Physics · Mechanics

Torque: reference point, lever arm and axis

Q: How do you calculate torque?

Vector torque about O is $\vec M_O=\vec r\times\vec F$, with r⃗ pointing from O to the force application point. Its magnitude is $rF\sin\theta$; alternatively $|\vec M|=Fd$, where d is shortest distance from O to the line of action. r equals d only for perpendicular force. An 80 N force perpendicular at 0.25 m gives $20 N\cdot{}m$; at 60° between r⃗ and F⃗ it gives approximately $17.3 N\cdot{}m$. For a selected axis, use the corresponding signed projection rather than automatically using full magnitude. In the xy plane, $M_z=r_xF_y-r_yF_x$. Define reference point, axis and positive turning sense before substituting values.

Q: What does the law of the lever state and how is it related to torque?

The ideal lever law $F_1d_1=F_2d_2$ describes two opposing moments about the same pivot. d₁ and d₂ are perpendicular distances to the force lines. For a horizontal lever under vertical loads, they equal the usual arm lengths. A 30 kg body at 2 m balances 40 kg at 1.5 m when other torques and self-weight are neglected. g cancels in this torque balance. Static equilibrium additionally requires zero net force, so the support reaction must be included. Zero net torque alone does not establish rest: a free body can rotate uniformly or accelerate linearly despite that torque balance.

Q: Why does a longer lever save force?

For a required torque magnitude, $F=\frac{|M|}{d}$, where d is perpendicular distance to the force line. With perpendicular force, $40 N\cdot{}m$ requires 200 N at a 20 cm wrench and 100 N at a 40 cm wrench. Oblique application requires accounting for angle; a longer rod alone does not guarantee the same effective arm. For an equal rotation angle, the farther-out handle travels along a longer arc. With the same ideal torque-angle balance, work is therefore unchanged. A pipe extension may also alter tool loading; the physics calculation does not replace checking permitted tool use.

The cross product of position and force describes turning effect. $Fr$ ·sinθ is magnitude; torque about a selected axis is the corresponding signed projection.

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Formula

Calculate

M = F \cdot r \cdot sin (θ)

M in N·m · F in N · r in m · θ in rad or degrees with matching calculation mode

Variables & units

Variables & units
Symbol	Meaning	Unit
$M$	Torque-vector magnitude in the illustrated planar case	N·m
$F$	Applied-force magnitude	N
$r$	Distance from reference point to application point; $d=r\sin\theta$ is perpendicular lever arm	m
$θ$	Angle between position and force vectors	rad or °

LaTeX

M = F \cdot r \cdot \sin(\theta)

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Diagram: a lever of length r about a pivot axis, at its end the force F acts at the angle θ, dashed the perpendicular component F·sin θ, at the axis a turning arrow M. — r is distance to the application point; perpendicular component F sinθ produces torque magnitude rFsinθ. The perpendicular lever arm is d = r sinθ.

Torque calculator

Find torque, force or lever arm, where the lever arm is the perpendicular distance from the pivot to the line of action of the force.

Find

Force magnitude F

Size of the force on the lever, e.g. 300 N when you push hard on a wheel wrench.

Perpendicular lever arm d

Perpendicular distance from the pivot to the line of action of the force, e.g. 40 cm on a long wrench.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for |M|, substitute, evaluate

1. Rearrange for |M|
$|M|=Fd$
2. Substitute values (in SI units)
- d = 40 cm = 0.4 m
$|M|=300\cdot 0.4=120$
3. Result
|M| = 120 N·m

Result: Torque magnitude |M|

|M| = 120 N·m

With F = 300 N and d = 40 cm, |M| = 120 N·m.

Quick tips

Formula triangle: M on top, F and d below. Cover the quantity you need and the rest is the calculation, e.g. F = M ÷ d.
If the force acts at a slant, only the perpendicular lever arm d = r · sin θ counts, with r the distance to the point of application and θ the angle between r and F. At θ = 30°, d = r ÷ 2.

Assumptions and limits

M = Fd with nonnegative magnitudes. d is perpendicular distance from the reference point to the force line, not arbitrary rod length. Rotation sense, general spatial direction and axial projection are not determined.

Derivation & background

Vector torque is $\vec M=\vec r\times\vec F$ . Only perpendicular force contributes to magnitude. Static equilibrium requires both zero net force and zero net torque. A fixed inertial reference gives $\sum\vec M=\frac{\mathrm d\vec L}{\mathrm dt}$ ; scalar $\sum M_z=I_z\alpha_z$ applies to a rigid fixed-distribution body about a fixed axis.

Exam blueprint

Validity range

Specify reference point/axis; $|\vec M|=rF\sin\theta$ is magnitude. Axis components require projection and signs. $\sum M_z=I_z\alpha_z$ requires a matching rigid body, constant I and an axis fixed in an inertial frame.

Derivation steps

Torque is a cross product; magnitude uses perpendicular lever arm, while an axis uses the relevant projection.

1Choose reference point O or an axis through O and its positive sense.
2Position vector r⃗ runs from O to the application point. Torque is $\vec M_O=\vec r\times\vec F$ .
3Split F⃗ parallel and perpendicular to r⃗; the perpendicular part is $F\sin\theta$ . The parallel part has zero cross product.
4Thus $|\vec M|=rF\sin\theta=Fd$ , where $d=r\sin\theta$ is shortest distance to the line of action.
5The right-hand rule determines vector direction. About z use signed $M_z=r_xF_y-r_yF_x$ , not necessarily $|\vec M|$ .
6Sum external torques about the same reference. Only a rigid constant-I body about a fixed axis simplifies dynamics to $\sum M_z=I_z\alpha_z$ .

