Torque vs Force: What Is the Difference?

A force is a thing. A torque is a relationship.

The question "torque or force?" is malformed, and spotting that is most of the answer. You never choose between them, because every torque is produced by a force. What you choose is the axis, and the torque is what you get when you ask how effectively that force turns things about that axis.

AP Physics 1 EK 5.3.A.1: torque results only from the force component perpendicular to the position vector from the axis of rotation to the point of application of the force. Every noun in that sentence except "torque" is either a force or a piece of geometry. Torque is the output.

Three consequences follow immediately, and they are the whole page.

1. A force has a value on its own. A torque does not. "A $50$ N force" is a complete statement. "A $50$ N force produces a torque of $12\ \text{N}\cdot\text{m}$" is only complete once you say about what.
2. One force gives many torques. Change the axis and the number changes, including its sign, including to zero. Worked example three gets $+15$, $0$ and $-45\ \text{N}\cdot\text{m}$ from a single unchanged $30$ N force.
3. A bigger force is not a bigger torque. Worked example two has a $5$ N push out-turning a $200$ N push, because the geometry beats the magnitude.

The torque entry defines the quantity and how to calculate torque has the procedure. What is here is the relationship between the two quantities, which neither of those pages is about.

Torque vs force, side by side

The "does the point of application matter" row is the source of every other row. Slide a force to a different point on a rigid body and the force sum is untouched. The torque sum changes completely. That single asymmetry is why the two need separate equations and separate diagrams.

The diagram row is a distinction the CED draws and most courses do not. EK 5.3.B.1.i says force diagrams are similar to free-body diagrams and are used to analyze the torques exerted on a rigid system, and EK 5.3.B.1.ii adds what makes them different: like free-body diagrams they show the relative magnitude and direction of the forces, and force diagrams also depict the location at which those forces are exerted relative to the axis of rotation. A free-body diagram collapses the body to a point, which throws away exactly the information a torque needs.

The two direction rows are a real course-scope difference and are covered below.

One force, many torques: the case that separates them

Put a $30$ N force on a rod, perpendicular to it, $0.50$ m from the left end of a $2.0$ m rod. Do not move it, do not change it. Now ask for the torque.

• About the left end: $\tau = (0.50)(30) = 15\ \text{N}\cdot\text{m}$, one rotational sense.
• About the point where the force is applied: $\tau = (0)(30) = 0$.
• About the right end: $\tau = (1.50)(30) = 45\ \text{N}\cdot\text{m}$, and the rod turns the other way, so $-45\ \text{N}\cdot\text{m}$ under the same sign convention.

The force was $30$ N in all three lines. Worked example three sets this out in full. A force sum could never behave like this, because a force sum has no axis to be about.

The zero case in the middle is the one worth studying, because it generalises. A force produces no torque about any axis on its own line of action, and "line of action" means the infinite straight line you get by extending the force arrow both ways. EK 5.3.A.2 defines the lever arm as the perpendicular distance from the axis of rotation to the line of action of the exerted force, so when the axis sits on that line the perpendicular distance is zero and so is the torque.

That is why pushing a door at its hinge does nothing, and why pushing the edge of a door straight toward the hinge does nothing either, however hard you push. Both have a lever arm of zero, by two different routes. Worked example two prices it: a $200$ N shove aimed at the hinge gives $0\ \text{N}\cdot\text{m}$, while a $5$ N nudge at the handle gives $4.0\ \text{N}\cdot\text{m}$ and the door swings.

The reverse statement is also useful. Sliding a force along its own line of action changes neither the force nor the torque, because the perpendicular distance from the axis to that line has not moved. So a torque does not depend on the exact point of application, only on which line the force acts along.

Why the two are not alternatives

Students often treat this as a choice: is the question a force question or a torque question? For a rigid body it is usually both, and the CED says so.

EK 5.6.A.3: to fully describe a rotating rigid system, linear and rotational analyses may need to be performed independently. Two analyses. Same object, same forces, run separately, because they answer different questions.

A wheel with a rope pulled off its rim shows it. The rope tension enters $\sum \vec{F}$ and helps decide how the wheel's centre of mass accelerates. The same tension enters $\sum \tau$ and helps decide how fast the wheel spins up. Delete either sum and half the motion is unaccounted for.

