Chapter 6 · System of Particles and Rotational Motion

6.2 Torque, Angular Momentum and Equilibrium of a Rigid Body

The turning effect of a force is its torque $\boldsymbol\tau=\mathbf{r}\times\mathbf{F}$; the rotational partner of momentum is angular momentum $\mathbf{l}=\mathbf{r}\times\mathbf{p}$. A rigid body is in equilibrium only when both the net force and the net torque vanish.

Try to open a heavy door by pushing right next to its hinges: you can push hard and it barely moves. Push with the same force at the handle and it swings open easily. Now grip a steering wheel with both hands and pull up with one, down with the other: the two forces add to zero, so by Newton's second law the wheel's centre of mass should not accelerate — and it doesn't — yet the wheel turns. Force alone does not decide whether, or how fast, a body starts to rotate. Where the force acts, and in which direction, matter just as much.

This section builds the quantities that describe turning. A new way of multiplying two vectors, the vector product, produces a vector perpendicular to both. With it we define angular velocity, the torque (moment of a force) and the angular momentum of a particle, and find that torque changes angular momentum just as force changes linear momentum. Finally we put the pieces together for a rigid body at rest: it is in equilibrium only if the forces balance and the torques balance — the physics behind see-saws, crowbars, cantilever balconies and ladders leaning against walls.

The vector product

In Section 5.1 the scalar product $\mathbf{A}\cdot\mathbf{B}=AB\cos\theta$ turned two vectors into a number. Rotation needs a product that produces a direction as well: the turning effect of a spanner on a nut is "about the axis of the bolt", perpendicular to both the spanner and the push.

📘 Definition 6.3 — Vector (cross) product

The vector product of vectors $\mathbf{A}$ and $\mathbf{B}$, with angle $\theta$ between them ($0\le\theta\le180^\circ$), is the vector $\mathbf{C}=\mathbf{A}\times\mathbf{B}$ with

$$|\mathbf{A}\times\mathbf{B}|=AB\sin\theta ,$$

directed perpendicular to the plane of $\mathbf{A}$ and $\mathbf{B}$ by the right-hand rule: point the fingers of your right hand along $\mathbf{A}$ and curl them towards $\mathbf{B}$ through the smaller angle; the thumb points along $\mathbf{A}\times\mathbf{B}$. (Equivalently, a right-handed screw turned from $\mathbf{A}$ towards $\mathbf{B}$ advances along $\mathbf{A}\times\mathbf{B}$.)

Geometrically, $|\mathbf{A}\times\mathbf{B}|$ is the area of the parallelogram with sides $\mathbf{A}$ and $\mathbf{B}$.

📐 Result 6.4 — Properties of the vector product

Not commutative: $\mathbf{B}\times\mathbf{A}=-\mathbf{A}\times\mathbf{B}$ (curling the other way reverses the thumb).
$\mathbf{A}\times\mathbf{A}=\mathbf{0}$, and $\mathbf{A}\times\mathbf{B}=\mathbf{0}$ whenever $\mathbf{A}$ and $\mathbf{B}$ are parallel or antiparallel.
Distributive: $\mathbf{A}\times(\mathbf{B}+\mathbf{C})=\mathbf{A}\times\mathbf{B}+\mathbf{A}\times\mathbf{C}$; and $(\lambda\mathbf{A})\times\mathbf{B}=\lambda(\mathbf{A}\times\mathbf{B})$.
Unit vectors (right-handed axes): $\hat{\mathbf{i}}\times\hat{\mathbf{j}}=\hat{\mathbf{k}}$, $\ \hat{\mathbf{j}}\times\hat{\mathbf{k}}=\hat{\mathbf{i}}$, $\ \hat{\mathbf{k}}\times\hat{\mathbf{i}}=\hat{\mathbf{j}}$; reversing the order gives a minus sign; $\hat{\mathbf{i}}\times\hat{\mathbf{i}}=\hat{\mathbf{j}}\times\hat{\mathbf{j}}=\hat{\mathbf{k}}\times\hat{\mathbf{k}}=\mathbf{0}$.

Using these, for $\mathbf{A}=A_x\hat{\mathbf{i}}+A_y\hat{\mathbf{j}}+A_z\hat{\mathbf{k}}$ and $\mathbf{B}=B_x\hat{\mathbf{i}}+B_y\hat{\mathbf{j}}+B_z\hat{\mathbf{k}}$,

$$\mathbf{A}\times\mathbf{B}=\begin{vmatrix}\hat{\mathbf{i}}&\hat{\mathbf{j}}&\hat{\mathbf{k}}\\A_x&A_y&A_z\\B_x&B_y&B_z\end{vmatrix}$$

that is, component by component,

$$(\mathbf{A}\times\mathbf{B})_x=A_yB_z-A_zB_y,$$ $$(\mathbf{A}\times\mathbf{B})_y=A_zB_x-A_xB_z,$$ $$(\mathbf{A}\times\mathbf{B})_z=A_xB_y-A_yB_x.$$

A memory aid for the unit vectors: write $\hat{\mathbf{i}}\to\hat{\mathbf{j}}\to\hat{\mathbf{k}}\to\hat{\mathbf{i}}$ round a circle. Going round the circle the "forward" way gives $+$ the third vector; going backwards gives $-$. If you have not met determinants yet in mathematics, simply use the expanded form — each component is "cross-multiply the other two components".

✏️ Example 1 — Computing a vector product

For $\mathbf{A}=3\hat{\mathbf{i}}-\hat{\mathbf{j}}+2\hat{\mathbf{k}}$ and $\mathbf{B}=\hat{\mathbf{i}}+2\hat{\mathbf{j}}-\hat{\mathbf{k}}$, find (a) $\mathbf{A}\times\mathbf{B}$, (b) the area of the parallelogram with sides $\mathbf{A}$ and $\mathbf{B}$, (c) the angle between $\mathbf{A}$ and $\mathbf{B}$, checking with the scalar product.

💡 Show solution▶

(a) $x$: $A_yB_z-A_zB_y=(-1)(-1)-(2)(2)=-3$. $\ y$: $A_zB_x-A_xB_z=(2)(1)-(3)(-1)=5$. $\ z$: $A_xB_y-A_yB_x=(3)(2)-(-1)(1)=7$. So $\mathbf{A}\times\mathbf{B}=-3\hat{\mathbf{i}}+5\hat{\mathbf{j}}+7\hat{\mathbf{k}}$.

Check ⟂ $\mathbf{A}\cdot(\mathbf{A}\times\mathbf{B})=-9-5+14=0$ and $\mathbf{B}\cdot(\mathbf{A}\times\mathbf{B})=-3+10-7=0$ ✓ — perpendicular to both.

