AP Physics 1 · Topic 5.3

Topic 5.3: Torque

Unit 5: Torque and Rotational Dynamics10-15% of the multiple-choice section

Torque measures how strongly a force turns a rigid system about an axis. Its magnitude is the distance from the axis to where the force acts, times the force, times the sine of the angle between them. Only the part of the force perpendicular to that distance produces torque.

AP Physics: Unit 5 (topics 5.3 Torque). AP Physics 1 Unit 5, Topic 5.3. Two learning objectives: 5.3.A, identify the torques exerted on a rigid system, and 5.3.B, describe the torques exerted on a rigid system. Four essential knowledge statements support them: 5.3.A.1 (torque results only from the force component perpendicular to the position vector from the axis to the point of application), 5.3.A.2 (the lever arm is the perpendicular distance from the axis to the line of action of the force), 5.3.B.1 with sub-statements i and ii (torques can be described using force diagrams, which unlike free-body diagrams also depict where each force is exerted relative to the axis), and 5.3.B.2 (the magnitude of torque is tau = rF-perpendicular = rF sin theta, with theta the angle between the force vector and the position vector). The boundary statement reads 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. The CED's suggested skills for this topic are 1.A, 2.A, 2.D, and 3.B. Unit 5 carries 10 to 15 percent of the multiple-choice section and about 15 to 20 class periods.

What Topic 5.3 requires

Topic 5.3 carries two learning objectives and four essential knowledge statements, one of which has two sub-statements. It sits in Unit 5, Torque and Rotational Dynamics, which the CED weights at 10 to 15 percent of the multiple-choice section and estimates at about 15 to 20 class periods.

Learning objective 5.3.A asks you to identify the torques exerted on a rigid system.

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.
5.3.A.2 The lever arm is the perpendicular distance from the axis of rotation to the line of action of the exerted force.

Learning objective 5.3.B asks you to describe the torques exerted on a rigid system.

5.3.B.1 Torques can be described using force diagrams.
5.3.B.1.i Force diagrams are similar to free-body diagrams and are used to analyze the torques exerted on a rigid system.
5.3.B.1.ii Similar to free-body diagrams, force diagrams represent the relative magnitude and direction of the forces exerted on a rigid system. Force diagrams also depict the location at which those forces are exerted relative to the axis of rotation.
5.3.B.2 The magnitude of the torque exerted on a rigid system by a force is described by the equation below, where $\theta$ is the angle between the force vector and the position vector from the axis of rotation to the point of application of the force.

\tau = rF_{\perp} = rF\sin\theta

The boundary statement for this topic is short, and both halves of it matter. 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. Read both halves. You are expected to handle signs and vector conventions in a plane. You are not expected to produce the direction of the torque vector itself.

The CED lists four suggested skills here: 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim. The calculation skill 2.B is not among them. Diagram, derive, and reason about ratios rather than plug and chug. Diagram, derive, and reason about ratios, rather than plug and chug.

Three equivalent readings of the same equation

One equation, three ways to say it out loud. They give identical numbers, and each one is the fastest route in a different kind of problem.

Reading one: full distance times the perpendicular component of the force.

\tau = r\left(F\sin\theta\right)

This is essential knowledge 5.3.A.1 written as arithmetic. Split the force into a piece along the position vector and a piece perpendicular to it. The along piece pulls the system toward or away from the axis and turns nothing. Only $F\sin\theta$ produces torque.

Reading two: lever arm times the full force.

\tau = \left(r\sin\theta\right)F = r_{\perp}F

This is essential knowledge 5.3.A.2. Group the sine with the distance instead of with the force and the quantity $r\sin\theta$ becomes the lever arm, the perpendicular distance from the axis to the line of action.

Reading three: the magnitude of a cross product. The full vector definition of torque is a cross product of the position vector and the force, and the magnitude of any cross product is the product of the two magnitudes times the sine of the angle between them, which is exactly $rF\sin\theta$ . AP Physics 1 uses the magnitude and stops there, because this topic's boundary statement puts the direction of the torque vector outside the course. Physics C: Mechanics keeps going and prints the vector form on its sheet.