Rearrangements

Required force

F=\frac{|\vec M|}{r\sin\theta}

For torque magnitude with positive perpendicular lever arm. Increasing effective arm reduces required force; changing angle can remove that benefit.

Law of the lever

F_1 \cdot r_1 = F_2 \cdot r_2

Here r₁ and r₂ denote perpendicular lever arms; the two load torques must have opposite signs. Static equilibrium additionally needs force balance.

Task variant

A bolt needs $40 N\cdot{}m$ . What force is required on a 20 cm lever (perpendicular)?

$F=\frac{M}{r}=\frac{40\,\mathrm{N\,m}}{0.2\,\mathrm m}=200\,\mathrm N$ .
A wrench twice as long halves the force to 100 N.

An ideal horizontal seesaw has negligible self-weight. A 30 kg child sits 2 m from its pivot. Where must a 40 kg child sit on the opposite side to balance load torques?

Law of the lever: $m_1r_1=m_2r_2$ ; g cancels.
$r_2=\frac{30\,\mathrm{kg}\cdot2\,\mathrm m}{40\,\mathrm{kg}}=1.5\,\mathrm m$ .

Common mistakes

Dropping the sin θ factor for an obliquely applied force.

Magnitude uses $F_\perp=F\sin\theta$ . At $90^\circ$ sine is one; oblique application gives a smaller effective component.

Confusing position distance r with perpendicular lever arm d in one calculation.

In $|M|=rF\sin\theta$ , r is distance to force application. In $|M|=Fd$ , d is perpendicular distance to the line of action. They coincide only for perpendicular application.

Equating $N\cdot{}m$ with joules.

Torque is turning effect, not stored energy. Rotational work requires $W=\int M_z\,\mathrm d\theta$ ; angle and displacement belong to the calculation.

Ignoring the sense of rotation and adding all torques.

Balance counterclockwise and clockwise torques with opposite signs.

Exam context

Tasks on levers, seesaws and torque wrenches as well as equilibrium conditions ( $\sum\vec F=0$ and $\sum\vec M=0$ ) for beams and cranes.

Formula cluster

Statics and rotation

Torque plays the role of force for rotational motion.

a⃗×b⃗ = (a₂b₃−a₃b₂ | a₃b₁−a₁b₃ | a₁b₂−a₂b₁)

Cross Product (Vector Product)

ΣF_ext = F_res = ma

Newton's Second Law

Worked example

A wrench receives $F=80\,\mathrm N$ at position distance $r=0.25\,\mathrm m$ . Perpendicular application gives $|M|=rF=20\,\mathrm{N\,m}$ . At $60^\circ$ between position vector and force, $|M|=rF\sin60^\circ\approx17.3\,\mathrm{N\,m}$ ; the perpendicular lever arm is then shorter than the position distance.

Applications

Bolted joints (torque wrench), engines (Nm rating), levers and pulleys, seesaw, statics of beams and cranes

How do you calculate with Torque?

Here is how to work through a typical Torque (M = F·r·sin(θ)) task step by step:

1
Task
A bolt needs $40 N\cdot{}m$ . What force is required on a 20 cm lever (perpendicular)?
Solution path
1. $F=\frac{M}{r}=\frac{40\,\mathrm{N\,m}}{0.2\,\mathrm m}=200\,\mathrm N$ .
2. A wrench twice as long halves the force to 100 N.
2
Task
An ideal horizontal seesaw has negligible self-weight. A 30 kg child sits 2 m from its pivot. Where must a 40 kg child sit on the opposite side to balance load torques?
Solution path
1. Law of the lever: $m_1r_1=m_2r_2$ ; g cancels.
2. $r_2=\frac{30\,\mathrm{kg}\cdot2\,\mathrm m}{40\,\mathrm{kg}}=1.5\,\mathrm m$ .

Understand and apply the equation

A door is easier to open at its handle than near its hinge. Force magnitude alone is not enough: distance and pulling direction also matter. Torque summarises this turning effect. Start with the perpendicular lever arm, then explore turning sense and, in the advanced discussion, the three-dimensional vector description.

How do you specify reference point, axis and sense of rotation?

Choose the pivot and draw the force’s straight line of action. Its shortest distance from the pivot is the lever arm. Pulling directly toward the hinge produces no turning effect about that hinge, although the force is not zero. The vectors below express this geometry precisely.[1]

\begin{aligned}\vec M_O&=\vec r\times\vec F\\|\vec M_O|&=rF\sin\theta\end{aligned}

r⃗ points from reference point O to the force application point; θ is between r⃗ and F⃗, from 0 to π.[1]

M denotes torque here; many textbooks use τ. Its vector is perpendicular to the plane of r⃗ and F⃗, with direction given by the right-hand rule. The nonnegative product $rF\sin\theta$ gives magnitude only, not a signed turning sense about a selected axis.[1]

M_z=\hat{\vec e}_z\cdot(\vec r\times\vec F)=r_xF_y-r_yF_x

For motion in the xy plane, Mz is positive for counterclockwise turning viewed from the positive z side.[1]

A force can have nonzero torque vector while contributing no torque about the selected z axis. That axis uses projection Mz, not total vector magnitude. Changing reference point generally changes the torque of an individual force.[1]

Calculating torque: which lever arm and angle count?