(b) Area $=|\mathbf{A}\times\mathbf{B}|=\sqrt{9+25+49}=\sqrt{83}=9.11$ square units.

(c) $A=\sqrt{14}$, $B=\sqrt6$, $AB=9.17$. $\sin\theta=\dfrac{9.11}{9.17}=0.994$, which allows $83.7^\circ$ or $96.3^\circ$. The scalar product decides: $\mathbf{A}\cdot\mathbf{B}=3-2-2=-1$, so $\cos\theta=-0.109$ and $\theta=96.3^\circ$.

Check $\sin^2\theta+\cos^2\theta=\dfrac{83+1}{84}=1$ ✓. In general $|\mathbf{A}\times\mathbf{B}|^2+(\mathbf{A}\cdot\mathbf{B})^2=A^2B^2$.

Angular velocity and torque

When a rigid body rotates about a fixed axis, every particle moves in a circle round the axis, and in a given time every line drawn perpendicular to the axis turns through the same angle $\theta$. So one angle describes the whole body, just as one position $x$ describes a body in pure translation. You met the angular speed of a single particle in Section 3.4; now it belongs to the whole body and gets a direction.

📘 Definition 6.4 — Angular velocity and angular acceleration

For a rigid body rotating about a fixed axis, with angular position $\theta$ (in radians) of any line perpendicular to the axis, the angular velocity and angular acceleration are

$$\omega=\frac{d\theta}{dt}\ \ (\text{rad/s}),\qquad \alpha=\frac{d\omega}{dt}\ \ (\text{rad/s}^2).$$

As vectors, $\boldsymbol\omega$ and $\boldsymbol\alpha$ lie along the axis: curl the fingers of your right hand in the sense of rotation and the thumb gives $\boldsymbol\omega$. A particle at perpendicular distance $r_\perp$ from the axis has speed and accelerations

$$v=\omega r_\perp,\qquad a_t=\alpha r_\perp,\qquad a_c=\omega^2r_\perp ,$$

(tangential and centripetal). In vector form $\mathbf{v}=\boldsymbol\omega\times\mathbf{r}$, where $\mathbf{r}$ is the particle's position measured from any point on the axis.

Every particle has the same $\omega$ but not the same speed: the tip of a fan blade moves much faster than a point near the hub. A body spinning anticlockwise as seen from above has $\boldsymbol\omega$ pointing up; clockwise, pointing down.

Torque. Now return to the door. Three things decide how strongly a force turns a body about a point $O$: the size of the force, how far from $O$ it acts, and its direction. A push at the handle along the door, straight towards the hinge, does nothing; only the part of the force perpendicular to the door turns it. All three are captured by one vector product.

📘 Definition 6.5 — Torque (moment of a force)

The torque about a point $O$ of a force $\mathbf{F}$ acting at a point with position vector $\mathbf{r}$ (measured from $O$) is

$$\boldsymbol\tau=\mathbf{r}\times\mathbf{F},\qquad \tau=rF\sin\theta ,$$

where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{F}$. Its magnitude can be read in two equivalent ways:

$$\tau=r_\perp F=rF_\perp ,$$

where $r_\perp=r\sin\theta$ is the lever arm (moment arm) — the perpendicular distance from $O$ to the line of action of $\mathbf{F}$ — and $F_\perp=F\sin\theta$ is the component of $\mathbf{F}$ perpendicular to $\mathbf{r}$. SI unit: $\text{N m}$; dimensions $[\text{M L}^2\text{T}^{-2}]$.

For forces in the plane of the page the torque points straight out of the page (anticlockwise turning, written ⊙ and counted positive) or into it (clockwise, ⊗, negative), so in plane problems torques can be added as signed numbers. A force whose line of action passes through $O$ has zero lever arm and no torque about $O$. Torque has the same units as work, but it is a vector and is never written in joules.

✏️ Example 2 — A spanner, and a torque in three dimensions

(a) A mechanic pulls on the end of a $25\ \text{cm}$ spanner with a force of $40\ \text{N}$ at $60^\circ$ to the handle, in the plane of rotation. Find the lever arm and the torque about the bolt. What is the largest torque she could get from this force? (b) A force $\mathbf{F}=(3\hat{\mathbf{i}}+2\hat{\mathbf{j}}-4\hat{\mathbf{k}})\ \text{N}$ acts at the point $\mathbf{r}=(\hat{\mathbf{i}}-\hat{\mathbf{j}}+2\hat{\mathbf{k}})\ \text{m}$. Find its torque about the origin and the magnitude of the torque.

💡 Show solution▶

(a) Lever arm $r_\perp=0.25\sin60^\circ=0.217\ \text{m}$; $\ \tau=r_\perp F=0.217\times40=8.7\ \text{N m}$.

Largest At $90^\circ$ to the handle: $\tau_\text{max}=rF=0.25\times40=10\ \text{N m}$.

(b) $\tau_x=r_yF_z-r_zF_y=(-1)(-4)-(2)(2)=0$; $\ \tau_y=r_zF_x-r_xF_z=(2)(3)-(1)(-4)=10$; $\ \tau_z=r_xF_y-r_yF_x=(1)(2)-(-1)(3)=5$. So $\boldsymbol\tau=(10\hat{\mathbf{j}}+5\hat{\mathbf{k}})\ \text{N m}$ and $\tau=\sqrt{125}=11.2\ \text{N m}$.

Check $\boldsymbol\tau\cdot\mathbf{r}=0-10+10=0$ and $\boldsymbol\tau\cdot\mathbf{F}=0+20-20=0$ ✓ — the torque is perpendicular to both $\mathbf{r}$ and $\mathbf{F}$. Part (a) is the default setting of Interactive 1.

A force $\mathbf{F}$ (red) acts on a spanner at distance $r$ from the bolt $O$, at angle $\theta$ to the handle. The green segment is the lever arm $r_\perp$ — the perpendicular from $O$ to the dashed line of action. Faint arrows split $\mathbf{F}$ into a part along the handle (no turning) and a part $F_\perp$ across it. Press Play to swing the force round and watch $\tau=rF\sin\theta$ trace a sine curve in the inset.

$r$:25 cm$F$:40 N$\theta$:60°

lever arm $r_\perp$—

$F_\perp=F\sin\theta$—

torque $\tau$—

largest possible $rF$—

Angular momentum and its rate of change

Linear momentum $\mathbf{p}=m\mathbf{v}$ measures "how much motion" a body has along a line, and force changes it: $\mathbf{F}=\dfrac{d\mathbf{p}}{dt}$. The rotational partner is built from $\mathbf{p}$ exactly as torque is built from $\mathbf{F}$.

