hat{k} = \mathbf{(10.0\hat{i} – 11.0\hat{j} – 13.0\hat{k})\text{ N}\cdot\text{m}}$$
(b) Position vector of $P$ relative to $A$:
$$\vec{r}’ = \vec{r}_P – \vec{r}_A = (2 – 1)\hat{i} + (3 – (-1))\hat{j} + (-1 – 2)\hat{k} = (\hat{i} + 4.0\hat{j} – 3.0\hat{k})\text{ m}$$
$$\vec{\tau}_A = \vec{r}’ \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1.0 & 4.0 & -3.0 \\ 3.0 & -2.0 & 4.0 \end{vmatrix} = \mathbf{(10.0\hat{i} – 13.0\hat{j} – 14.0\hat{k})\text{ N}\cdot\text{m}}$$
(c) Verification: $|\vec{\tau}_O| = \sqrt{10^2 + (-11)^2 + (-13)^2} = \sqrt{100 + 121 + 169} = \sqrt{390}\text{ N}\cdot\text{m} \approx 19.748\text{ N}\cdot\text{m}$.
$r = \sqrt{4 + 9 + 1} = \sqrt{14}\text{ m}$, $F = \sqrt{9 + 4 + 16} = \sqrt{29}\text{ N}$.
$\vec{r}_P \cdot \vec{F} = 6 – 6 – 4 = -4\text{ N}\cdot\text{m} \implies \cos\theta = \frac{-4}{\sqrt{406}} \implies \sin\theta = \sqrt{\frac{390}{406}}$.
$r F \sin\theta = \sqrt{14}\sqrt{29}\sqrt{\frac{390}{406}} = \sqrt{390}\text{ N}\cdot\text{m}$. Verified!
Example 2 (Mathematical Derivation – Lever Arm & Line of Action Method):
A force $\vec{F} = (4.0\hat{i} + 3.0\hat{j})\text{ N}$ acts along the line $3x – 4y + 12 = 0$ in the $xy$-plane.
(a) Determine the perpendicular distance (lever arm $r_{\perp}$) from origin $O(0, 0)$ to the line of action.
(b) Compute the magnitude and direction of torque about the origin using $\tau = F r_{\perp}$.
(c) Verify using an arbitrary point on the line of action.
Solution:
(a) Distance from origin $(0, 0)$ to $3x – 4y + 12 = 0$:
$$r_{\perp} = \frac{|3(0) – 4(0) + 12|}{\sqrt{3^2 + (-4)^2}} = \frac{12}{5} = 2.40\text{ meters}$$
(b) Force magnitude: $F = \sqrt{4^2 + 3^2} = 5.0\text{ N}$.
Torque magnitude: $\tau = F \cdot r_{\perp} = 5.0 \times 2.40 = 12.0\text{ N}\cdot\text{m}$.
Direction: Line cuts axes at $(-4, 0)$ and $(0, 3)$. $\vec{F} = 4\hat{i} + 3\hat{j}$ produces clockwise rotation about origin: $\mathbf{\vec{\tau} = -12.0\hat{k}\text{ N}\cdot\text{m}}$.
(c) Choose $y$-intercept $(0, 3.0\text{ m})$ on line: $\vec{r} = 3.0\hat{j}\text{ m}$.
$$\vec{\tau} = (3.0\hat{j}) \times (4.0\hat{i} + 3.0\hat{j}) = 12.0(\hat{j} \times \hat{i}) = -12.0\hat{k}\text{ N}\cdot\text{m}$$. Exactly matches!
Example 3 (Standard JEE Advanced Scenario – Ladder Equilibrium on Rough Floor):
A uniform ladder of mass $M = 20.0\text{ kg}$ and length $L = 5.0\text{ m}$ rests against a smooth vertical wall at an angle $\theta = 53^\circ$ to the horizontal rough floor ($\cos 53^\circ = 0.60, \sin 53^\circ = 0.80$). Take $g = 10.0\text{ m/s}^2$.
(a) Write the translational equilibrium equations.
(b) Taking torques about the base, find the normal reaction $N_w$ exerted by the wall.
(c) Find the normal force $N_f$ and static friction force $f_s$ exerted by the floor.
(d) Find the minimum static friction coefficient $\mu_s$ to prevent slipping.
Solution:
(a) Equilibrium: $\Sigma F_x = 0 \implies f_s = N_w$. $\Sigma F_y = 0 \implies N_f = M g = 20.0(10.0) = 200.0\text{ N}$.
(b) Torque about base eliminates $N_f$ and $f_s$:
$$\Sigma \tau_{\text{base}} = N_w (L\sin 53^\circ) – M g \left(\frac{L}{2}\cos 53^\circ\right) = 0$$
$$N_w = \frac{1}{2} M g \cot 53^\circ = \frac{1}{2}(200.0)\left(\frac{0.60}{0.80}\right) = 75.0\text{ N}$$
(c) $N_f = 200.0\text{ N}, \quad f_s = N_w = 75.0\text{ N}$.