Resolve force into components parallel and perpendicular to r⃗. The parallel component has zero cross product; only $F_\perp=F\sin\theta$ contributes to magnitude. Thus $|\vec M|=rF_\perp$ . Alternatively, measure shortest distance d from O to the force’s line of action.[1]

d=r\sin\theta,\qquad |\vec M|=Fd

Perpendicular lever arm d is not generally the rod length r.[1]

At $\theta=90^\circ$ , $d=r$ and magnitude is maximal for specified r and F. At $\theta=0^\circ$ or 180°, the line of action passes through O and torque is zero. The force can still translate, compress or stretch the body. Zero torque does not mean zero force.[1]

In the lever diagram, r is position distance to the application point and $F\sin\theta$ is the perpendicular force component. It represents part of the same force, not an additional force in the balance. Turning-arrow sense belongs to the selected view of the axis.[1]

View diagram: How angle changes turning effect

How angle changes turning effect

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Magnitude from the first example: position distance and force stay constant. At 0° and 180°, the line of action passes through the reference point. This curve does not specify signed axial turning sense.[1]

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Rearranging torque and distinguishing newton metres from energy

\begin{aligned}[M]&=\mathrm{N\,m}\\F&=\frac{|\vec M|}{r\sin\theta}\\d&=\frac{|\vec M|}{F}\end{aligned}

Denominators must be positive; with zero lever arm this rearrangement cannot determine a required force.[1]

Convert 25 cm to 0.25 m. Divide N·cm by 100 to obtain N·m. Degree angles require degree-mode trigonometry; derivatives with respect to rotation angle use radians. An angle of 35° equals $\frac{35\pi}{180}\,\mathrm{rad}$ .[1]

Torque and energy share an SI dimension but are different quantities. Energy is a scalar; torque describes directional turning effect. Rotational work requires integrating axial torque over angle: $W=\int M_z\,\mathrm d\theta$ . A stationary applied torque can therefore be positive while its instantaneous mechanical power is zero.[4]

How do you calculate torque magnitude, vector and axial component?

The first example calculates magnitude from distance, force and angle. In the spatial second example, the cross product also determines torque-vector direction; its projection then gives the contribution about the chosen axis.[1]

An oblique applied force

Given: $r=0.32\,\mathrm m$ ; force magnitude 45 N; $\theta=35^\circ$ between position and force. Both vectors lie in the xy plane. Viewed from the positive z side, the torque acts counterclockwise.
Find: Find perpendicular lever arm and signed axial torque.

Approach: Use the perpendicular lever arm; the specified counterclockwise action gives positive z torque.
$\begin{aligned}d&=r\sin\theta\\|\mathbf M|&=Fd\\M_z&=+|\mathbf M|\end{aligned}$
Find perpendicular distance to the line of action.
$d=0.32\sin35^\circ\approx0.183544\,\mathrm m$
Multiply force by perpendicular lever arm.
$|\vec M|=45\cdot0.183544\approx8.25950\,\mathrm{N\,m}$
Assign axial sign from the turning sense.
$M_z\approx+8.26\,\mathrm{N\,m}$

Result: Perpendicular lever arm is approximately 18.35 cm; axial torque is +8.26 N·m.

Check: Check: $rF=14.4\,\mathrm{N\,m}$ is the maximum magnitude; 8.26 N·m is smaller. The perpendicular component is $F_\perp=45\sin35^\circ\,\mathrm N$ . Multiplication by 0.32 m gives the same torque magnitude.

Distinguishing vector torque and axial torque

Given: $\vec r=(0.12,0.08,0)\,\mathrm m$ relative to O; $\vec F=(0,30,10)\,\mathrm N$ . The selected axis is positive z.
Find: Find the full torque vector and its contribution about z.

Approach: Form the complete cross product about O, then project onto the chosen positive z-axis.
$\begin{aligned}\mathbf M_O&=\mathbf r\times\mathbf F\\M_z&=\mathbf M_O\cdot\hat{\mathbf z}\end{aligned}$
Calculate cross-product components.
$\vec M=(r_yF_z-r_zF_y,\;r_zF_x-r_xF_z,\;r_xF_y-r_yF_x)$
Insert values component by component.
$\vec M=(0.08\cdot10,\;-0.12\cdot10,\;0.12\cdot30)=(0.80,\;-1.20,\;3.60)\,\mathrm{N\,m}$
Calculate axial projection and total magnitude separately.
$\begin{aligned}M_z&=+3.60\,\mathrm{N\,m}\\{} |\vec M|&=\sqrt{0.80^2+1.20^2+3.60^2}\approx3.87814\,\mathrm{N\,m}\end{aligned}$

Result: Torque about z is +3.60 N·m; full vector magnitude is approximately 3.88 N·m.

Check: Check: $\vec M\cdot\vec r=(0.8\cdot0.12-1.2\cdot0.08)\,\mathrm{N\,m^2}=0$ and $\vec M\cdot\vec F=(-1.2\cdot30+3.6\cdot10)\,\mathrm{N^2\,m}=0$ . The cross product is perpendicular to both original vectors.

When does the lever law apply in static equilibrium?

Static equilibrium of a rigid body requires both zero net force and zero net torque. Zero net torque alone does not prevent linear acceleration. For an ideal horizontal lever with perpendicular loads, torque balance gives $F_1d_1=F_2d_2$ ; the support reaction must also appear in force balance.[5]

Original check: downward loads of 100 N at 0.40 m left and 80 N at 0.50 m right contribute +40 and −40 N·m about the pivot. Their moments cancel. The support must nevertheless exert 180 N upward if there are no other forces or self-weight.[1]

Equal opposite parallel forces on distinct lines form a force couple. Net force is zero while net torque can be nonzero. Couple torque is independent of reference point because shifting the origin adds a term proportional to the already zero net force.[1]

How are torque and angular acceleration related?