📘 Definition 6.6 — Angular momentum of a particle

The angular momentum about a point $O$ of a particle of momentum $\mathbf{p}=m\mathbf{v}$ at position $\mathbf{r}$ (from $O$) is

$$\mathbf{l}=\mathbf{r}\times\mathbf{p},\qquad l=rp\sin\theta=r_\perp p ,$$

where $r_\perp$ is the perpendicular distance from $O$ to the line along which the particle is moving. SI unit: $\text{kg m}^2\text{/s}$ (equivalently $\text{J s}$). For a particle moving in a circle of radius $r$ about $O$, $\mathbf{r}\perp\mathbf{v}$ and $l=mvr=m\omega r^2$.

How does $\mathbf{l}$ change? Differentiate $\mathbf{r}\times\mathbf{p}$ with the product rule, which works for vector products as long as the order of each pair is kept (you will meet the product rule in mathematics):

$$\frac{d\mathbf{l}}{dt}=\frac{d\mathbf{r}}{dt}\times\mathbf{p}+\mathbf{r}\times\frac{d\mathbf{p}}{dt}=\mathbf{v}\times m\mathbf{v}+\mathbf{r}\times\mathbf{F}.$$

The first term is zero because $\mathbf{v}$ is parallel to itself, and the second is the torque. For a system of particles, add up: the internal forces come in equal and opposite pairs acting along the line joining each pair, so each pair has equal and opposite torques and the internal torques cancel.

📐 Result 6.5 — Torque and angular momentum

For a particle, and for a system of particles (with $\mathbf{L}=\sum\mathbf{l}_i$ and torques taken about the same fixed point $O$),

$$\frac{d\mathbf{l}}{dt}=\boldsymbol\tau,\qquad \frac{d\mathbf{L}}{dt}=\boldsymbol\tau_\text{ext}.$$

The internal forces are assumed to act along the lines joining the particles (true for gravity, contact forces between particles and spring forces). In particular, if the net external torque about $O$ is zero, the total angular momentum about $O$ is constant — the conservation of angular momentum, put to work in Section 6.3.

The parallel with the linear case is exact: $\mathbf{F}\leftrightarrow\boldsymbol\tau$, $\mathbf{p}\leftrightarrow\mathbf{L}$, $\mathbf{F}_\text{ext}=\dfrac{d\mathbf{P}}{dt}\leftrightarrow\boldsymbol\tau_\text{ext}=\dfrac{d\mathbf{L}}{dt}$. A force directed always towards a fixed point (a central force — the tension in a whirled string, the Sun's gravity on a planet) has zero torque about that point, so it leaves the angular momentum about that point unchanged. This is why a planet sweeps out equal areas in equal times, as you will see in Chapter 7.

✏️ Example 3 — Angular momentum of a particle and of a car

(a) A $0.20\ \text{kg}$ puck slides at a constant $5.0\ \text{m/s}$ along the line $y=3.0\ \text{m}$ in the $+x$ direction on frictionless ice. Find its angular momentum about the origin. Does it change as the puck moves? (b) A $1200\ \text{kg}$ car goes round a circular track of radius $50\ \text{m}$ at $20\ \text{m/s}$. Find its angular momentum about the centre. (c) The car then speeds up uniformly to $25\ \text{m/s}$ in $10\ \text{s}$. Find the average torque about the centre, and check it with the tangential force.

💡 Show solution▶

(a) $\mathbf{r}=x\hat{\mathbf{i}}+3.0\hat{\mathbf{j}}$, $\mathbf{p}=1.0\hat{\mathbf{i}}\ \text{kg m/s}$. $\ l_z=xp_y-yp_x=0-3.0\times1.0=-3.0$, so $\mathbf{l}=-3.0\hat{\mathbf{k}}\ \text{kg m}^2\text{/s}$ (clockwise sense, into the page) — the same for every $x$. Equivalently $l=r_\perp p=3.0\times1.0$.

Why constant No net force, so no torque: $\dfrac{d\mathbf{l}}{dt}=0$ ✓. Angular momentum is not only for things that go round.

(b) $L=mvr=1200\times20\times50=1.2\times10^6\ \text{kg m}^2\text{/s}$, vertical (up if the car goes anticlockwise seen from above).

Check Tangential acceleration $0.50\ \text{m/s}^2$ needs a tangential force $1200\times0.50=600\ \text{N}$, with lever arm $50\ \text{m}$: $\tau=3.0\times10^4\ \text{N m}$ ✓. The centripetal force passes through the centre and adds no torque.

Equilibrium of a rigid body and the centre of gravity

In Section 4.3 a particle was in equilibrium when the forces on it added to zero. A rigid body needs more: two equal and opposite forces applied at different points of a steering wheel add to zero but still turn it. By Result 6.2, zero net force keeps the centre of mass from accelerating; by Result 6.5, zero net torque keeps the angular momentum from changing. A body at rest stays at rest only if both hold.

📐 Result 6.6 — Conditions for equilibrium of a rigid body

A rigid body is in mechanical equilibrium — its centre of mass has constant velocity and its angular momentum is constant (in particular, a body at rest stays at rest) — if and only if

$$\textstyle\sum\mathbf{F}_\text{ext}=\mathbf{0}\quad(\text{translational equilibrium}),$$ $$\textstyle\sum\boldsymbol\tau_\text{ext}=\mathbf{0}\quad(\text{rotational equilibrium}).$$

When the net force is zero, the net torque is the same about every point, so the torque condition may be applied about any point you choose. For forces in one plane these are three scalar equations: $\sum F_x=0$, $\sum F_y=0$, $\sum\tau=0$.

Why any point? Move the reference point from $O$ to $O'$ at position $\mathbf{a}$. Each torque becomes $(\mathbf{r}_i-\mathbf{a})\times\mathbf{F}_i$, and the total changes by $-\mathbf{a}\times\sum\mathbf{F}_i$, which is zero when the forces balance. In practice, take torques about a point where an unknown force acts — that force then drops out of the equation.

📘 Definition 6.7 — Couple

A couple is a pair of forces of equal magnitude $F$, opposite directions and different (parallel) lines of action a perpendicular distance $d$ apart. Its net force is zero, so it causes no translation; its torque has magnitude $Fd$, the same about every point, and it produces pure rotation.

You apply a couple when you turn a tap, a steering wheel, a screwdriver or a jar lid; the Earth's magnetic field applies a couple to a compass needle. A single force can never be balanced by a couple — a couple is only balanced by an equal and opposite couple.

The oldest application of these ideas is the lever: a rigid bar that can turn about a fixed point, the fulcrum. A load $F_1$ acts at distance $d_1$ from the fulcrum and an effort $F_2$ at distance $d_2$. Taking torques about the fulcrum removes the unknown force there.

