Exercises — Torque τ = r × F — definition, physical meaning

What we know: perpendicular means $\theta =90^{\circ }$, and $\sin 90^{\circ }=1$. Why $\sin$: only the sideways part of the push spins the bolt (the mint arrow in the figure); at $90^{\circ }$ all of the force is sideways, so we get the maximum possible twist. $\tau =rF\sin \theta =0.25\times 20\times 1=5\text{N⋅m}.$ Answer: $\tau =5\text{N⋅m}$.

What we know: "along the spanner toward the bolt" means $\vec{F}$ points the same way (or opposite way) as $\vec{r}$, so $\theta =0^{\circ }$ (or $180^{\circ }$). Why zero: a push aimed at the pivot cannot circulate anything — like pushing a door into its hinges. Picture the parallelogram of $\vec{r}$ and $\vec{F}$: when they line up it is flat, area $=0$. $\tau =0.25\times 20\times \sin 0^{\circ }=0.25\times 20\times 0=0\text{N⋅m}.$ Answer: $\tau =0\text{N⋅m}$.

Unit: newton-metre, $\text{N⋅m}$. Kind: torque is a vector — it has both a size and a rotation axis / sense (right-hand rule). It shares dimensions with the joule but is not energy.

Step 1 — pick the tool: we know $r$, $F$, and the angle between them, so $\tau =rF\sin \theta$ is the direct route. Step 2 — why $\sin 30^{\circ }$: only the perpendicular slice of the force, $F\sin \theta$, twists (the mint arrow in the figure). $\sin 30^{\circ }=0.5$, so exactly half the force is useful. $\tau =0.40\times 60\times 0.5=12\text{N⋅m}.$ Answer: $\tau =12\text{N⋅m}$.

Step 1 — pick the tool: the vectors are given in components and lie in the $xy$-plane, so the 2D shortcut is fastest: ${\tau }_z=x{F}_y-y{F}_x$. Step 2 — plug in: here $x=0,y=2,F_x=3,F_y=4$. ${\tau }_z=(0)(4)-(2)(3)=-6.$ Step 3 — read the sign (convention from the top): negative ${\tau }_z$ means clockwise, i.e. the torque vector points into the page ($-\hat{k}$). Answer: $\vec{\tau }=-6\hat{k}\text{N⋅m}$ (clockwise).

Why no angle needed: the two readings of $rF\sin \theta$ are (full distance) × (sideways force) and (perpendicular distance) × (full force) — the two ways of grouping the figure. If someone already handed you the perpendicular distance $d_{\perp }=r\sin \theta$ (the butter arrow), the $\sin \theta$ is baked into it — so you simply multiply by the whole force. $\tau =d_{\perp }F=0.15\times 80=12\text{N⋅m}.$ Answer: $\tau =12\text{N⋅m}$ (magnitude; sign depends on which way this force would spin the body).

Step 1: A's torque: ${\tau }_A=0.90\times 50\times \sin 90^{\circ }=45\text{N⋅m}$. Step 2: B's torque: ${\tau }_B=0.90\times 50\times \sin 50^{\circ }$. With $\sin 50^{\circ }\approx 0.766$, ${\tau }_B\approx 34.47\text{N⋅m}$. Step 3 — compare: extra fraction $=\frac{{\tau }_A-{\tau }_B}{{\tau }_B}=\frac{45-34.47}{34.47}\approx 0.305$. Answer: A produces about $30.5\%$ more torque than B, purely from pushing at the better angle.

(a) Component route: ${\tau }_z=x{F}_y-y{F}_x=(1)(-2)-(3)(5)=-2-15=-17$. So $\vec{\tau }=-17\hat{k}\text{N⋅m}$. By the sign convention from the top, negative ${\tau }_z$ = clockwise. (b) Magnitude cross-check.