\sum\vec M_O=\frac{\mathrm d\vec L_O}{\mathrm dt}

For a fixed inertial reference O, net external torque determines total angular-momentum change of a closed system.[3]

The simple scalar equation $\sum M_z=I_z\alpha_z$ applies to a rigid body with fixed mass distribution rotating about a fixed axis. Iz is inertia about that same axis. Torque alone therefore does not determine angular acceleration; matching inertia and net torque are also needed. Newton’s second law for rotation uses that same fixed axis and the complete net external torque.[2]

Explore further: Time-dependent inertia

\frac{\mathrm d}{\mathrm dt}(I_z\omega_z)=I_z\alpha_z+\dot I_z\omega_z

When I varies in a suitable fixed-axis description, the second term cannot be discarded.[1]

Explore further: Spatial rotation and moving references

Spatial motion may involve an inertia tensor and gyroscopic effects. Reference point also matters: a moving accelerating reference requires a more careful balance. $M=I\alpha$ is therefore not a universal vector law for every body and axis.[3]

What is the difference between torque and mechanical power?

A longer perpendicular lever arm reduces required force for equal torque. The application point also travels farther for the same rotation angle, so an ideal lossless lever does not save work. With oblique pulling, a longer rod does not automatically increase effective lever arm by the same factor; angle matters.[1]

A torque wrench controls tightening torque but does not determine bolt preload alone. Thread geometry, contact and friction affect conversion. A torque value in N·m is not power either; mechanical power requires angular velocity, $P=M_z\omega_z$ for an axial torque.[4][6]

Five problems with solutions

Calculate · Foundation

A 60 N force acts downward, perpendicular to a lever 0.15 m right of its pivot. Find Mz with positive z out of the page.

Show hint

Check turning sense and magnitude separately.

Learning goal: Find torque component with rotation sign.

Show solution and calculation

Approach: For positive z out of the page, $M_z=r_xF_y-r_yF_x$ .
To the right, $r_x=+0.15\,\mathrm m$ ; downward, $F_y=-60\,\mathrm N$ .
Torque is therefore −9 N·m.

Check: clockwise turning matches the negative sign; the magnitude agrees with force times perpendicular lever arm.

M_z=(0.15\,\mathrm m)(-60\,\mathrm N)=-9\,\mathrm{N\,m}

Rearrange · Foundation

A torque magnitude of 12 N·m is required with perpendicular lever arm $d=0.24\,\mathrm m$ . Find force.

Show hint

Solve $|M|=Fd$ for F.

Learning goal: Infer force from perpendicular lever arm.

Show solution and calculation

Approach: The specified distance is already the perpendicular lever arm.
From $|M|=Fd$ , obtain $F=\frac{|M|}{d}$ , giving 50 N.

Check: $(50\,\mathrm N)(0.24\,\mathrm m)=12\,\mathrm{N\,m}$ . An arbitrary rod length cannot replace this distance; the applied force must follow the specified line of action.

F=\frac{12\,\mathrm{N\,m}}{0.24\,\mathrm m}=50\,\mathrm N

Transfer · Application

A rigid fixed-axis rotor has $I_z=0.40\,\mathrm{kg\,m^2}$ . Driving torque is +3 N·m and opposing torque −1 N·m. Find αz.

Show hint

Sum torques first; constant I is explicitly specified.

Learning goal: Convert net torque to angular acceleration.

Show solution and calculation

Approach: The matching inertia is constant for this rigid fixed-axis rotor.
First sum signed external torques, then use $\alpha_z=\frac{\sum M_z}{I_z}$ .
The result is +5 rad/s².

Check: $I_z\alpha_z=+2\,\mathrm{N\,m}$ , reproducing net torque, rather than driving torque alone. Positive acceleration does not specify rotation direction without an initial value.

\begin{aligned}\sum M_z&=(3-1)\,\mathrm{N\,m}=2\,\mathrm{N\,m}\\\alpha_z&=\frac{2}{0.40}\,\mathrm{rad\,s^{-2}}=+5\,\mathrm{rad\,s^{-2}}\end{aligned}

Interpret · Application

In a plane, +30 N upward acts at $x=+0.10\,\mathrm m$ and −30 N downward at $x=-0.10\,\mathrm m$ , with $y=0$ . Positive z points out of the plane. Find net force and Mz. Does shifting the reference point change total torque?

Show hint

Both xFy products have the same sign.

Learning goal: Recognize a couple with zero resultant but finite torque.

Show solution and calculation

Approach: forces cancel, while torques are +3 N·m each, totaling +6 N·m.
This is a force couple.
Changing reference point changes total torque by a cross product with net force, which is zero here.

Check: force times perpendicular line separation gives $(30\,\mathrm N)(0.20\,\mathrm m)=6\,\mathrm{N\,m}$ . Zero net force alone does not establish full static equilibrium.

\begin{aligned}\sum F_y&=30-30=0\\M_z&=0.10\cdot30+(-0.10)(-30)=6\,\mathrm{N\,m}\end{aligned}

[1][5]

Error analysis · Application

Position relative to the origin is $\vec r=(1,0,0)\,\mathrm m$ and force $\vec F=(0,0,5)\,\mathrm N$ . A solution writes “ $M_z=rF=5\,\mathrm{N\,m}$ ”. Find the torque vector and Mz.

Show hint

Take the cross product first, then project onto z.

Learning goal: Separate total torque magnitude from torque about a chosen axis.