Two consequences follow from all three readings at once, and both are worth checking for before you compute anything:

A force applied at the axis produces no torque, because $r = 0$ .
A force pointing along the position vector, directly toward or directly away from the axis, produces no torque, because $\sin 0^\circ = \sin 180^\circ = 0$ .

Neither of those forces is zero. Both produce zero torque. A force can be large, real, and completely irrelevant to rotation.

Angles, and why the sine is measured where it is

The $\theta$ in $rF\sin\theta$ sits inside a sine, so degrees are the natural unit for it and a calculator in degree mode returns the right torque. That is not true of every angle in Unit 5: essential knowledge 5.1.A.1 measures angular displacement in radians, and the moment an angle multiplies a radius, as in $v = r\omega$ , only radians work. The exam's Table of Information carries a trigonometry table that lists sine, cosine, and tangent at $0^\circ$ , $30^\circ$ , $37^\circ$ , $45^\circ$ , $53^\circ$ , $60^\circ$ , and $90^\circ$ , which is a strong hint about the angles AP questions tend to choose.

The angle $\theta$ in $rF\sin\theta$ is defined in 5.3.B.2 as the angle between the force vector and the position vector from the axis to the point of application. Not the angle to the horizontal. Not the angle to the ground. The angle between those two specific vectors.

That definition has three practical consequences.

A perpendicular force is the maximum. At $\theta = 90^\circ$ , $\sin\theta = 1$ and $\tau = rF$ . Every other angle gives less torque for the same force at the same distance.
Obtuse angles behave like their supplements. $\sin 150^\circ = \sin 30^\circ$ , and $\sin 120^\circ = \sin 60^\circ$ . Pulling a wrench handle at $120^\circ$ gives exactly the same torque magnitude as pulling at $60^\circ$ . If your calculator is in radian mode you will not notice this; check the mode before the exam.
Contrast it with work. Work uses $F d\cos\theta$ and torque uses $rF\sin\theta$ . Work keeps the component of the force along the displacement; torque keeps the component perpendicular to the position vector. Students who learned work first sometimes carry the cosine across. The two are complementary on purpose: a force that does maximum work along a line does zero torque about a point on that line.

The unit of torque is the newton meter. A newton meter has the same base units as a joule, but torque is not energy: it measures a force's turning effect about an axis rather than energy transferred along a path. Keep torque in N·m.

The lever arm and the line of action

Essential knowledge 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. Two phrases in there do real work.

Line of action means the infinite straight line running through the force arrow in both directions, not just the arrow you drew. Perpendicular distance means the shortest distance from the axis to that line, which is always measured at a right angle to it.

The useful theorem that falls out: sliding a force anywhere along its own line of action does not change the torque it produces. Both $r$ and $\theta$ change, but the product $r\sin\theta$ does not. This is why a rope's torque about a pivot depends on where the rope's line runs, not on which knot you call the point of application.

To find a lever arm on a diagram, draw the force arrow at the point where the force acts, extend it into a full line forward and backward, then drop a perpendicular from the axis of rotation onto that line. The length of that perpendicular is $r_{\perp}$ , and $\tau = r_{\perp}F$ .

The door question the CED lists among its essential questions for this unit ("Why does it matter where a door handle is placed?") is exactly this construction. A handle far from the hinge gives a long lever arm for a push perpendicular to the door. A handle next to the hinge gives a short one. Pushing along the plane of the door aims the line of action straight through the hinge, the perpendicular distance is zero, and the door does not budge no matter how hard you push. The unit's other essential question, "Why are long wrenches more effective?", has the same one-line answer: more $r_{\perp}$ for the same $F$ .

Declaring a sign convention: counterclockwise is positive

For the rest of this page, counterclockwise torque is positive and clockwise torque is negative. Every number below is signed under that convention.

