AP Physics C: Mechanics · Topic 5.3

Topic 5.3: Torque

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

In AP Physics C: Mechanics, torque is the cross product of the position vector and the force. Its magnitude is the two magnitudes times the sine of the angle between them, and its direction is perpendicular to both, by the right-hand rule. AP Physics 1 calls that direction out of scope.

AP Physics: Unit 5 (topics 5.3 Torque). AP Physics C: Mechanics Unit 5, Topic 5.3. Two learning objectives. 5.3.A, identify the torques exerted on a rigid system, supported by 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) and 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). 5.3.B, describe the torques exerted on a rigid system, supported by 5.3.B.1 (torques can be described using force diagrams) with 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) and 5.3.B.1.ii (force diagrams represent the relative magnitude and direction of the forces exerted on a rigid system and also depict the location at which those forces are exerted relative to the axis of rotation); and by 5.3.B.2 (the torque exerted on a rigid system about a chosen pivot point by a given force is described by tau-vector = r-vector cross F-vector) with 5.3.B.2.i (the cross-product between two vectors A and B results in a vector quantity of magnitude AB sin theta), 5.3.B.2.ii (the direction of the vector resulting from the cross-product is perpendicular to both vectors and therefore is normal to the plane defined by them) and 5.3.B.2.iii (that direction can be qualitatively determined by applying the appropriate right-hand rule). TOPIC 5.3 PRINTS NO BOUNDARY STATEMENT in this course; it is one of two Unit 5 topics with none, along with 5.6. AP Physics 1's Topic 5.3 does print one: '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.' AP Physics 1's 5.3.A.1, 5.3.A.2, 5.3.B.1, 5.3.B.1.i and 5.3.B.1.ii are identical to this course's; its 5.3.B.2 gives the magnitude as tau = r F-perpendicular = r F sin theta and it has no 5.3.B.2.i, .ii or .iii. The phrase 'cross-product' appears three times in the C: Mechanics CED and zero times in the AP Physics 1 CED; 'right-hand' appears once in the C: Mechanics CED and zero times in the AP Physics 1 CED. Suggested skills 1.A, 2.B, 2.C, 3.C; AP Physics 1's Topic 5.3 suggests 1.A, 2.A, 2.D, 3.B, sharing only 1.A. Sheets: the C: Mechanics mechanics table prints the cross-product form with arrows and its Table of Information carries a VECTORS box printing the cross-product magnitude with absolute-value bars and the dot product; the AP Physics 1 sheet prints tau = r-perpendicular F = r F sin theta and has no VECTORS box. Note two CED-versus-sheet discrepancies: 5.3.B.2.i prints the cross-product magnitude without absolute-value bars while the sheet prints it with them, and AP Physics 1's 5.3.B.2 puts the perpendicular subscript on F while its sheet puts it on r. CED sample multiple-choice question 9 aligns to skill 2.B, LO 5.3.B and EK 5.3.B.2: a uniform 0.60 m, 4.0 kg beam pivoted at a wall, held horizontal by a string, with a 1.0 kg block at the end, asking for the torque about the pivot exerted by the weight of the block; keyed answer B, 6 N times metres. Sample instructional activities 2 and 3 are assigned to Topic 5.3: designing a suspended walkway and its two support forces, and a bike wheel released from rest that peels out, explained with a force diagram.

What Topic 5.3 requires

Topic 5.3 is one of only two topics in Unit 5 with two learning objectives, the other being Topic 5.4. Its suggested skills are 1.A (create diagrams, tables, charts, or schematics to represent physical situations), 2.B (calculate or estimate an unknown quantity with units from known quantities), 2.C (compare physical quantities between two or more scenarios or at different times and/or locations within a single scenario) and 3.C (justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws).

5.3.A: 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.

5.3.B: Describe the torques exerted on a rigid system.