Show solution and calculation

Approach: $\vec r\times\vec F=(0,-5,0)\,\mathrm{N\,m}$ , so $M_z=0$ .
Total magnitude 5 N·m is correct but was incorrectly assigned to the z axis.
Component check: $r_xF_y-r_yF_x=0$ .
The force may drive rotation about another suitable axis; zero z component does not make the entire torque vector zero.

\begin{aligned}\mathbf M&=\mathbf r\times\mathbf F\\&=\begin{pmatrix}0\\-5\\0\end{pmatrix}\,\mathrm{N\,m}\\M_z&=0\end{aligned}

[1]

Key points

Specify reference point, axis and positive turning sense. Position distance r needs the force angle; perpendicular arm d already includes that geometry. Calculate the cross product or its axial component, then add all signed external torques.

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

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This explanation, 2 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Made lever-arm/angle calculation, axis and rotation sense, cross product, lever equilibrium and unit distinction accessible through questions; retained spatial/planar examples and linked cross product, rotational balance and mechanical power in context.
October 1, 2026Clarified the entry and learning transitions; checked and refined existing exercise solutions, typeset mathematical prose and added an original quantitative article figure.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Quanta. Torque. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/drehmoment

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Frequently asked questions about Torque

How do you calculate torque?+

Vector torque about O is $\vec M_O=\vec r\times\vec F$ , with r⃗ pointing from O to the force application point. Its magnitude is $rF\sin\theta$ ; alternatively $|\vec M|=Fd$ , where d is shortest distance from O to the line of action. r equals d only for perpendicular force. An 80 N force perpendicular at 0.25 m gives $20 N\cdot{}m$ ; at 60° between r⃗ and F⃗ it gives approximately $17.3 N\cdot{}m$ . For a selected axis, use the corresponding signed projection rather than automatically using full magnitude. In the xy plane, $M_z=r_xF_y-r_yF_x$ . Define reference point, axis and positive turning sense before substituting values.

What does the law of the lever state and how is it related to torque?+

The ideal lever law $F_1d_1=F_2d_2$ describes two opposing moments about the same pivot. d₁ and d₂ are perpendicular distances to the force lines. For a horizontal lever under vertical loads, they equal the usual arm lengths. A 30 kg body at 2 m balances 40 kg at 1.5 m when other torques and self-weight are neglected. g cancels in this torque balance. Static equilibrium additionally requires zero net force, so the support reaction must be included. Zero net torque alone does not establish rest: a free body can rotate uniformly or accelerate linearly despite that torque balance.

Why does a longer lever save force?+

For a required torque magnitude, $F=\frac{|M|}{d}$ , where d is perpendicular distance to the force line. With perpendicular force, $40 N\cdot{}m$ requires 200 N at a 20 cm wrench and 100 N at a 40 cm wrench. Oblique application requires accounting for angle; a longer rod alone does not guarantee the same effective arm. For an equal rotation angle, the farther-out handle travels along a longer arc. With the same ideal torque-angle balance, work is therefore unchanged. A pipe extension may also alter tool loading; the physics calculation does not replace checking permitted tool use.

Is torque the same as energy, since both are measured in

N\cdot{}m

No. The quantities share an SI dimension but differ mathematically and physically. Work is the path integral of $\vec F\cdot\mathrm d\vec r$ , a dot product, and is expressed in joules. Torque is $\vec r\times\vec F$ , a cross product describing turning effect, and is expressed in $N\cdot{}m$ . The original vectors need not be perpendicular; the sine factor accounts for their angle. For fixed-axis rotation, $W=\int M_z\,\mathrm d\theta$ produces work. Constant axial torque simplifies this to $M_z\Delta\theta$ with angle in radians. A torque of $20 N\cdot{}m$ therefore does not imply 20 J of stored energy. Mechanical power may be zero at a stationary application point despite nonzero torque.

What is a torque wrench for?+

A torque wrench helps apply a specified tightening torque in a controlled way. Torque is not the same as bolt preload: conversion depends on factors including thread geometry and friction in the threads and bearing surface. Equal torque can therefore produce different preloads. Tools use different measuring or release principles, including indicating, electronic and releasing designs; not all produce an audible click. Specified grip position and accessory geometry are part of the measuring arrangement. $Fr$ does not mean arbitrary gripping or extending leaves the indication correct. Generic example values cannot replace a connection’s actual manufacturer specification. Torque, effective lever arm and application conditions all matter.

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Curated exam set for "Torque":

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 1, §10.6 Torque

Related formulas

W = F \cdot s \cdot \cos(\alpha)

Mechanical Work

\Delta\vec p=\int_{t_1}^{t_2}\vec F_{\mathrm{ext}}dt, \qquad \Delta\vec L=\int_{t_1}^{t_2}\vec\tau_{\mathrm{ext}}dt

Linear and angular impulse–momentum balance

\sum\tau=I\alpha

Newton’s second law for rotation

More formulas: Mechanics

F_G = G \cdot \frac{m_1 \cdot m_2}{r^2}

Newton's Law of Universal Gravitation

E_{kin} = \frac{1}{2} m v^2

Kinetic Energy

E_{pot} = m \cdot g \cdot h

Potential Energy (Gravitational)

F = k \cdot x

Hooke's Law (Spring Force)

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Quanta Study editorial team

Torque: reference point, lever arm and axis

The cross product of position and force describes turning effect.

Fr

·sinθ is magnitude; torque about a selected axis is the corresponding signed projection.