This is the vector convention the boundary statement is talking about. Topic 5.1 sets it up: essential knowledge 5.1.A.1.ii says that one direction of angular displacement about an axis of rotation, clockwise or counterclockwise, is typically indicated as mathematically positive, with the other direction becoming mathematically negative. Topics 5.1 and 5.2 both carry a boundary statement restricting descriptions of the directions of rotation for a point or object to clockwise and counterclockwise with respect to a given axis of rotation. So the entire directional vocabulary you need in AP Physics 1 rotation is two words and two signs.

Choosing counterclockwise as positive is a convention, not a law. Clockwise-positive is equally valid, gives every torque the opposite sign, and produces an identical physical answer. What is not optional is holding one choice for a whole problem. Assign each torque its sign at the moment you write it down, before any algebra runs, and the error of adding two opposing torques as though they cooperated cannot happen.

The net torque is the signed sum:

\tau_{\text{net}} = \Sigma\tau

Under counterclockwise-positive, a positive $\tau_{\text{net}}$ means the system's angular velocity is being driven counterclockwise, a negative one means clockwise, and zero means the angular velocity is not changing at all. That last case is Topic 5.5, and the nonzero cases are Topic 5.6.

One thing the sign does not tell you: which way the torque vector points in three dimensions. The right-hand rule and the vector $\vec{\tau}$ are Physics C material. In AP Physics 1 the sign is the whole directional story.

Force diagrams are not free-body diagrams

Essential knowledge 5.3.B.1 introduces a representation that skill 1.A then asks you to draw. Sub-statement 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. Sub-statement 5.3.B.1.ii says that, similar to free-body diagrams, force diagrams represent the relative magnitude and direction of the forces exerted on a rigid system, and that force diagrams also depict the location at which those forces are exerted relative to the axis of rotation.

That last clause is the entire difference, and it is worth stating plainly against what you already know from Topic 2.2.

	Free-body diagram	Force diagram
Object drawn as	a dot or a box	the extended shape, with its axis marked
Force arrows start	all at one point	at the actual point of application
Relative magnitudes shown	yes	yes
Answers the question	what is $\Sigma\vec{F}$ ?	what is $\Sigma\tau$ ?

Collapsing an object to a point is exactly the move you must not make here. Topic 5.1's essential knowledge 5.1.A.1.i is blunt about why: a rigid system is one that holds its shape but in which different points of the system move in different directions during rotation, and a rigid system cannot be modeled as an object. The moment you shrink a beam to a dot, every force acts at $r = 0$ , every torque is zero, and the diagram has destroyed the information you needed.

Practical rules for drawing one that scores:

Draw the body to a rough scale and mark the axis of rotation with a clear symbol.
Place the tail of each force arrow at the point where that force is applied, and put a uniform body's weight at its center of mass.
Label distances from the axis along the body, since those are the $r$ values that will enter every torque.
Keep relative lengths honest. A 200 N arrow drawn shorter than a 50 N arrow loses credit for the representation even when the algebra is right.

Functional dependence: what the exam does with 2.D

Skill 2.D, predicting new values or factors of change using functional dependence, is attached to this topic, and the CED's own unit overview names torque as its worked illustration: students might be asked to determine the torque exerted on a system if the force exerted is doubled. The unit overview also says that the analysis of functional relationships is assessed on the fourth free-response question, the Qualitative/Quantitative Translation question, as well as in the multiple-choice section.

So read $\tau = rF\sin\theta$ as three separate dependencies rather than as one calculation. Torque is directly proportional to $F$ with $r$ and $\theta$ fixed, and directly proportional to $r$ with $F$ and $\theta$ fixed. The third one is where the marks go: torque varies as $\sin\theta$ , which is not proportional to $\theta$ . Halving the angle from $90^\circ$ to $45^\circ$ does not halve the torque, it multiplies it by $\sin 45^\circ \approx 0.707$ , and going from $60^\circ$ to $30^\circ$ multiplies it by $\sin 30^\circ / \sin 60^\circ \approx 0.577$ .

The efficient way to answer these is to build the ratio and never compute either torque:

\frac{\tau_2}{\tau_1} = \frac{r_2}{r_1}\cdot\frac{F_2}{F_1}\cdot\frac{\sin\theta_2}{\sin\theta_1}

Every quantity that did not change cancels. Worked example three below runs this both ways and confirms the two agree.