• $r=\sqrt{1^2+3^2}=\sqrt{10}$.
• $F=\sqrt{5^2+(-2{)}^2}=\sqrt{29}$.
• The dot product gives $\cos \theta$: $\vec{r}\cdot \vec{F}=(1)(5)+(3)(-2)=5-6=-1$, so $\cos \theta =\frac{-1}{\sqrt{10}\sqrt{29}}$.
• Then $\sin \theta =\sqrt{1-{\cos }^2\theta }$, and $rF\sin \theta =\sqrt{10}\sqrt{29}\sin \theta$.

Numerically $\sqrt{290}\approx 17.03$, $\cos \theta \approx -0.0587$, $\sin \theta \approx 0.9983$, so $rF\sin \theta \approx 17.03\times 0.9983\approx 17.0$. ✓ Answer: $|\vec{\tau }|=17\text{N⋅m}$, matching the component method. (The magnitude formula loses the sign; the component formula keeps it — hence we know it is clockwise.)

Set-up: $\vec{r}$ points along the rod (length $1.2\text{m}$, at $60^{\circ }$ above horizontal). $\vec{F}$ points straight down. The angle between an arrow tilted $60^{\circ }$ above horizontal and a downward arrow is $\theta =90^{\circ }+60^{\circ }=150^{\circ }$. Why the lever arm is neat here: the perpendicular distance from the pivot to the vertical line of the weight (the butter arrow in the figure) is the horizontal reach, $d_{\perp }=r\cos \varphi =1.2\cos 60^{\circ }=0.6\text{m}$. $\tau =d_{\perp }F=0.6\times 40=24\text{N⋅m}.$ Cross-check with $\sin \theta$: $\sin 150^{\circ }=0.5$, so $\tau =1.2\times 40\times 0.5=24\text{N⋅m}$. ✓ Answer: $\tau =24\text{N⋅m}$ (the weight spins the rod clockwise down, so as a signed quantity $\tau =-24\text{N⋅m}$).

Why we can just add: torque about a fixed axis is a signed number; the net twist is the sum of the individual signed twists. Using the convention from the top (anticlockwise $=+$):

• Downward force at $0.5\text{m}$: it tends to rotate the beam clockwise → negative. Magnitude $=0.5\times 30=15\text{N⋅m}$. Contribution $=-15$.
• Upward force at $2.0\text{m}$: it tends to rotate the beam anticlockwise → positive. Magnitude $=2.0\times 50=100\text{N⋅m}$. Contribution $=+100$.
• (Both forces are perpendicular to the horizontal beam, so $\sin \theta =1$ throughout.) ${\tau }_{\text{net}}=+100-15=+85\text{N⋅m}.$ Answer: ${\tau }_{\text{net}}=85\text{N⋅m}$, anticlockwise.

(a) Torque — why the lever arm is exactly $R$: the rope leaves the rim tangentially, i.e. along the circle's edge. The position vector $\vec{r}$ from the centre to that point of the rim is a radius, which always meets the tangent at a right angle. So $\theta =90^{\circ }$, $\sin \theta =1$, and the perpendicular distance from the centre to the rope's line of action equals the full radius $R$ (the butter arrow lies right along the radius). Hence: $\tau =RF\sin 90^{\circ }=RF=0.10\times 8=0.8\text{N⋅m}.$ (b) Angular acceleration — new tool: just as $F=ma$ links force to linear acceleration, Newton's Second Law for Rotation links torque to angular acceleration through the Moment of Inertia: $\tau =I\alpha ⟹\alpha =\frac{\tau }{I}=\frac{0.8}{0.02}=40{\text{rad/s}}^2.$ Answer: $\tau =0.8\text{N⋅m}$, $\alpha =40{\text{rad/s}}^2$.

Why these steps: angular velocity grows at a steady rate $\alpha$ from rest, so $\omega =\alpha t$. Then Angular Momentum is $L=I\omega$ — the rotational cousin of linear momentum $p=mv$.