(d) $f_s \le \mu_s N_f \implies 75.0 \le \mu_s (200.0) \implies \mathbf{\mu_{s,\text{min}} = \frac{75.0}{200.0} = 0.375}$.
Example 4 (Edge Case – Gravitational Torque in Projectile Motion & Angular Momentum):
A particle of mass $m = 0.50\text{ kg}$ is fired from origin $O$ with speed $u = 20.0\text{ m/s}$ at $\alpha = 45^\circ$ to the horizontal. Take $g = 10.0\text{ m/s}^2$.
(a) Express the instantaneous torque $\vec{\tau}(t)$ of gravity about $O$ as a function of time.
(b) Evaluate torque at the apex and at the landing point.
(c) Find angular momentum $\vec{L}(t)$ by integrating torque and verify via $\vec{L} = \vec{r} \times \vec{p}$.
Solution:
(a) $x(t) = (u\cos 45^\circ)t = 10\sqrt{2}\,t\text{ m}$. $\vec{F}_g = -m g\hat{j} = -5.0\hat{j}\text{ N}$.
$$\vec{\tau}(t) = \vec{r}(t) \times \vec{F}_g = (10\sqrt{2}\,t\hat{i} + y\hat{j}) \times (-5.0\hat{j}) = \mathbf{-50\sqrt{2}\,t\,\hat{k}\text{ N}\cdot\text{m}}$$
(b) Time to apex: $t_{\text{apex}} = \frac{u\sin 45^\circ}{g} = \sqrt{2}\text{ s} \implies \vec{\tau}_{\text{apex}} = -50\sqrt{2}(\sqrt{2})\hat{k} = -100.0\hat{k}\text{ N}\cdot\text{m}$.
Time of flight: $T = 2\sqrt{2}\text{ s} \implies \vec{\tau}_{\text{landing}} = -50\sqrt{2}(2\sqrt{2})\hat{k} = -200.0\hat{k}\text{ N}\cdot\text{m}$.
(c) Integrating torque: $\vec{L}(t) = \int_0^t (-50\sqrt{2}\,t’\hat{k})\,dt’ = \mathbf{-25\sqrt{2}\,t^2\,\hat{k}\text{ kg}\cdot\text{m}^2/\text{s}}$.
*Direct verification:* $\vec{r} \times \vec{p} = (x p_y – y p_x)\hat{k} = [(10\sqrt{2}t)(5\sqrt{2}-5t) – (10\sqrt{2}t-5t^2)(5\sqrt{2})]\hat{k} = -25\sqrt{2}t^2\hat{k}$. Exactly verified!
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
The dimensional formula of torque (moment of a force) is:
(A) $[M L T^{-2}]$
(B) $[M L^2 T^{-2}]$
(C) $[M L^2 T^{-1}]$
(D) $[M^0 L^2 T^{-2}]$
Problem 2 (JEE Main – Single Correct):
A force $\vec{F} = (2.0\hat{i} + 3.0\hat{j})\text{ N}$ acts at a point with position vector $\vec{r} = (4.0\hat{i} + 5.0\hat{j})\text{ m}$. The torque of the force about the origin is:
(A) $+2.0\hat{k}\text{ N}\cdot\text{m}$
(B) $-2.0\hat{k}\text{ N}\cdot\text{m}$
(C) $+22.0\hat{k}\text{ N}\cdot\text{m}$
(D) $-22.0\hat{k}\text{ N}\cdot\text{m}$
Problem 3 (JEE Main – Single Correct):
A couple consists of two forces of magnitude $F$ each acting along parallel lines separated by a perpendicular distance $d$. The moment of this couple about any point in the plane:
(A) Depends on the coordinates of the chosen point
(B) Is identically zero
(C) Is independent of the chosen point and equals $F d$
(D) Equals $2 F d$
Problem 4 (JEE Main – Single Correct):
A uniform horizontal beam $AB$ of length $L$ and weight $W$ is hinged to a wall at end $A$ and supported by a vertical cable at end $B$. The vertical reaction force exerted by the hinge at $A$ on the beam is:
(A) $W$
(B) $\frac{W}{2}$
(C) $\frac{W}{4}$
(D) $2W$
Problem 5 (JEE Main – Single Correct):
A force $\vec{F} = F_0\hat{k}$ acts at a point with position vector $\vec{r} = x_0\hat{i} + y_0\hat{j}$. The torque of this force about the $z$-axis is:
(A) $F_0 \sqrt{x_0^2 + y_0^2}$
(B) $F_0 x_0$
(C) $F_0 y_0$
(D) $0$
Problem 6 (JEE Advanced – One or More Correct):
For an extended rigid body subjected to a system of coplanar forces:
(A) If $\Sigma \vec{F} = \vec{0}$, the net torque is independent of the choice of reference origin.
(B) If $\Sigma \vec{\tau} = \vec{0}$ about three non-collinear points in the plane, the body is in complete mechanical equilibrium.
(C) A force whose line of action passes through the reference point generates zero torque about that point.