BasicUpper secondaryUniversity/college

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Symbol

Meaning

Unit

M

Torque-vector magnitude in the illustrated planar case

N·m

F

Applied-force magnitude

r

Distance from reference point to application point;

d=r\sin\theta

is perpendicular lever arm

θ

Angle between position and force vectors

rad or °

Derivation & background

Exam blueprint

Validity range

Derivation steps

Torque is a cross product; magnitude uses perpendicular lever arm, while an axis uses the relevant projection.

1Choose reference point O or an axis through O and its positive sense.
2Position vector r⃗ runs from O to the application point. Torque is $\vec M_O=\vec r\times\vec F$ .
3Split F⃗ parallel and perpendicular to r⃗; the perpendicular part is $F\sin\theta$ . The parallel part has zero cross product.
4Thus $|\vec M|=rF\sin\theta=Fd$ , where $d=r\sin\theta$ is shortest distance to the line of action.
5The right-hand rule determines vector direction. About z use signed $M_z=r_xF_y-r_yF_x$ , not necessarily $|\vec M|$ .
6Sum external torques about the same reference. Only a rigid constant-I body about a fixed axis simplifies dynamics to $\sum M_z=I_z\alpha_z$ .

How do you calculate with Torque?

Here is how to work through a typical Torque (M = F·r·sin(θ)) task step by step:

Task

A bolt needs $40 N\cdot{}m$ . What force is required on a 20 cm lever (perpendicular)?

Solution path

$F=\frac{M}{r}=\frac{40\,\mathrm{N\,m}}{0.2\,\mathrm m}=200\,\mathrm N$ .
A wrench twice as long halves the force to 100 N.

Task

An ideal horizontal seesaw has negligible self-weight. A 30 kg child sits 2 m from its pivot. Where must a 40 kg child sit on the opposite side to balance load torques?

Solution path

Law of the lever: $m_1r_1=m_2r_2$ ; g cancels.
$r_2=\frac{30\,\mathrm{kg}\cdot2\,\mathrm m}{40\,\mathrm{kg}}=1.5\,\mathrm m$ .

Understand and apply the equation

How do you specify reference point, axis and sense of rotation?

\begin{aligned}\vec M_O&=\vec r\times\vec F\\|\vec M_O|&=rF\sin\theta\end{aligned}

r⃗ points from reference point O to the force application point; θ is between r⃗ and F⃗, from 0 to π.[1]

M_z=\hat{\vec e}_z\cdot(\vec r\times\vec F)=r_xF_y-r_yF_x

For motion in the xy plane, Mz is positive for counterclockwise turning viewed from the positive z side.[1]

Calculating torque: which lever arm and angle count?

d=r\sin\theta,\qquad |\vec M|=Fd

Perpendicular lever arm d is not generally the rod length r.[1]

View diagram: How angle changes turning effect

How angle changes turning effect

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Rearranging torque and distinguishing newton metres from energy

\begin{aligned}[M]&=\mathrm{N\,m}\\F&=\frac{|\vec M|}{r\sin\theta}\\d&=\frac{|\vec M|}{F}\end{aligned}

Denominators must be positive; with zero lever arm this rearrangement cannot determine a required force.[1]

How do you calculate torque magnitude, vector and axial component?

An oblique applied force

Given: $r=0.32\,\mathrm m$ ; force magnitude 45 N; $\theta=35^\circ$ between position and force. Both vectors lie in the xy plane. Viewed from the positive z side, the torque acts counterclockwise.
Find: Find perpendicular lever arm and signed axial torque.

Approach: Use the perpendicular lever arm; the specified counterclockwise action gives positive z torque.
$\begin{aligned}d&=r\sin\theta\\|\mathbf M|&=Fd\\M_z&=+|\mathbf M|\end{aligned}$
Find perpendicular distance to the line of action.
$d=0.32\sin35^\circ\approx0.183544\,\mathrm m$
Multiply force by perpendicular lever arm.
$|\vec M|=45\cdot0.183544\approx8.25950\,\mathrm{N\,m}$
Assign axial sign from the turning sense.
$M_z\approx+8.26\,\mathrm{N\,m}$

Result: Perpendicular lever arm is approximately 18.35 cm; axial torque is +8.26 N·m.

Distinguishing vector torque and axial torque

Given: $\vec r=(0.12,0.08,0)\,\mathrm m$ relative to O; $\vec F=(0,30,10)\,\mathrm N$ . The selected axis is positive z.
Find: Find the full torque vector and its contribution about z.

Approach: Form the complete cross product about O, then project onto the chosen positive z-axis.
$\begin{aligned}\mathbf M_O&=\mathbf r\times\mathbf F\\M_z&=\mathbf M_O\cdot\hat{\mathbf z}\end{aligned}$
Calculate cross-product components.
$\vec M=(r_yF_z-r_zF_y,\;r_zF_x-r_xF_z,\;r_xF_y-r_yF_x)$
Insert values component by component.
$\vec M=(0.08\cdot10,\;-0.12\cdot10,\;0.12\cdot30)=(0.80,\;-1.20,\;3.60)\,\mathrm{N\,m}$
Calculate axial projection and total magnitude separately.
$\begin{aligned}M_z&=+3.60\,\mathrm{N\,m}\\{} |\vec M|&=\sqrt{0.80^2+1.20^2+3.60^2}\approx3.87814\,\mathrm{N\,m}\end{aligned}$

Result: Torque about z is +3.60 N·m; full vector magnitude is approximately 3.88 N·m.

When does the lever law apply in static equilibrium?

How are torque and angular acceleration related?