What makes the pairing tractable is that they take different inputs from the same force:

• The force sum takes the whole vector, magnitude and direction, and ignores where it acts.
• The torque sum takes the perpendicular component and the distance, and ignores everything about the force that is aimed at the axis.

So the radial part of a force is invisible to the torque equation and fully present in the force equation. A tyre pressed straight down onto an axle adds to the force balance and contributes nothing to the spin.

The rest of the analogy runs clean and is worth having as a table of correspondences: force pairs with torque, mass with rotational inertia, acceleration with angular acceleration, momentum with angular momentum, impulse with angular impulse. The place it stops is the axis. Every rotational quantity in that list is defined relative to an axis and every translational one is not, which is why the analogy is a memory aid and not a proof.

Direction: in scope for one course and not the other

This is a genuine and checkable difference between the two AP mechanics courses, and answering as general physics will mislead an AP Physics 1 student.

The AP Physics 1 Topic 5.3 boundary statement, in full: while AP Physics 1 expects students to mathematically manipulate the magnitude of torque using vector conventions, the direction of torque is beyond the scope of the course. So there is no torque vector on that exam. What you do instead is pick a rotational sense as positive, usually counterclockwise, and give every torque a sign. The Topic 5.5 boundary statement backs this up by ruling out simultaneous analysis of rotation in multiple planes, so one plane and one sign convention is always enough.

The AP Physics 1 equation sheet matches: it prints $\tau = r_{\perp}F = rF\sin\theta$, a magnitude, with no vector arrows on the torque.

AP Physics C: Mechanics goes the other way and has no such boundary statement. Its EK 5.3.B.2 gives the torque exerted on a rigid system about a chosen pivot point by a given force as $\vec{\tau} = \vec{r} \times \vec{F}$, and three sub-points develop it: EK 5.3.B.2.i gives the cross product magnitude as $AB\sin\theta$; EK 5.3.B.2.ii says the direction of the resulting vector is perpendicular to both vectors and therefore normal to the plane they define; EK 5.3.B.2.iii says that direction can be qualitatively determined by applying the appropriate right-hand rule. That sheet prints $\vec{\tau} = \vec{r} \times \vec{F}$ with the arrows.

So the honest summary is:

The force side does not split like this. Force is a vector in both courses, with components, in Unit 2 of each. So the asymmetry between force and torque is larger in AP Physics 1 than in C: Mechanics, and a student moving between the two courses meets torque as a genuinely different object.

One thing the boundary statement does not do is excuse you from signs. It removes the vector and leaves the bookkeeping, and a sign error in $\sum \tau$ is as fatal as a direction error in $\sum \vec{F}$.

Where the confusion costs a mark

• Calling a torque a force. "The torque force on the beam" is not a phrase that can be marked correct. Torques are measured in $\text{N}\cdot\text{m}$ and forces in $\text{N}$, and a reader cannot tell which quantity you meant.
• Adding a torque into a force sum, or the reverse. They have different units. A term in $\text{N}$ inside $\sum \tau = 0$ is wrong before the arithmetic.
• Quoting a torque with no axis. LO 5.3.A asks you to identify the torques exerted on a rigid system, and every one of them is about something. A bare number is an incomplete answer and makes later parts unmarkable.
• Assuming the largest force gives the largest torque. A $200$ N force through the axis gives zero; a $5$ N force at a good lever arm does not.
• Dropping the $\sin\theta$. If the angle between $\vec{r}$ and $\vec{F}$ is $37^\circ$, only $60$ percent of the force turns anything, using the value from the trigonometry table printed on the sheet. Worked example one shows a $50$ N force delivering $12.0$, $9.6$, $7.2$ and $0\ \text{N}\cdot\text{m}$ at four different angles.
• Using the angle to the horizontal instead of the angle between $\vec{r}$ and $\vec{F}$. EK 5.3.B.2 defines $\theta$ as the angle between the force vector and the position vector from the axis of rotation to the point of application of the force. For a horizontal beam with vertical forces, that angle is $90^\circ$ and $\sin\theta = 1$.
• Collapsing the body to a point before computing torques. A free-body diagram is the wrong diagram here; EK 5.3.B.1.ii wants a force diagram that shows where each force acts relative to the axis.
• Concluding that zero net force means zero net torque. It does not. Two equal and opposite forces at the ends of a rod give exactly zero net force and a large net torque, which is the subject of translational vs rotational equilibrium.
• Assigning a direction to a torque on the AP Physics 1 exam. The boundary statement puts that beyond scope. Use signs, and state your convention.
• Writing a torque in joules. The base units match and the quantities do not. That confusion has its own page: torque vs work.