📐 Result 6.7 — Principle of moments

For a body in equilibrium under coplanar forces, about any point,

$$\text{sum of anticlockwise moments}=\text{sum of clockwise moments}.$$

For a lever with a load $F_1$ at lever arm $d_1$ and an effort $F_2$ at lever arm $d_2$ about the fulcrum, $F_1d_1=F_2d_2$. The ratio $\dfrac{F_1}{F_2}=\dfrac{d_2}{d_1}$ is the mechanical advantage; it exceeds 1 when the effort arm is longer than the load arm (a crowbar, a bottle opener, a wheelbarrow).

Centre of gravity. Every particle of a body has a weight $m_i\mathbf{g}$, and each of these forces exerts a torque. Is there a single point where the whole weight can be imagined to act?

📘 Definition 6.8 — Centre of gravity

The centre of gravity of a body is the point about which the total gravitational torque on the body is zero. Supported at that point alone, the body balances in any orientation.

📐 Result 6.8 — Centre of gravity and centre of mass

If $\mathbf{g}$ is the same in magnitude and direction at every part of the body, the gravitational torque about a point $O$ is $\sum\mathbf{r}_i\times m_i\mathbf{g}=\left(\sum m_i\mathbf{r}_i\right)\times\mathbf{g}=M\mathbf{R}_\text{cm}\times\mathbf{g}$. It vanishes when $O$ is the centre of mass, so the centre of gravity coincides with the centre of mass, and the weight $M\mathbf{g}$ may be treated as acting there.

For bodies so large that $\mathbf{g}$ varies across them (a mountain, a very tall tower) the two points differ slightly; for everyday objects the difference is negligible.

To find the centre of gravity of an irregular flat card, hang it from a pin near its edge: it comes to rest with its centre of gravity directly below the pin (only then is the torque of its weight about the pin zero). Mark the vertical line with a plumb line, hang it from a second point and mark again: the centre of gravity is where the lines cross.

🧭 Problem-solving strategy — rigid body in equilibrium

Draw the body (not a dot!) with every external force at the point where it acts: weight at the centre of gravity, normal forces and friction at contacts, tensions along strings, hinge forces (unknown direction: use two components).
Choose axes and write $\sum F_x=0$, $\sum F_y=0$.
Choose the point for torques where the most unknown forces act, so they drop out. Find each lever arm (perpendicular distance from the point to the line of action) and the sense (anticlockwise $+$, clockwise $-$).
Solve, then check by taking torques about a different point — it must also give zero.

✏️ Example 4 — Balancing a see-saw

A uniform see-saw $4.0\ \text{m}$ long, of mass $20\ \text{kg}$, is pivoted at its centre. A $30\ \text{kg}$ child sits $1.6\ \text{m}$ to the left of the pivot. (a) Where must a $40\ \text{kg}$ child sit to balance it? (b) Find the force exerted by the pivot.

💡 Show solution▶

Forces Weights $30g$ at $1.6\ \text{m}$ left, $40g$ at $d$ right, $20g$ at the centre (the pivot); pivot force $N$ upward.

(a) Torques about the pivot (the plank's weight and $N$ have zero lever arm): $30g\times1.6=40g\times d$, so $d=1.2\ \text{m}$ to the right.

(b) $\sum F_y=0$: $N=(30+40+20)\times9.8=882\ \text{N}\approx8.8\times10^2\ \text{N}$, upward.

Check Torques about the left end ($2.0\ \text{m}$ from the pivot): anticlockwise $N\times2.0=1764$; clockwise $30g(0.40)+20g(2.0)+40g(3.2)=117.6+392+1254.4=1764\ \text{N m}$ ✓. This is the default of Interactive 2.

A uniform $20\ \text{kg}$ plank $4.0\ \text{m}$ long rests on a pivot. Set each child's mass and distance from the plank's centre. Moments are taken about the pivot (anticlockwise positive); if they do not balance, the plank turns towards the larger moment until one end hits the ground. Move the pivot off-centre and the plank's own weight joins in — it acts at the centre of gravity.

A:30 kg1.6 mB:40 kg1.2 mpivot:

anticlockwise moments—

clockwise moments—

net torque—

pivot force (if balanced)—

✏️ Example 5 — A loaded beam on two supports

A uniform beam $4.0\ \text{m}$ long and of mass $50\ \text{kg}$ rests on supports A and B at its two ends. A $30\ \text{kg}$ crate sits on it $1.0\ \text{m}$ from A. Find the forces exerted by the supports.

💡 Show solution▶

Forces Upward $R_A$ (at $0$) and $R_B$ (at $4.0\ \text{m}$); downward $50g=490\ \text{N}$ at $2.0\ \text{m}$ (centre of gravity) and $30g=294\ \text{N}$ at $1.0\ \text{m}$.

Torques about A $R_B(4.0)=490(2.0)+294(1.0)=1274$, so $R_B=318.5\ \text{N}\approx3.2\times10^2\ \text{N}$.

Forces $R_A=490+294-318.5=465.5\ \text{N}\approx4.7\times10^2\ \text{N}$.

Check Torques about B: $R_A(4.0)=490(2.0)+294(3.0)=1862$, $R_A=465.5\ \text{N}$ ✓. The support nearer the crate carries more of the load, as you would expect.

✏️ Example 6 — A ladder against a smooth wall

A uniform ladder $5.0\ \text{m}$ long, of mass $20\ \text{kg}$, leans against a smooth vertical wall with its foot on rough ground $3.0\ \text{m}$ from the wall. Find (a) the force of the wall on the ladder, (b) the normal force and the friction force at the ground, (c) the least coefficient of static friction for the ladder not to slip.

💡 Show solution▶

Geometry The top is $\sqrt{5.0^2-3.0^2}=4.0\ \text{m}$ up the wall. Forces: weight $196\ \text{N}$ at the midpoint ($1.5\ \text{m}$ horizontally from the foot); wall force $F_w$, horizontal (smooth wall: no friction); at the ground, normal $N$ up and friction $f$ towards the wall.

(a) Torques about the foot ($N$ and $f$ drop out): $F_w\times4.0=196\times1.5$, so $F_w=73.5\ \text{N}\approx74\ \text{N}$.

(b) $\sum F_y=0$: $N=196\ \text{N}\approx2.0\times10^2\ \text{N}$. $\ \sum F_x=0$: $f=F_w=74\ \text{N}$.

Check Torques about the top of the ladder: $N(3.0)=196(1.5)+f(4.0)$, i.e. $588=294+294$ ✓. A person climbing the ladder moves the load's lever arm outward, so a ladder that is safe when empty may slip under a climber.