5.3.B.1 Torques can be described using force diagrams. 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.
5.3.B.2 The torque exerted on a rigid system about a chosen pivot point by a given force is described by

\vec{\tau} = \vec{r} \times \vec{F}

5.3.B.2.i The cross-product between two vectors, $\vec{A}$ and $\vec{B}$ , results in a vector quantity of magnitude $\vec{A} \times \vec{B} = AB\sin\theta$ .
5.3.B.2.ii The direction of the vector resulting from the cross-product of vectors $\vec{A}$ and $\vec{B}$ is perpendicular to both vectors $\vec{A}$ and $\vec{B}$ and therefore is normal to the plane defined by vectors $\vec{A}$ and $\vec{B}$ .
5.3.B.2.iii The direction of the vector resulting from the cross-product of vectors $\vec{A}$ and $\vec{B}$ can be qualitatively determined by applying the appropriate right-hand rule.

Topic 5.3 prints no boundary statement. It is one of two topics in Unit 5 that does not, along with Topic 5.6. That absence is the subject of the next section, and it is the sharpest single difference between this course and AP Physics 1.

The one place the two frameworks openly contradict each other

AP Physics 1 has a Topic 5.3 called Torque, with the same two learning objectives and the same codes. Count the numbered items under each and the arithmetic is this: AP Physics C: Mechanics prints nine, AP Physics 1 prints six, five of those six are word for word identical, and the sixth is where the two courses part. Under its version, College Board prints a boundary statement. Quoted whole:

"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."

AP Physics C: Mechanics prints no boundary statement under Topic 5.3 at all. What it prints instead is statement 5.3.B.2 with the cross product, and then three sub-statements that do nothing except explain how to get a direction out of it: the magnitude, the fact that the result is perpendicular to both inputs and therefore normal to the plane they define, and the right-hand rule.

One framework declares the direction of torque out of scope. The other spends three consecutive sub-statements on it. This is not a difference of emphasis that an author has inferred; it is two sentences in two published documents that point in opposite directions, and you can check both.

The vocabulary counts confirm it. The phrase "cross-product" appears three times in the AP Physics C: Mechanics course and exam description, all three inside 5.3.B.2.i, 5.3.B.2.ii and 5.3.B.2.iii, and zero times in the AP Physics 1 course and exam description. "Right-hand" appears exactly once in the C: Mechanics document, in 5.3.B.2.iii, and zero times in the AP Physics 1 document.

The equation sheets agree with the frameworks. The AP Physics C: Mechanics sheet prints $\vec{\tau} = \vec{r} \times \vec{F}$ , arrows over all three symbols. The AP Physics 1 sheet prints $\tau = r_{\perp}F = rF\sin\theta$ , a magnitude with no arrow anywhere in it. And the C: Mechanics Table of Information carries a VECTORS box, printing $\lvert\vec{A} \times \vec{B}\rvert = AB\sin\theta$ , $\vec{A} \cdot \vec{B} = AB\cos\theta$ , unit-vector notation and component addition. The AP Physics 1 Table of Information has no VECTORS box; it is two pages long, and neither page has one.

Even the suggested skills diverge here. AP Physics C: Mechanics lists 1.A, 2.B, 2.C and 3.C for Topic 5.3. AP Physics 1 lists 1.A, 2.A, 2.D and 3.B. They share exactly one.

Getting a direction out of the cross product

Statement 5.3.B.2 says the torque is exerted "about a chosen pivot point". Choose it first and write it down, because $\vec{r}$ is measured from there and a torque with no stated axis is not a number.

Then there are three things to extract.

Magnitude. Statement 5.3.B.2.i gives $\lvert\vec{r} \times \vec{F}\rvert = rF\sin\theta$ , where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$ when the two are drawn tail to tail. Two readings of that same product are worth having in hand, and both are in 5.3.A:

$rF_{\perp}$ , from 5.3.A.1: only the component of the force perpendicular to $\vec{r}$ produces torque.
$r_{\perp}F$ , from 5.3.A.2: the lever arm is the perpendicular distance from the axis to the line of action, and the whole force acts through it.