\sum\vec M_O=\frac{\mathrm d\vec L_O}{\mathrm dt}

For a fixed inertial reference O, net external torque determines total angular-momentum change of a closed system.[3]

Explore further: Time-dependent inertia

\frac{\mathrm d}{\mathrm dt}(I_z\omega_z)=I_z\alpha_z+\dot I_z\omega_z

When I varies in a suitable fixed-axis description, the second term cannot be discarded.[1]

Explore further: Spatial rotation and moving references

What is the difference between torque and mechanical power?

Five problems with solutions

Calculate · Foundation

A 60 N force acts downward, perpendicular to a lever 0.15 m right of its pivot. Find Mz with positive z out of the page.

Show hint

Check turning sense and magnitude separately.

Learning goal: Find torque component with rotation sign.

Show solution and calculation

Approach: For positive z out of the page, $M_z=r_xF_y-r_yF_x$ .
To the right, $r_x=+0.15\,\mathrm m$ ; downward, $F_y=-60\,\mathrm N$ .
Torque is therefore −9 N·m.

Check: clockwise turning matches the negative sign; the magnitude agrees with force times perpendicular lever arm.

M_z=(0.15\,\mathrm m)(-60\,\mathrm N)=-9\,\mathrm{N\,m}

Rearrange · Foundation

A torque magnitude of 12 N·m is required with perpendicular lever arm $d=0.24\,\mathrm m$ . Find force.

Show hint

Solve $|M|=Fd$ for F.

Learning goal: Infer force from perpendicular lever arm.

Show solution and calculation

Approach: The specified distance is already the perpendicular lever arm.
From $|M|=Fd$ , obtain $F=\frac{|M|}{d}$ , giving 50 N.

Check: $(50\,\mathrm N)(0.24\,\mathrm m)=12\,\mathrm{N\,m}$ . An arbitrary rod length cannot replace this distance; the applied force must follow the specified line of action.

F=\frac{12\,\mathrm{N\,m}}{0.24\,\mathrm m}=50\,\mathrm N

Transfer · Application

A rigid fixed-axis rotor has $I_z=0.40\,\mathrm{kg\,m^2}$ . Driving torque is +3 N·m and opposing torque −1 N·m. Find αz.

Show hint

Sum torques first; constant I is explicitly specified.

Learning goal: Convert net torque to angular acceleration.

Show solution and calculation

Approach: The matching inertia is constant for this rigid fixed-axis rotor.
First sum signed external torques, then use $\alpha_z=\frac{\sum M_z}{I_z}$ .
The result is +5 rad/s².

Check: $I_z\alpha_z=+2\,\mathrm{N\,m}$ , reproducing net torque, rather than driving torque alone. Positive acceleration does not specify rotation direction without an initial value.

\begin{aligned}\sum M_z&=(3-1)\,\mathrm{N\,m}=2\,\mathrm{N\,m}\\\alpha_z&=\frac{2}{0.40}\,\mathrm{rad\,s^{-2}}=+5\,\mathrm{rad\,s^{-2}}\end{aligned}

Interpret · Application

In a plane, +30 N upward acts at $x=+0.10\,\mathrm m$ and −30 N downward at $x=-0.10\,\mathrm m$ , with $y=0$ . Positive z points out of the plane. Find net force and Mz. Does shifting the reference point change total torque?

Show hint

Both xFy products have the same sign.

Learning goal: Recognize a couple with zero resultant but finite torque.

Show solution and calculation

Approach: forces cancel, while torques are +3 N·m each, totaling +6 N·m.
This is a force couple.
Changing reference point changes total torque by a cross product with net force, which is zero here.

Check: force times perpendicular line separation gives $(30\,\mathrm N)(0.20\,\mathrm m)=6\,\mathrm{N\,m}$ . Zero net force alone does not establish full static equilibrium.

\begin{aligned}\sum F_y&=30-30=0\\M_z&=0.10\cdot30+(-0.10)(-30)=6\,\mathrm{N\,m}\end{aligned}

[1][5]

Error analysis · Application

Position relative to the origin is $\vec r=(1,0,0)\,\mathrm m$ and force $\vec F=(0,0,5)\,\mathrm N$ . A solution writes “ $M_z=rF=5\,\mathrm{N\,m}$ ”. Find the torque vector and Mz.

Show hint

Take the cross product first, then project onto z.

Learning goal: Separate total torque magnitude from torque about a chosen axis.

Show solution and calculation

Approach: $\vec r\times\vec F=(0,-5,0)\,\mathrm{N\,m}$ , so $M_z=0$ .
Total magnitude 5 N·m is correct but was incorrectly assigned to the z axis.
Component check: $r_xF_y-r_yF_x=0$ .
The force may drive rotation about another suitable axis; zero z component does not make the entire torque vector zero.

\begin{aligned}\mathbf M&=\mathbf r\times\mathbf F\\&=\begin{pmatrix}0\\-5\\0\end{pmatrix}\,\mathrm{N\,m}\\M_z&=0\end{aligned}

[1]

Key points

Sources for this explanation

Quanta · Updated October 1, 2026

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An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Made lever-arm/angle calculation, axis and rotation sense, cross product, lever equilibrium and unit distinction accessible through questions; retained spatial/planar examples and linked cross product, rotational balance and mechanical power in context.
October 1, 2026Clarified the entry and learning transitions; checked and refined existing exercise solutions, typeset mathematical prose and added an original quantitative article figure.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Frequently asked questions about Torque

How do you calculate torque?+

What does the law of the lever state and how is it related to torque?+

Why does a longer lever save force?+

Is torque the same as energy, since both are measured in

N\cdot{}m

What is a torque wrench for?+

Retain Torque for exams

Create a curated FSRS exam set for M = F·r·sin(θ): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Torque: reference point, lever arm and axis

Variables & units

Torque calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Statics and rotation

Worked example

Applications

How do you calculate with Torque?