A related derivation question (skill 2.A) asks for symbols only: show that the force needed to produce a given torque $\tau$ at a distance $r$ and angle $\theta$ is $F = \tau/(r\sin\theta)$ , then state what happens to $F$ as $\theta$ approaches zero. The answer is that $F$ grows without bound, which is the algebraic version of the fact that you cannot open a door by pushing toward the hinge.

How Topic 5.3 is tested

The AP Physics 1 exam is 3 hours long: 42 multiple-choice questions in 85 minutes for 50 percent of the score, then 4 free-response questions in 95 minutes for the other 50 percent. Those four are Mathematical Routines, Translation Between Representations, Experimental Design and Analysis, and Qualitative/Quantitative Translation. A four-function, scientific, or graphing calculator is allowed on both sections. Unit 5 supplies 10 to 15 percent of the multiple-choice section.

Question shapes that follow from this topic's four suggested skills:

Compute a torque with a non-perpendicular force (skill 2.A applied numerically). The difficulty is the sine and the sign.
Draw or critique a force diagram (skill 1.A). Expect a prompt that asks where each arrow's tail belongs relative to the axis, which is 5.3.B.1.ii read as a rubric.
Rank torques (skill 3.B). Several forces of different magnitudes at different points and angles on one body, ordered by turning effect. Compute $r_{\perp}$ for each and compare those; do not compare forces.
Predict a factor of change (skill 2.D). Answer as a ratio.
Identify the zero-torque force (skill 3.B). One listed force acts at the axis or along the position vector and contributes nothing regardless of its size.

The CED's sample instructional activities for Topic 5.3 sit at the qualitative end of the same ideas: predicting what happens to a spun bike wheel's linear and angular velocity as it is released and rolls, planning how much force two end supports must provide for a suspended walkway, and telling a hard-boiled egg from a raw one by spinning each and touching it to a stop.

Where this goes next: Topic 5.4 supplies the rotational inertia that torque has to work against, Topic 5.5 sets the net torque to zero, and Topic 5.6 puts them together as $\alpha = \tau_{\text{net}}/I$ . For the step-by-step routine on equilibrium problems, use the how to calculate torque guide; to check a torque or a lever arm against worked numbers, use the torque calculator.

A wrench pulled at an obtuse angle, computed three ways

You pull on a wrench handle at a point $r = 0.32$ m from the bolt with a force of magnitude $F = 55$ N. The angle between the position vector (bolt to your hand) and the force is $\theta = 120^\circ$ , and the pull turns the bolt counterclockwise. Find the torque about the bolt using all three readings of the formula, then state what fraction of the maximum possible torque you are getting.

List the values and fix the convention: $r = 0.32$ m, $F = 55$ N, $\theta = 120^\circ$ , counterclockwise positive, so the answer should come out positive.
Reading one, perpendicular component of the force (5.3.A.1): $F_{\perp} = F\sin\theta = (55\ \text{N})(\sin 120^\circ) = (55)(0.8660) = 47.63$ N. Then $\tau = rF_{\perp} = (0.32\ \text{m})(47.63\ \text{N}) = 15.24$ N·m.
Reading two, lever arm (5.3.A.2): $r_{\perp} = r\sin\theta = (0.32\ \text{m})(0.8660) = 0.2771$ m. Then $\tau = r_{\perp}F = (0.2771\ \text{m})(55\ \text{N}) = 15.24$ N·m.
Reading three, straight through 5.3.B.2: $\tau = rF\sin\theta = (0.32)(55)(0.8660) = 15.24$ N·m. All three agree to every digit, since they are the same three factors multiplied in a different order.
Apply the sign: the pull turns the bolt counterclockwise, so $\tau = +15$ N·m to two significant figures.
Maximum possible torque at this $r$ and $F$ is at $\theta = 90^\circ$ : $\tau_{\max} = rF = (0.32)(55) = 17.6$ N·m. The fraction obtained is $15.24/17.6 = 0.866$ , which is $\sin 120^\circ$ again.
Note the obtuse angle: $\sin 120^\circ = \sin 60^\circ = 0.8660$ . Pulling at $120^\circ$ and pulling at $60^\circ$ produce the same torque magnitude, because the perpendicular component of the force has the same size either way.