(D) The SI unit of torque is $\text{N}\cdot\text{m}$, which is dimensionally equivalent to Joule but physically distinct.
Problem 7 (JEE Advanced – One or More Correct):
A particle of mass $m$ is launched from the ground at angle $\theta$ to the horizontal. As it moves along its parabolic flight path under uniform gravity:
(A) The magnitude of the torque of gravity about the launch point increases linearly with time $t$.
(B) The magnitude of the torque of gravity about the launch point is maximum at the moment of landing.
(C) The torque of gravity about the apex of the trajectory is zero at the instant the particle passes the apex.
(D) The rate of change of angular momentum about the launch origin is equal to the gravitational torque about that origin.
Problem 8 (JEE Advanced – One or More Correct):
A uniform heavy rod of mass $M$ and length $L$ rests horizontally on two knife edges located at distances $L/4$ and $3L/4$ from one end:
(A) The upward reaction forces at both knife edges are equal to $\frac{Mg}{2}$.
(B) The net torque about the centre of mass of the rod is zero.
(C) If the knife edges are moved to the extreme ends of the rod, the upward reactions remain equal to $\frac{Mg}{2}$.
(D) If an additional mass $M$ is placed at one extreme end, the rod will topple off the supports.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A force $\vec{F} = (2.0\hat{i} + 3.0\hat{j} – 4.0\hat{k})\text{ N}$ acts at point $P(3.0\text{ m}, 0, 2.0\text{ m})$. The magnitude of the torque of this force about the origin is $\sqrt{N}\text{ N}\cdot\text{m}$. Find the integer value of $N$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A uniform ladder of mass $M = 30.0\text{ kg}$ rests against a smooth vertical wall at an inclination of $45^\circ$ to the horizontal floor. Taking $g = 10.0\text{ m/s}^2$, calculate the magnitude of the horizontal normal force (in Newtons) exerted by the wall on the ladder.
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (A) | 3. (C) | 4. (B) | 5. (D) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 373 | 10. 150
Solution 1:
$[\tau] = [r][F] = [L][M L T^{-2}] = [M L^2 T^{-2}]$.
Correct Option: (B)
Solution 2:
$\vec{\tau} = (4\hat{i} + 5\hat{j}) \times (2\hat{i} + 3\hat{j}) = (12 – 10)\hat{k} = +2.0\hat{k}\text{ N}\cdot\text{m}$.
Correct Option: (A)
Solution 3:
The moment of a couple is $\vec{\tau} = \vec{r}_{12} \times \vec{F}$, which depends solely on the separation between the lines of action and is independent of the reference origin.
Correct Option: (C)
Solution 4:
Torque about end $B$: $N_A(L) – W(L/2) = 0 \implies N_A = W/2$.
Correct Option: (B)
Solution 5:
Because the force is parallel to the $z$-axis ($\vec{F} \parallel \hat{k}$), its moment about the $z$-axis is identically zero ($\tau_z = 0$).
Correct Option: (D)
Solution 6:
All four statements (A, B, C, D) are verified fundamental theorems of rigid body mechanics.
Correct Options: (A, B, C, D)
Solution 7:
– (A) True: $\tau_O(t) = -m g (u\cos\theta)t\hat{k} \propto t$.
– (B) True: Maximum at flight time $T$ when $x$ is maximum.
– (C) True: At apex, $\vec{r}_{\text{rel}} = \vec{0}$, so torque about apex is zero.
– (D) True: Rotational second law $\vec{\tau} = \frac{d\vec{L}}{dt}$.
Correct Options: (A, B, C, D)
Solution 8:
– (A) True: Knife edges placed symmetrically at $L/4$ from either side of CM ($L/2$) give $R_1 = R_2 = M g/2$.
– (B) True: Net torque about CM is $R_1(L/4) – R_2(L/4) = 0$.
: Net torque about CM is $R_1(L/4) – R_2(L/4) = 0$.
– (C) True: By symmetry, supports at ends give $R_1 = R_2 = M g/2$.
– (D) True: Placing mass $M$ at end $x = 0$ gives $M g(L/4) – M g(L/4) = 0$; any additional perturbation topples the rod.
Correct Options: (A, B, C, D)
Solution 9:
$\vec{\tau} = (3\hat{i} + 2\hat{k}) \times (2\hat{i} + 3\hat{j} – 4\hat{k}) = -6.0\hat{i} + 16.0\hat{j} + 9.0\hat{k}\text{ N}\cdot\text{m}$.
$\tau^2 = (-6)^2 + 16^2 + 9^2 = 36 + 256 + 81 = 373 \implies \tau = \sqrt{373}\text{ N}\cdot\text{m}$.
Correct Answer: 373
Solution 10:
Torque about base: $N_w (L\sin 45^\circ) – M g (L/2 \cos 45^\circ) = 0 \implies N_w = \frac{1}{2} M g \cot 45^\circ = \frac{1}{2}(30.0)(10.0)(1.0) = 150.0\text{ N}$.
Correct Answer: 150