These are the same number written two ways, because $r(F\sin\theta) = (r\sin\theta)F$ . Use whichever the geometry hands you.

Direction. Statement 5.3.B.2.ii says the result is perpendicular to both $\vec{r}$ and $\vec{F}$ , and therefore normal to the plane they define. That is a strong claim and it is worth verifying once, which worked example 2 does: the dot product of the computed torque with each input comes out exactly zero.

The right-hand rule. Statement 5.3.B.2.iii says the direction can be qualitatively determined by applying the appropriate right-hand rule. Point the fingers of your right hand along $\vec{r}$ , curl them toward $\vec{F}$ through the smaller angle, and your thumb points along $\vec{\tau}$ . Note the order: $\vec{r}$ first, then $\vec{F}$ . The cross product is not commutative, and $\vec{F} \times \vec{r}$ points the other way.

Two consequences you can read straight off the definition:

If $\vec{r}$ and $\vec{F}$ are parallel or antiparallel, $\sin\theta = 0$ and the torque is zero regardless of how large the force is. Push a door straight at its hinge and nothing turns. Statement 5.3.A.1 says the same thing in words.
If the point of application is on the axis, $r = 0$ and the torque is zero. A force applied at the pivot never turns anything about that pivot.

Most of the time you want one signed component, not three

A full three-dimensional torque vector is not what most AP Physics C: Mechanics problems call for, and the unit says so. Topic 5.5's boundary statement reads: "AP Physics C: Mechanics does not expect students to simultaneously analyze rotation in multiple planes." Almost every problem you will meet has $\vec{r}$ and $\vec{F}$ both lying in one plane, and then the whole torque points along the one axis perpendicular to it.

Set up that case once and reuse it. Put the plane of the problem in the $xy$ plane with $\hat{k}$ out of the page. Then

\tau_z = xF_y - yF_x

and there is nothing else. A positive $\tau_z$ points out of the page, which by the right-hand rule is a counterclockwise turning tendency as you look at the page; a negative $\tau_z$ points into the page and turns things clockwise. Net torque is then a plain algebraic sum of signed numbers.

This is what "vector conventions" means in practice, and it is the sense in which AP Physics 1 also uses signs. The difference is that in AP Physics 1 the sign is a bookkeeping convention with no physical direction behind it, while here it is the $z$ component of a vector whose direction the right-hand rule fixes and whose perpendicularity 5.3.B.2.ii asserts.

Be careful in the other direction too. Topic 5.1's and Topic 5.2's boundary statements both say the directions of angular displacement, angular velocity and angular acceleration as vectors will not be assessed, and limit descriptions of those directions to clockwise and counterclockwise. So within one unit, torque has a vector direction you can be asked about and $\omega$ does not. The asymmetry is real, and it is the most precise thing you can know about Unit 5's scope.

What is identical, so you know exactly where the difference is

It would be easy to overstate the gap. Topic 5.3 in AP Physics C: Mechanics prints nine numbered items and AP Physics 1's prints six. Five of the six are word for word identical in the two documents. Here is every one of the nine, item by item.

Item	AP Physics C: Mechanics	AP Physics 1
5.3.A.1 torque from the perpendicular force component	printed	identical
5.3.A.2 lever arm as the perpendicular distance to the line of action	printed	identical
5.3.B.1 torques can be described using force diagrams	printed	identical
5.3.B.1.i force diagrams analyze torques on a rigid system	printed	identical
5.3.B.1.ii they also depict where each force is exerted	printed	identical
5.3.B.2	$\vec{\tau} = \vec{r} \times \vec{F}$ , about a chosen pivot point	the magnitude, $\tau = rF_{\perp} = rF\sin\theta$
5.3.B.2.i cross-product magnitude	printed	does not exist
5.3.B.2.ii perpendicular to both, normal to the plane	printed	does not exist
5.3.B.2.iii the right-hand rule	printed	does not exist
Boundary statement	none	the direction of torque is beyond the scope

So the definition of a lever arm, the rule that only the perpendicular force component matters, and the whole treatment of force diagrams are common ground. Statement 5.3.B.1.ii is worth reading in either course, because it is the sentence that separates a force diagram from a free-body diagram: a force diagram also depicts the location at which each force is exerted relative to the axis of rotation. A free-body diagram collapses everything to a point, which is exactly the information a torque problem cannot afford to lose.