$\tau = +15$ N·m (counterclockwise), or 15.24 N·m before rounding. That is 86.6 percent of the 17.6 N·m you would get by pulling perpendicular to the handle, and the three readings of $\tau = rF\sin\theta$ return identical numbers.

Net torque on a hinged beam, with one force that contributes nothing

A rigid beam is hinged at its left end, and the hinge is the axis of rotation. Three forces act on it. $F_1 = 24$ N points straight down at a point 1.8 m from the hinge, perpendicular to the beam, turning it clockwise. $F_2 = 60$ N acts at a point 0.60 m from the hinge, at $30^\circ$ to the beam, turning it counterclockwise. $F_3 = 90$ N pulls horizontally outward along the beam, directly away from the hinge, at a point 1.2 m from the hinge. Find the net torque about the hinge, and then find the single perpendicular force at the far end (1.8 m) that would bring the beam to rotational equilibrium.

Convention: counterclockwise positive, clockwise negative.
$F_1$ is perpendicular to the beam, so $\theta_1 = 90^\circ$ and $\sin\theta_1 = 1$ . It turns the beam clockwise, so it gets a minus sign: $\tau_1 = -(1.8\ \text{m})(24\ \text{N})(1) = -43.2$ N·m.
$F_2$ acts at $30^\circ$ to the beam, and the equation sheet's trigonometry table gives $\sin 30^\circ = 1/2$ exactly: $\tau_2 = +(0.60\ \text{m})(60\ \text{N})(0.5) = +18.0$ N·m. Cross-check with the lever arm: $r_{\perp} = (0.60)(0.5) = 0.30$ m, and $(0.30)(60) = 18.0$ N·m.
$F_3$ points directly away from the hinge, so it lies along the position vector: $\theta_3 = 0^\circ$ , $\sin 0^\circ = 0$ , and $\tau_3 = 0$ regardless of its 90 N magnitude. Its line of action passes straight through the axis, so its lever arm is zero. This force is the largest of the three and it turns the beam not at all.
Sum the signed torques: $\tau_{\text{net}} = -43.2 + 18.0 + 0 = -25.2$ N·m.
The sign says clockwise. Under the declared convention a negative net torque drives the beam clockwise, which matches the fact that the clockwise contribution (43.2 N·m) beat the counterclockwise one (18.0 N·m).
For equilibrium, the added torque must be $+25.2$ N·m. Applied perpendicular at $r = 1.8$ m, $\sin\theta = 1$ , so $F = (25.2\ \text{N·m})/(1.8\ \text{m}) = 14.0$ N, directed so that it turns the beam counterclockwise (upward at the far end).
Check the size: 14.0 N is much less than $F_1 = 24$ N even though both act at 1.8 m, because $F_2$ was already supplying part of the counterclockwise torque. Removing $F_2$ would require the full $43.2/1.8 = 24$ N, which is $F_1$ again, as expected.

$\tau_{\text{net}} = -25.2$ N·m, that is 25.2 N·m clockwise. A 14.0 N force applied perpendicular to the beam at the far end, turning it counterclockwise, restores rotational equilibrium. The 90 N force along the beam contributes exactly zero torque.

Predicting a factor of change without computing a torque

A student tightens a bolt by pulling perpendicular to a wrench handle at $r = 0.36$ m with $F = 45$ N. On the next attempt the student pulls twice as hard, but slides the hand in to one third of the original distance and pulls at $\theta = 25^\circ$ to the handle instead of perpendicular. By what factor does the torque change? Confirm by computing both torques.