How do you specify reference point, axis and sense of rotation?

Calculating torque: which lever arm and angle count?

How angle changes turning effect

Rearranging torque and distinguishing newton metres from energy

How do you calculate torque magnitude, vector and axial component?

An oblique applied force

Distinguishing vector torque and axial torque

When does the lever law apply in static equilibrium?

How are torque and angular acceleration related?

What is the difference between torque and mechanical power?

Five problems with solutions

A 60 N force acts downward, perpendicular to a lever 0.15 m right of its pivot. Find Mz with positive z out of the page.

A torque magnitude of 12 N·m is required with perpendicular lever arm d=0.24 md=0.24\,\mathrm md=0.24m. Find force.

A rigid fixed-axis rotor has Iz=0.40 kg m2I_z=0.40\,\mathrm{kg\,m^2}Iz​=0.40kgm2. Driving torque is +3 N·m and opposing torque −1 N·m. Find αz.

In a plane, +30 N upward acts at x=+0.10 mx=+0.10\,\mathrm mx=+0.10m and −30 N downward at x=−0.10 mx=-0.10\,\mathrm mx=−0.10m, with y=0y=0y=0. Positive z points out of the plane. Find net force and Mz. Does shifting the reference point change total torque?

Position relative to the origin is r⃗=(1,0,0) m\vec r=(1,0,0)\,\mathrm mr=(1,0,0)m and force F⃗=(0,0,5) N\vec F=(0,0,5)\,\mathrm NF=(0,0,5)N. A solution writes “Mz=rF=5 N mM_z=rF=5\,\mathrm{N\,m}Mz​=rF=5Nm”. Find the torque vector and Mz.

Key points

Sources for this explanation

Frequently asked questions about Torque

Subject references and further reading

Related formulas

More formulas: Mechanics

Retain Torque for exams

Torque: reference point, lever arm and axis

Variables & units

Torque calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Statics and rotation

Worked example

Applications

How do you calculate with Torque?

How do you specify reference point, axis and sense of rotation?

Calculating torque: which lever arm and angle count?

How angle changes turning effect

Rearranging torque and distinguishing newton metres from energy

How do you calculate torque magnitude, vector and axial component?

An oblique applied force

Distinguishing vector torque and axial torque

When does the lever law apply in static equilibrium?

How are torque and angular acceleration related?

What is the difference between torque and mechanical power?

Five problems with solutions

A 60 N force acts downward, perpendicular to a lever 0.15 m right of its pivot. Find Mz with positive z out of the page.

A torque magnitude of 12 N·m is required with perpendicular lever arm d=0.24 md=0.24\,\mathrm md=0.24m. Find force.

A rigid fixed-axis rotor has Iz=0.40 kg m2I_z=0.40\,\mathrm{kg\,m^2}Iz​=0.40kgm2. Driving torque is +3 N·m and opposing torque −1 N·m. Find αz.

In a plane, +30 N upward acts at x=+0.10 mx=+0.10\,\mathrm mx=+0.10m and −30 N downward at x=−0.10 mx=-0.10\,\mathrm mx=−0.10m, with y=0y=0y=0. Positive z points out of the plane. Find net force and Mz. Does shifting the reference point change total torque?

Position relative to the origin is r⃗=(1,0,0) m\vec r=(1,0,0)\,\mathrm mr=(1,0,0)m and force F⃗=(0,0,5) N\vec F=(0,0,5)\,\mathrm NF=(0,0,5)N. A solution writes “Mz=rF=5 N mM_z=rF=5\,\mathrm{N\,m}Mz​=rF=5Nm”. Find the torque vector and Mz.

Key points

Sources for this explanation

Frequently asked questions about Torque

Subject references and further reading

Related formulas

More formulas: Mechanics

Retain Torque for exams

A torque magnitude of 12 N·m is required with perpendicular lever arm $d=0.24\,\mathrm m$ . Find force.

A rigid fixed-axis rotor has $I_z=0.40\,\mathrm{kg\,m^2}$ . Driving torque is +3 N·m and opposing torque −1 N·m. Find αz.

In a plane, +30 N upward acts at $x=+0.10\,\mathrm m$ and −30 N downward at $x=-0.10\,\mathrm m$ , with $y=0$ . Positive z points out of the plane. Find net force and Mz. Does shifting the reference point change total torque?

Position relative to the origin is $\vec r=(1,0,0)\,\mathrm m$ and force $\vec F=(0,0,5)\,\mathrm N$ . A solution writes “ $M_z=rF=5\,\mathrm{N\,m}$ ”. Find the torque vector and Mz.

A torque magnitude of 12 N·m is required with perpendicular lever arm $d=0.24\,\mathrm m$ . Find force.

A rigid fixed-axis rotor has $I_z=0.40\,\mathrm{kg\,m^2}$ . Driving torque is +3 N·m and opposing torque −1 N·m. Find αz.

In a plane, +30 N upward acts at $x=+0.10\,\mathrm m$ and −30 N downward at $x=-0.10\,\mathrm m$ , with $y=0$ . Positive z points out of the plane. Find net force and Mz. Does shifting the reference point change total torque?

Position relative to the origin is $\vec r=(1,0,0)\,\mathrm m$ and force $\vec F=(0,0,5)\,\mathrm N$ . A solution writes “ $M_z=rF=5\,\mathrm{N\,m}$ ”. Find the torque vector and Mz.