⚠️ Common mistakes

1. Order in a vector product. $\mathbf{r}\times\mathbf{F}$, not $\mathbf{F}\times\mathbf{r}$ — the second has the opposite sign. Check every component with the cyclic pattern $x\to y\to z\to x$.

2. Distance instead of lever arm. The lever arm is the perpendicular distance from the point to the line of action of the force, not the distance to the point where the force acts (unless the force is perpendicular to $\mathbf{r}$).

3. "Zero net force means equilibrium." Not for an extended body: a couple has zero net force but turns the body. Check both conditions.

4. Forgetting the body's own weight. A beam, plank or ladder has a weight acting at its centre of gravity — include it (it vanishes from the torque equation only if you take torques about the centre of gravity).

5. Torque in joules. $\text{N m}$ for torque, $\text{J}$ for work and energy — same dimensions, different physical quantities.

6. Mixing senses. Decide once which sense is positive (anticlockwise, say) and give every torque its sign before adding.

🚀 Beyond the board syllabus (JEE Advanced) — rolling without slipping: the kinematics

Rolling motion is listed in JEE Advanced 2026 only; it is not in the CBSE, JEE Main or NEET 2026 syllabi (it has been removed from the NCERT chapter).

A wheel or ball rolling on the ground combines a translation of its centre with a rotation about the centre. It rolls without slipping (pure rolling) when the point in contact with the ground is momentarily at rest — no sliding. In one turn the centre moves forward one circumference, $2\pi R$, so

$$v_\text{cm}=R\omega .$$

The velocity of any point is $\mathbf{v}_\text{cm}+\boldsymbol\omega\times\mathbf{r}$ ($\mathbf{r}$ from the centre). At the bottom the rotational part $R\omega$ points backward and cancels $v_\text{cm}$: speed $0$. At the top both point forward: speed $2v_\text{cm}$. At the ends of the horizontal diameter the two parts are perpendicular: speed $\sqrt2\,v_\text{cm}$. At each instant the wheel is turning about the contact point, and every point's velocity is perpendicular to the line joining it to that point.

Example. A bicycle wheel of radius $0.35\ \text{m}$ rolls without slipping at $7.0\ \text{m/s}$. Then $\omega=\dfrac{7.0}{0.35}=20\ \text{rad/s}$, the top of the tyre moves at $14\ \text{m/s}$, the point touching the road is at rest, and the wheel turns $\dfrac{1000}{2\pi(0.35)}=455$ times per kilometre. (The energy of rolling bodies is in the extension box of Section 6.3.)

Exercises

Take $g=9.8\ \text{m/s}^2$. Anticlockwise torques are positive unless stated otherwise.

The vector product.

1 Write down $\hat{\mathbf{i}}\times\hat{\mathbf{j}}$, $\hat{\mathbf{j}}\times\hat{\mathbf{i}}$, $\hat{\mathbf{k}}\times\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}\times\hat{\mathbf{j}}$.
2 Find $\mathbf{A}\times\mathbf{B}$ for $\mathbf{A}=2\hat{\mathbf{i}}+3\hat{\mathbf{j}}$ and $\mathbf{B}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$.
3 Find $\mathbf{C}=\mathbf{A}\times\mathbf{B}$ for $\mathbf{A}=\hat{\mathbf{i}}+2\hat{\mathbf{j}}+3\hat{\mathbf{k}}$ and $\mathbf{B}=2\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$, and verify that $\mathbf{C}$ is perpendicular to both.
4 Two sides of a triangle are $\mathbf{A}=3\hat{\mathbf{i}}+4\hat{\mathbf{j}}$ and $\mathbf{B}=-2\hat{\mathbf{i}}+5\hat{\mathbf{j}}$ (metres), drawn from a common vertex. Find the area of the triangle.
5 $|\mathbf{A}|=4$, $|\mathbf{B}|=5$ and $|\mathbf{A}\times\mathbf{B}|=10$. Find the possible angles between $\mathbf{A}$ and $\mathbf{B}$, and the possible values of $\mathbf{A}\cdot\mathbf{B}$.
6 Using the component formula, show that the $z$-components of $\mathbf{A}\times\mathbf{B}$ and $\mathbf{B}\times\mathbf{A}$ are equal and opposite.
7 Find a unit vector perpendicular to both $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\hat{\mathbf{j}}+\hat{\mathbf{k}}$.

💡 Show solutions▶

1 $\hat{\mathbf{k}}$, $\ -\hat{\mathbf{k}}$, $\ \hat{\mathbf{j}}$, $\ \mathbf{0}$.

2 Only a $z$-component: $A_xB_y-A_yB_x=2(-1)-3(1)=-5$, so $\mathbf{A}\times\mathbf{B}=-5\hat{\mathbf{k}}$.

3 $C_x=2(1)-3(-1)=5$, $C_y=3(2)-1(1)=5$, $C_z=1(-1)-2(2)=-5$: $\mathbf{C}=5\hat{\mathbf{i}}+5\hat{\mathbf{j}}-5\hat{\mathbf{k}}$. $\mathbf{A}\cdot\mathbf{C}=5+10-15=0$, $\mathbf{B}\cdot\mathbf{C}=10-5-5=0$ ✓.

4 $\mathbf{A}\times\mathbf{B}=(3\times5-4\times(-2))\hat{\mathbf{k}}=23\hat{\mathbf{k}}$; area $=\tfrac12(23)=11.5\ \text{m}^2$.

5 $\sin\theta=\dfrac{10}{20}=0.5$: $\theta=30^\circ$ or $150^\circ$. $\mathbf{A}\cdot\mathbf{B}=20\cos\theta=\pm17.3$.

6 $(\mathbf{A}\times\mathbf{B})_z=A_xB_y-A_yB_x$; $(\mathbf{B}\times\mathbf{A})_z=B_xA_y-B_yA_x=-(A_xB_y-A_yB_x)$ ✓.

7 $(\hat{\mathbf{i}}+\hat{\mathbf{j}})\times(\hat{\mathbf{j}}+\hat{\mathbf{k}})=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$, of magnitude $\sqrt3$: unit vector $\dfrac{1}{\sqrt3}(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})$ (or its negative).

Angular velocity, torque and angular momentum.