Set up the ratio and let everything unchanged cancel: $\dfrac{\tau_2}{\tau_1} = \dfrac{r_2}{r_1}\cdot\dfrac{F_2}{F_1}\cdot\dfrac{\sin\theta_2}{\sin\theta_1}$ .
Substitute the three factors of change: $\dfrac{r_2}{r_1} = \dfrac{1}{3}$ , $\dfrac{F_2}{F_1} = 2$ , and $\dfrac{\sin\theta_2}{\sin\theta_1} = \dfrac{\sin 25^\circ}{\sin 90^\circ} = \dfrac{0.4226}{1} = 0.4226$ .
Multiply: $\dfrac{\tau_2}{\tau_1} = \left(\dfrac{1}{3}\right)(2)(0.4226) = 0.2817$ . The torque drops to about 28 percent of its original value.
Now confirm the long way. Original: $\tau_1 = (0.36\ \text{m})(45\ \text{N})(\sin 90^\circ) = (0.36)(45)(1) = 16.2$ N·m.
New values: $r_2 = 0.36/3 = 0.12$ m, $F_2 = 2(45) = 90$ N, $\theta_2 = 25^\circ$ . So $\tau_2 = (0.12)(90)(0.4226) = (10.8)(0.4226) = 4.564$ N·m.
Ratio check: $4.564/16.2 = 0.2817$ , matching the ratio method exactly.
Interpret it: pulling twice as hard bought a factor of 2, but the shorter grip cost a factor of 3 and the shallow angle cost another factor of 2.37, and the losses won. What the two losses did together was shrink the lever arm from $r_{\perp} = (0.36)(\sin 90^\circ) = 0.360$ m to $r_{\perp} = (0.12)(\sin 25^\circ) = 0.0507$ m, a factor of $0.1409$ , or about one seventh. Doubling a force does not rescue a lever arm cut to a seventh.

The torque falls to 0.28 of its original value, about 28 percent. Numerically $\tau_1 = 16.2$ N·m and $\tau_2 = 4.56$ N·m, and $4.56/16.2 = 0.28$ , confirming the ratio.

Frequently asked questions

What is torque in AP Physics 1?

Torque is the measure of how strongly a force turns a rigid system about an axis of rotation. Essential knowledge 5.3.A.1 states 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. Its magnitude is given by tau = r times F-perpendicular = rF sin theta, and it is measured in newton meters.

What is the lever arm?

The lever arm is the perpendicular distance from the axis of rotation to the line of action of the exerted force, which is essential knowledge 5.3.A.2. On a diagram you find it by extending the force arrow into a full straight line and then dropping a perpendicular from the axis onto that line. Because it equals r sin theta, torque can be written as the lever arm times the full force. Sliding a force along its own line of action leaves the lever arm, and therefore the torque, unchanged.

Do I need the right-hand rule for torque in AP Physics 1?

No. The Topic 5.3 boundary statement says that 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. You do need signs: pick clockwise or counterclockwise as positive and hold that choice through the whole problem. The right-hand rule and the vector cross product form of torque belong to AP Physics C: Mechanics.

When is the torque from a force zero?

In two cases, and neither requires the force itself to be zero. First, when the force is applied at the axis of rotation, so r = 0. Second, when the force points directly toward or directly away from the axis, so the angle between the position vector and the force is 0 or 180 degrees and sin theta = 0. In the second case the force's line of action passes through the axis, giving a lever arm of zero. A very large force can produce no torque at all.

Why does torque use sine when work uses cosine?

Because they keep opposite components of the force. Work, W = Fd cos theta, keeps the component along the displacement. Torque, tau = rF sin theta, keeps the component perpendicular to the position vector from the axis, because only that perpendicular part turns the system. The two are complementary: a force aimed straight along a line through the axis does maximum work moving an object along that line and zero torque about a point on it.

What is the difference between a force diagram and a free-body diagram?

A free-body diagram collapses the object to a point and shows only the magnitudes and directions of the forces, which is all you need for the net force. A force diagram, described in essential knowledge 5.3.B.1, keeps the object's extended shape and also depicts the location at which each force is exerted relative to the axis of rotation. That location information is what makes torque calculable, so torque problems require the force diagram version.

Is torque measured in joules?

No. Torque is measured in newton meters. A newton meter has the same base units as a joule, but torque and energy are different physical quantities: torque measures a force's turning effect about an axis, while a joule measures energy transferred. Write torque answers as N·m and reserve joules for energy.