• $\omega =\alpha t=40\times 3=120\text{rad/s}$.
• $L=I\omega =0.02\times 120=2.4{\text{kg⋅m}}^2/\text{s}$. Answer: $L=2.4{\text{kg⋅m}}^2/\text{s}$.

Why torque about the foot: at the foot both the ground's normal force and its friction act — taking torques there makes those two unknowns vanish (their lever arm is zero), leaving one equation in one unknown. This is the standard Equilibrium of Rigid Bodies trick. Weight's torque (clockwise, tends to slide the top down):

• The weight acts at the midpoint, $2\text{m}$ up the ladder (Center of Mass & Gravity).
• Its horizontal lever arm from the foot $=2\cos 60^{\circ }=1.0\text{m}$.
• ${\tau }_W=W\times 1.0=200\text{N⋅m}$ (clockwise → $-$ in our convention). Wall force's torque (anticlockwise):
• The wall pushes horizontally at the top, height $=L\sin 60^{\circ }=4\sin 60^{\circ }=3.464\text{m}$.
• Its vertical lever arm from the foot is that height: ${\tau }_N=N\times 3.464$ (anticlockwise → $+$). Balance (net torque $=0$): $N\times 3.464-200=0⟹N=\frac{200}{3.464}\approx 57.7\text{N}.$ Answer: $N\approx 57.7\text{N}$.

Why torque about the hinge: the hinge's reaction force acts there with zero lever arm, so it drops out — leaving only $W$ and $T$. Weight's torque (clockwise → $-$): the weight pulls down at the free end, $1.5\text{m}$ out, perpendicular to the horizontal rod ($\sin 90^{\circ }=1$): ${\tau }_W=1.5\times 120=180\text{N⋅m (clockwise)}.$ Wire's torque (anticlockwise → $+$): the wire pulls at the free end at $40^{\circ }$ to the rod, so only the perpendicular slice $T\sin 40^{\circ }$ twists the rod; its lever arm is the full rod length $1.5\text{m}$: ${\tau }_T=1.5\times T\sin 40^{\circ }\text{(anticlockwise)}.$ Balance (net torque about the hinge $=0$): $1.5T\sin 40^{\circ }-180=0⟹1.5T\sin 40^{\circ }=180.$ Solve for $T$: $T=\frac{180}{1.5\sin 40^{\circ }}=\frac{120}{\sin 40^{\circ }}\approx \frac{120}{0.6428}\approx 186.7\text{N}.$ Answer: $T\approx 186.7\text{N}$.

Step 1 — angular acceleration: $\alpha =\tau /I=4/0.5=8{\text{rad/s}}^2$ (from Newton's Second Law for Rotation). Step 2 — angle turned: using the rotational analogue of $v^2=2as$, namely ${\omega }^2=2\alpha \vartheta$: $\vartheta =\frac{{\omega }^2}{2\alpha }=\frac{20^2}{2\times 8}=\frac{400}{16}=25\text{rad}.$ Step 3 — rotational work: the Work-Energy Theorem (Rotational) gives work $=\tau \vartheta$ for a constant torque: $W=\tau \vartheta =4\times 25=100\text{J}.$ Step 4 — check against kinetic energy: rotational KE $=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times 0.5\times 20^2=0.25\times 400=100\text{J}$. ✓ Answer: $\vartheta =25\text{rad}$, $W=100\text{J}$, equal to the KE gained.

Which formula gives torque directly from components? ::: ${\tau }_z=x{F}_y-y{F}_x$ (2D, signed). Which formula gives torque from magnitudes and angle? ::: $\tau =rF\sin \theta$. Why take torques about a hinge/pivot in equilibrium problems? ::: The reaction force there has zero lever arm and drops out, leaving fewer unknowns. Link torque to angular acceleration? ::: $\tau =I\alpha$. Rotational work of a constant torque? ::: $W=\tau \vartheta$, with $\vartheta$ in radians.