1 A ceiling fan turns at $180$ rpm. Find its angular velocity in rad/s, and the speed and centripetal acceleration of a blade tip $0.60\ \text{m}$ from the axis.
2 The fan of part 1 reaches $180$ rpm from rest in $4.0\ \text{s}$ with constant angular acceleration. Find $\alpha$ and the tangential acceleration of the blade tip.
3 Find the angular velocity of the Earth's daily spin (take one turn in $24\ \text{h}$) and the speed of a point on the equator ($R_E=6.37\times10^6\ \text{m}$).
4 A door is $0.80\ \text{m}$ wide. Find the torque about the hinge line of a $20\ \text{N}$ horizontal push (a) at the outer edge, perpendicular to the door; (b) $0.20\ \text{m}$ from the hinge, perpendicular; (c) at the outer edge, at $30^\circ$ to the plane of the door; (d) at the outer edge, directed straight at the hinge.
5 A force $\mathbf{F}=(2\hat{\mathbf{i}}+3\hat{\mathbf{j}})\ \text{N}$ acts at $\mathbf{r}=(4\hat{\mathbf{i}}+\hat{\mathbf{j}})\ \text{m}$. Find its torque about the origin.
6 A nut needs a torque of $12\ \text{N m}$ to loosen it. What is the least force that will do it with a spanner $0.30\ \text{m}$ long, and in which direction must it act?
7 A $2.0\ \text{kg}$ particle at $\mathbf{r}=3.0\hat{\mathbf{i}}\ \text{m}$ has velocity $4.0\hat{\mathbf{j}}\ \text{m/s}$. Find its angular momentum about the origin.
8 Show that a particle moving with constant velocity has constant angular momentum about any fixed point. What is its angular momentum about a point on its line of motion?

💡 Show solutions▶

1 $\omega=\dfrac{180\times2\pi}{60}=18.8\ \text{rad/s}$ ($6\pi$). $v=\omega r=11.3\ \text{m/s}$; $a_c=\omega^2r=2.1\times10^2\ \text{m/s}^2$ ($213$).

2 $\alpha=\dfrac{18.85}{4.0}=4.7\ \text{rad/s}^2$ ($4.71$); $a_t=\alpha r=2.8\ \text{m/s}^2$.

3 $\omega=\dfrac{2\pi}{86\,400}=7.27\times10^{-5}\ \text{rad/s}$; $v=\omega R_E=463\ \text{m/s}$.

4 (a) $0.80\times20=16\ \text{N m}$; (b) $0.20\times20=4.0\ \text{N m}$; (c) $0.80\times20\sin30^\circ=8.0\ \text{N m}$; (d) $0$ — the line of action passes through the hinge.

5 $\tau_z=xF_y-yF_x=4(3)-1(2)=10$: $\boldsymbol\tau=10\hat{\mathbf{k}}\ \text{N m}$ (anticlockwise).

6 Largest lever arm when $\mathbf{F}\perp$ handle: $F_\text{min}=\dfrac{12}{0.30}=40\ \text{N}$, applied at the end, perpendicular to the spanner.

7 $\mathbf{l}=\mathbf{r}\times m\mathbf{v}=3.0\hat{\mathbf{i}}\times8.0\hat{\mathbf{j}}=24\hat{\mathbf{k}}\ \text{kg m}^2\text{/s}$.

8 $l=r_\perp\,mv$: both $r_\perp$ (the distance from the point to the fixed line of motion) and $mv$ are constant, and the direction (perpendicular to the plane of the point and the line) does not change. Equivalently, $\mathbf{F}=0$ so $\boldsymbol\tau=0$ and $\dfrac{d\mathbf{l}}{dt}=0$. About a point on the line, $r_\perp=0$, so $\mathbf{l}=\mathbf{0}$.

III

Equilibrium of rigid bodies.

1 A crowbar lifts a $600\ \text{N}$ load that acts $0.20\ \text{m}$ from the fulcrum. The effort is applied $1.5\ \text{m}$ from the fulcrum on the other side. Find the effort needed and the mechanical advantage.
2 A metre stick balances on a knife-edge at its $50\ \text{cm}$ mark. A $100\ \text{g}$ mass hangs at the $20\ \text{cm}$ mark. Where must a $150\ \text{g}$ mass hang to restore balance?
3 A uniform $120\ \text{g}$ metre stick is supported at its $40\ \text{cm}$ mark and balances when a mass $m$ hangs at the $10\ \text{cm}$ mark. Find $m$.
4 A non-uniform plank $1.5\ \text{m}$ long rests horizontally on two scales, one under each end. They read $60\ \text{N}$ (end A) and $90\ \text{N}$ (end B). Find the weight of the plank and the position of its centre of gravity.
5 A uniform beam $6.0\ \text{m}$ long, of mass $80\ \text{kg}$, rests on supports $1.0\ \text{m}$ from each end. A $40\ \text{kg}$ load hangs from the right-hand end. Find the forces of the two supports.
6 A driver turns a steering wheel of diameter $0.40\ \text{m}$ by pushing up with $15\ \text{N}$ on one side and pulling down with $15\ \text{N}$ on the other, both forces tangent to the rim. Find the net force and the torque.
7 A uniform ladder $4.0\ \text{m}$ long, of mass $15\ \text{kg}$, rests against a smooth wall at $60^\circ$ to the horizontal ground. Find the force of the wall, the normal and friction forces at the ground, and the least coefficient of static friction.
8 A $20\ \text{kg}$ shop sign hangs from the end of a light horizontal rod $1.5\ \text{m}$ long, hinged to a wall. A cable runs from the end of the rod to a point on the wall $1.0\ \text{m}$ above the hinge. Find the tension in the cable and the force of the hinge.

💡 Show solutions▶

1 $F\times1.5=600\times0.20$: $F=80\ \text{N}$. Mechanical advantage $\dfrac{600}{80}=7.5$ $\left(=\dfrac{1.5}{0.20}\right)$.

2 $100\times30=150\times d$: $d=20\ \text{cm}$ right of the knife-edge, i.e. at the $70\ \text{cm}$ mark.

3 The stick's weight acts at $50\ \text{cm}$, $10\ \text{cm}$ right of the support; $m$ is $30\ \text{cm}$ left: $m(30)=120(10)$, $m=40\ \text{g}$.

4 $W=60+90=150\ \text{N}$. Torques about A: $150x=90\times1.5$, $x=0.90\ \text{m}$ from A (nearer the end with the larger reading).

5 Supports at $x=1.0$ and $5.0\ \text{m}$. About the left support: $R_2(4.0)=80g(2.0)+40g(5.0)=360g$, $R_2=90g=882\ \text{N}$; $R_1=120g-90g=30g=294\ \text{N}$.

6 Net force zero; the forces form a couple of torque $Fd=15\times0.40=6.0\ \text{N m}$.

7 About the foot: $F_w(4.0\sin60^\circ)=147(2.0\cos60^\circ)$, $F_w=42\ \text{N}$ ($42.4$). $N=147\ \text{N}$, $f=42\ \text{N}$, $\mu_s\ge\dfrac{42.4}{147}=0.29$ ($=\tfrac12\cot60^\circ$).