When they behave alike, and why that lulls you

Three situations let you treat torque as though it were just a force with a longer name, and all three are common enough to build a bad habit on.

A single fixed axis with everything perpendicular. In a seesaw or beam problem the axis never moves and every force is at $90^\circ$ to the beam, so $\tau = rF$ and the torque is just the force scaled by a fixed distance. Under those conditions the two quantities really are proportional, and it takes a question that tilts a force or moves the pivot to break the habit.

The two second laws look identical. $\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}}$ and $\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}}$ have the same shape, and every symbol in one has a counterpart in the other. The analogy is real, EK 5.1.A.4 endorses the corresponding kinematic version, and it still hides the fact that the right-hand side of the second equation depends on a choice the first one never has to make.

Both are summed, both have an equilibrium condition, both point somewhere. For a stationary beam, you write two force equations and one torque equation and they all look like the same kind of statement.

The distinction reappears the moment any of the following happens, and one of them happens in nearly every real question:

• The pivot moves. Worked example three: three axes, three answers.
• A force stops being perpendicular. Worked example one: four angles, four answers, one force.
• A force is aimed at the axis. It vanishes from one sum and not the other.
• The question asks about the centre of mass and the spin at once. EK 5.6.A.3's two independent analyses.

The reflex worth building: write the axis next to every torque, in words, before the number. It costs one phrase and it is the entire difference between the two quantities.

What the CED asks, and how the exam frames it

The two quantities are introduced three units apart in AP Physics 1, which is part of why the relationship between them gets lost.

Force is Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the multiple-choice section over about 22 to 27 class periods. Topic 2.2 covers forces and free-body diagrams, Topic 2.4 gives translational equilibrium and Newton's first law, and Topic 2.5 gives the second law at EK 2.5.A.2: the acceleration of a system's centre of mass has a magnitude proportional to the magnitude of the net force exerted on the system and is in the same direction as that net force.

Torque is Unit 5, Torque and Rotational Dynamics, weighted at 10 to 15 percent. Topic 5.3, Torque, has two learning objectives. LO 5.3.A, identify the torques exerted on a rigid system, carries EK 5.3.A.1, that torque results only from the force component perpendicular to the position vector from the axis of rotation to the point of application of the force, and EK 5.3.A.2, the definition of the lever arm as the perpendicular distance from the axis of rotation to the line of action of the exerted force. LO 5.3.B, describe the torques exerted on a rigid system, carries EK 5.3.B.1 on force diagrams with its two sub-points, and EK 5.3.B.2 with the equation $\tau = rF_{\perp} = rF\sin\theta$ and the definition of $\theta$. Suggested skills for Topic 5.3 are 1.A, 2.A, 2.D and 3.B. The topic carries the one boundary statement, on direction.

On the AP Physics 1 equation sheet, force and torque sit in the two different columns of the mechanics block. The translational column prints $\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}}$ and the individual force laws; the rotational column prints $\tau = r_{\perp}F = rF\sin\theta$ and $\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}}$. The symbol key lists $F$ as force in both columns and $\tau$ as torque in the rotational one only.

AP Physics C: Mechanics covers torque in its own Topic 5.3 with the same two learning objectives and the same EK 5.3.A.1 and 5.3.A.2, then replaces the magnitude equation with the cross product at EK 5.3.B.2 and adds three sub-points on the cross product, its direction and the right-hand rule. Suggested skills there are 1.A, 2.B, 2.C and 3.C.

The CED framing is at Topic 5.3 in AP Physics 1 and Topic 5.3 in C: Mechanics, and the numbers can be checked with the torque calculator or worked in the torque and rotational motion practice set.