The difference is contained in one restated statement, three that AP Physics 1 does not have, and one absent boundary statement. That is a smaller surface than "the C course is harder", and it is a more useful thing to know.

What each sheet prints, and where a CED disagrees with its own sheet

Item	C: Mechanics sheet	AP Physics 1 sheet
torque	$\vec{\tau} = \vec{r} \times \vec{F}$	$\tau = r_{\perp}F = rF\sin\theta$
cross-product magnitude	$\lvert\vec{A} \times \vec{B}\rvert = AB\sin\theta$ , in the VECTORS box	not printed, there is no VECTORS box
dot product	$\vec{A} \cdot \vec{B} = AB\cos\theta$ , in the VECTORS box	not printed
unit vectors and component addition	printed in the VECTORS box	not printed
calculus rules	printed in the CALCULUS box	not printed
trigonometric values for common angles	printed	printed

Two details on this table are the kind that get miscopied, so they are worth stating exactly.

The cross-product magnitude is printed, and you should not conclude otherwise from a partial transcription of the sheet. It lives in the VECTORS box on the appendix page that also carries the geometry, calculus and identity boxes, not in the mechanics equation table. The same page prints $s = r\theta$ in its geometry box.

The C: Mechanics CED and the C: Mechanics sheet write the cross-product magnitude differently. Statement 5.3.B.2.i prints $\vec{A} \times \vec{B} = AB\sin\theta$ with no absolute-value bars, in a sentence that says the cross product "results in a vector quantity of magnitude" that expression. The sheet prints $\lvert\vec{A} \times \vec{B}\rvert = AB\sin\theta$ with the bars. The sheet's version is the careful one: a vector does not equal a scalar. Read the CED's line as the sentence intends it.

AP Physics 1's own CED and sheet also disagree with each other, in the mirror image of the same way. Its statement 5.3.B.2 prints $\tau = rF_{\perp} = rF\sin\theta$ , putting the perpendicular subscript on the force. Its equation sheet prints $\tau = r_{\perp}F = rF\sin\theta$ , putting it on the position. Both are correct and both equal $rF\sin\theta$ ; they are 5.3.A.1's reading and 5.3.A.2's reading of one product.

Traps

No axis, no torque. Statement 5.3.B.2 says "about a chosen pivot point". The same force gives different torques about different points, and one of them is usually zero. Declare the axis in the first line of every solution.

A large force can exert zero torque. If the force's line of action passes through the axis, the lever arm is zero and so is the torque, no matter the magnitude. The cross product says it with $\sin 0 = 0$ .

The angle in $rF\sin\theta$ is between the vectors, tail to tail. It is not the angle to the horizontal, not the angle to the object, and not the angle at the far end of the triangle you happened to draw. Statement 5.3.B.2 in AP Physics 1 spells the definition out; the C: Mechanics version leaves it to the cross product, where it is the same angle.

Order matters. $\vec{r} \times \vec{F}$ and $\vec{F} \times \vec{r}$ have equal magnitudes and opposite directions. Point along $\vec{r}$ first.

Do not import the direction of torque into rotational kinematics. Topics 5.1 and 5.2 carry boundary statements that limit descriptions of the directions of angular displacement, angular velocity and angular acceleration to clockwise and counterclockwise. The freedom Topic 5.3 gives you is about torque.

Do not import the AP Physics 1 boundary statement into this course. It says "While AP Physics 1 expects" and it appears only in the AP Physics 1 document. Nothing like it appears under this course's Topic 5.3.