8 Cable at angle $\phi$ to the rod with $\tan\phi=\dfrac{1.0}{1.5}$, $\sin\phi=0.555$. Torques about the hinge: $T\sin\phi\,(1.5)=196(1.5)$, $T=\dfrac{196}{0.555}=3.5\times10^2\ \text{N}$ ($353$). Hinge: horizontal $T\cos\phi=294\ \text{N}$ (pushing the rod away from the wall), vertical $196-T\sin\phi=0$. So the hinge force is $2.9\times10^2\ \text{N}$, horizontal.

Applications and harder problems.

1 (Modelling) A forearm is held horizontal with a $3.0\ \text{kg}$ ball in the hand, $35\ \text{cm}$ from the elbow. The forearm and hand ($1.5\ \text{kg}$) have their centre of gravity $15\ \text{cm}$ from the elbow. The biceps pulls vertically upward on the forearm $4.0\ \text{cm}$ from the elbow. Find the biceps force and the force at the elbow joint.
2 A uniform plank $2.0\ \text{m}$ long, of mass $10\ \text{kg}$, lies on a table with $0.60\ \text{m}$ sticking out over the edge. How far beyond the edge can an $8.0\ \text{kg}$ dog walk before the plank tips? Can it reach the end?
3 Prove that if the net force on a body is zero, the net torque is the same about every point.
4 Estimate the angular momentum of the Earth in its orbit about the Sun ($m=5.97\times10^{24}\ \text{kg}$, orbit radius $1.50\times10^{11}\ \text{m}$, period $3.16\times10^7\ \text{s}$), treating the orbit as a circle.
5 A $50\ \text{g}$ ball on a string is whirled in a horizontal circle of radius $0.80\ \text{m}$ at $2.0$ revolutions per second. Find its angular momentum about the centre.
6 (Modelling) A tower crane's jib holds a $2.0\ \text{t}$ load $30\ \text{m}$ from the tower; its counterweight is $10\ \text{m}$ from the tower on the other side. Ignoring the jib's own weight, what counterweight makes the net torque about the tower zero? With that counterweight, what is the largest load at $15\ \text{m}$? Why do crane drivers work from a chart of "load × distance"?

💡 Show solutions▶

1 About the elbow: $F(0.040)=29.4(0.35)+14.7(0.15)=10.29+2.21=12.50$, $F=3.1\times10^2\ \text{N}$ ($312$) — more than seven times the ball's weight. Joint: $312-29.4-14.7=268\ \text{N}\approx2.7\times10^2\ \text{N}$, pushing down on the forearm.

2 Tipping about the table edge. The plank's centre of gravity is $1.0-0.60=0.40\ \text{m}$ inside the edge. Tips when $8.0x=10(0.40)$: $x=0.50\ \text{m}$. The overhang is $0.60\ \text{m}$, so the plank tips before the dog reaches the end.

3 About $O'$ at $\mathbf{a}$: $\boldsymbol\tau'=\sum(\mathbf{r}_i-\mathbf{a})\times\mathbf{F}_i=\sum\mathbf{r}_i\times\mathbf{F}_i-\mathbf{a}\times\sum\mathbf{F}_i=\boldsymbol\tau-\mathbf{0}$ ✓.

4 $v=\dfrac{2\pi r}{T}=2.98\times10^4\ \text{m/s}$; $L=mvr=2.67\times10^{40}\ \text{kg m}^2\text{/s}$.

5 $\omega=4\pi=12.6\ \text{rad/s}$; $L=m\omega r^2=0.050\times12.57\times0.64=0.40\ \text{kg m}^2\text{/s}$.

6 $M_c(10)=2.0(30)$: $M_c=6.0\ \text{t}$. At $15\ \text{m}$: $6.0(10)=m(15)$, $m=4.0\ \text{t}$. What matters for tipping and for the stress in the tower is the load's moment, $60\ \text{t m}$ here, so the safe load falls as the distance grows.

🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks Define the torque of a force about a point. Give its SI unit and dimensional formula. Torque and work have the same dimensions; why is torque never expressed in joules?

💡 Show solution▶

Definition $\boldsymbol\tau=\mathbf{r}\times\mathbf{F}$, where $\mathbf{r}$ is the position of the point of application relative to the point; magnitude $rF\sin\theta$ = force × perpendicular distance of its line of action. [1]

Units $\text{N m}$; $[\text{M L}^2\text{T}^{-2}]$. Torque is a vector (a turning effect about an axis) and the $\text{m}$ is a lever arm perpendicular to the force; work is a scalar (energy transferred) with the displacement along the force. The joule is reserved for energy. [1]

Exam tip Equal dimensions do not make two quantities the same — frequency and angular velocity, for instance, both have dimensions $[\text{T}^{-1}]$.

E2. NEET For two non-zero vectors, $\mathbf{A}\times\mathbf{B}=\mathbf{B}\times\mathbf{A}$. The angle between $\mathbf{A}$ and $\mathbf{B}$ could be (a) $\dfrac{\pi}{2}$ (b) $\dfrac{\pi}{4}$ (c) $\dfrac{\pi}{3}$ (d) $\pi$

💡 Show solution▶

Idea $\mathbf{B}\times\mathbf{A}=-\mathbf{A}\times\mathbf{B}$, so the condition means $\mathbf{A}\times\mathbf{B}=-\mathbf{A}\times\mathbf{B}$, i.e. $\mathbf{A}\times\mathbf{B}=\mathbf{0}$: $\sin\theta=0$, $\theta=0$ or $\pi$.

Answer (d).

Exam tip (a) is the condition for $\mathbf{A}\cdot\mathbf{B}=0$ — do not confuse the two products.

E3. JEE Main (Numerical value) A force $\mathbf{F}=(2\hat{\mathbf{i}}+3\hat{\mathbf{j}}-\hat{\mathbf{k}})\ \text{N}$ acts at the point $(1,\,-1,\,2)\ \text{m}$. Find the magnitude of its torque, in N m to two decimal places, about the point $(0,\,1,\,1)\ \text{m}$.

💡 Show solution▶

Set-up $\mathbf{r}$ from the reference point: $(1-0,\,-1-1,\,2-1)=(1,\,-2,\,1)\ \text{m}$.

Product $\tau_x=(-2)(-1)-(1)(3)=-1$, $\ \tau_y=(1)(2)-(1)(-1)=3$, $\ \tau_z=(1)(3)-(-2)(2)=7$.

Answer $\tau=\sqrt{1+9+49}=\sqrt{59}=7.68\ \text{N m}$.

Exam tip Using the position vector of the point of application from the origin ($1,-1,2$) is the classic slip: it gives $\sqrt{75}=8.66\ \text{N m}$.

E4. JEE Advanced (One or more correct) A uniform ladder of weight $W$ and length $L$ rests against a smooth vertical wall, making an angle $\theta$ with the rough horizontal floor, and is on the point of slipping. Which statements are correct? (A) The force of the wall is $\tfrac12W\cot\theta$. (B) The normal force of the floor is $W$. (C) The friction force at the floor is $\tfrac12W\tan\theta$. (D) The coefficient of friction at the floor is $\tfrac12\cot\theta$.

💡 Show solution▶

(A) Torques about the foot: $F_w(L\sin\theta)=W\left(\tfrac{L}{2}\cos\theta\right)$, so $F_w=\tfrac12W\cot\theta$ ✓.

(B) The only vertical forces are $N$ and $W$ (the smooth wall pushes horizontally): $N=W$ ✓.

(D) On the point of slipping $f=\mu N$: $\mu=\dfrac{\tfrac12W\cot\theta}{W}=\tfrac12\cot\theta$ ✓.

Answer (A), (B), (D). Check with Example 6: $\theta$ with $\cot\theta=\tfrac34$ gives $\mu=0.375$ ✓.

Exam tip A steeper ladder (larger $\theta$) needs less friction — limiting case $\theta=90^\circ$: no friction needed.

E5. JEE Main NEET A uniform metre stick balances horizontally on a knife-edge at its $40\ \text{cm}$ mark when a $50\ \text{g}$ mass hangs at its $10\ \text{cm}$ mark. The mass of the metre stick is (a) $50\ \text{g}$ (b) $100\ \text{g}$ (c) $150\ \text{g}$ (d) $200\ \text{g}$

💡 Show solution▶

Idea The stick's weight acts at $50\ \text{cm}$, $10\ \text{cm}$ right of the knife-edge; the $50\ \text{g}$ is $30\ \text{cm}$ left. $50\times30=M\times10$.

Answer $M=150\ \text{g}$: (c).

Exam tip (a) forgets the stick's weight acts at its centre; measuring arms from the end of the stick instead of the knife-edge leads to the other options.

E6. Board · Case study A uniform diving board $4.0\ \text{m}$ long, of mass $30\ \text{kg}$, is held by a bolt at its left end A and rests on a support B $1.0\ \text{m}$ from A. A $60\ \text{kg}$ diver stands at the right-hand end. (i) Find the force of support B. [2] (ii) Find the force of the bolt at A, with its direction. [1] (iii) Explain why the board must be bolted at A rather than simply resting on two supports. [1]

💡 Show solution▶

(i) Torques about A: $F_B(1.0)=30g(2.0)+60g(4.0)=300g$, $F_B=2940\ \text{N}\approx2.9\times10^3\ \text{N}$ upward. [2]

(ii) $\sum F_y=0$: $F_A+2940=90g=882$, so $F_A=-2058\ \text{N}$: $2.1\times10^3\ \text{N}$ downward. [1]

(iii) About B, both the board's weight ($1.0\ \text{m}$ beyond B) and the diver ($3.0\ \text{m}$ beyond B) give clockwise torques; nothing on the A side could balance them by pushing up. The bolt must pull down at A to supply the anticlockwise torque. [1]

Exam tip A negative answer for an assumed-upward force simply means it acts downward. Check about B: $2058(1.0)=30g(1.0)+60g(3.0)=294+1764$ ✓.

🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. Why do workplace safety guidelines tell us to "lift with the legs, not the back" — how large are the forces in the lower back when lifting a box?

Why it matters. Lower-back injury is one of the commonest causes of lost working days worldwide, among warehouse workers, nurses, farm labourers and parents carrying children. Occupational-safety agencies in many countries publish lifting limits based on exactly the kind of torque model you can build with Result 6.7. The key fact: back muscles attach very close to the spine, so they work at a huge mechanical disadvantage.

What to do. Model the upper body as a rigid lever pivoting at the base of the spine. The back muscles pull along the spine with a lever arm of about $5.0\ \text{cm}$. Take a $70\ \text{kg}$ person whose upper body (head, arms and trunk) is $60\%$ of body mass, lifting a $15\ \text{kg}$ box.

Stooping (back bent forward, knees straight): the upper body's centre of gravity is $0.25\ \text{m}$ (horizontally) in front of the pivot and the box $0.50\ \text{m}$. Find the total torque of the two weights about the pivot and the muscle force needed to balance it.
Squatting (back nearly upright, box held close): take $0.10\ \text{m}$ and $0.25\ \text{m}$. Repeat.
Compare the muscle forces with the box's weight. Roughly what multiple of the box's weight must the muscles supply in each case?
The spine is squeezed by the muscle force plus the weight above. Estimate the compressive force in each case and compare with a typical injury threshold of a few kilonewtons quoted in the ergonomics literature.
Draw a graph of muscle force against the horizontal distance of the box from the pivot. What practical advice follows?

Tools: Calculator or spreadsheet; photographs of someone lifting (to estimate distances); ergonomics guidance from a national occupational-safety body.

Hand in: A labelled force diagram for each posture, your calculations, the graph, and a short "lifting safety" poster with one physics sentence explaining each tip.

Going further. The muscles do not pull exactly along the spine, and the trunk is not a single rigid lever. Find out how biomechanists refine this model (several vertebrae, muscle angles) and whether the refinements raise or lower the estimate.

Instructor note: upper-body weight $0.60\times70\times9.8=412\ \text{N}$, box $147\ \text{N}$. Stooping: $412(0.25)+147(0.50)=176\ \text{N m}$, muscle force $\approx3.5\ \text{kN}$ (about $24$ times the box's weight). Squatting: $412(0.10)+147(0.25)=77.9\ \text{N m}$, muscle force $\approx1.6\ \text{kN}$. Compression roughly muscle force plus $0.56\ \text{kN}$: about $4\ \text{kN}$ versus $2\ \text{kN}$, so stooping approaches commonly quoted limits of about $3.4\ \text{kN}$. The dominant variable is the horizontal distance of the load — keep it close.

🧪 Self-check quiz

0/5 correct

Q1. For right-handed axes, $\hat{\mathbf{k}}\times\hat{\mathbf{j}}$ equals

Q2. A $30\ \text{N}$ force acts at the end of a $0.20\ \text{m}$ spanner at $30^\circ$ to the handle. The torque about the bolt is

Q3. A couple acting on a rigid body at rest

Q4. A particle moves in a straight line at constant velocity. Its angular momentum about a fixed point not on the line

Q5. A rigid body is in mechanical equilibrium when

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