A distributed weight is not automatically at the geometric center. Replacing an object's weight with a single force at its center of mass is legitimate, but the center of mass of a nonuniform body is not its midpoint. Worked example 3 integrates the torque directly and finds it 10/9 of what the uniform assumption would give.

Watch the sign convention across a whole problem. Declare $\hat{k}$ out of the page and positive $\tau_z$ counterclockwise, then hold it to the end. An axis that flips mid-solution is a defect even when each line is individually right.

How Topic 5.3 is assessed

Two of the fifteen sample multiple-choice questions in the AP Physics C: Mechanics course and exam description align to Unit 5, and one of them is a Topic 5.3 question.

Sample question 9 is aligned to skill 2.B, learning objective 5.3.B and essential knowledge 5.3.B.2. A uniform beam of length 0.60 m and mass 4.0 kg is fixed to a wall by a pivot, held horizontal by a string fixed to the beam and to the wall, with a block of mass 1.0 kg attached at the end of the beam. It asks which value is nearest to the magnitude of the torque about the pivot exerted by the weight of the block. The keyed answer is B, 6 N times metres.

That question rewards reading. The beam's own 4.0 kg mass and the string are both in the figure and neither belongs in the answer, because the question names one force. The arithmetic is $(1.0)(9.8)(0.60) = 5.9$ N times metres, nearest to 6. This is what skill 2.B, calculate or estimate an unknown quantity, looks like when the difficulty is in the parsing rather than in the algebra.

Skill 3.C, justify or support a claim using evidence, is also suggested for this topic, and it is the skill behind the CED's two Topic 5.3 sample instructional activities. Activity 2 asks students to design a walkway of given mass to be suspended from a ceiling and to determine the force each of two supports must provide as a person of given mass walks across it. Activity 3 asks students to spin a bike wheel, release it from rest on the floor, predict what happens to its linear velocity and its angular velocity as it "peels out", and then explain why using a force diagram. Note the phrase: a force diagram, in the 5.3.B.1 sense, not a free-body diagram.

Skill 2.A, deriving symbolic expressions, is not among the four suggested for Topic 5.3 in this course, although it carries 25 to 30% of the multiple-choice section overall. Skills 2.B and 2.C, which are suggested here, carry 20 to 25% and 10 to 15% respectively, and 3.C carries 5 to 10%.

If you are taking AP Physics 1, this is not your page

This page teaches torque as a vector with a direction, using a cross product and a right-hand rule. Your course description states in a boundary statement that the direction of torque is beyond the scope of AP Physics 1. Everything in the sections above about $\hat{k}$ , right-hand rules and perpendicularity is outside your framework, and you will not be assessed on it.

If you are in AP Physics 1, read the [AP Physics 1 Topic 5.3 page](/ap-physics-1/unit-5-torque-and-rotational-dynamics/5-3-torque) instead. It is written to your framework, it treats torque as a signed magnitude, and it stops where your boundary statement stops. This page is for AP Physics C: Mechanics students. The two pages share a title because College Board gave the two topics the same one, not because the material is the same.

What you do share is the physics of 5.3.A.1 and 5.3.A.2: only the perpendicular force component matters, and the lever arm is the perpendicular distance from the axis to the line of action. Both courses build on that.

From here in this course, Topic 5.4 turns rotational inertia into an integral, Topic 5.5 sets the sum of torques to zero, and Topic 5.6 sets it equal to $I\alpha$ . For step-by-step procedure rather than framework, the how to calculate torque guide owns the routine and the torque calculator checks arithmetic. The Unit 5 hub lists all six topics.

Net torque on a hinged beam, as a vector with a direction

A uniform beam of length $L = 2.40$ m and mass 12.0 kg is hinged to a wall at its left end and lies horizontally. A lamp of mass 6.0 kg hangs from the beam 2.00 m from the hinge. A cable attached to the far end pulls with a force of 250 N directed at $35^\circ$ above the beam. Using $g = 9.8\ \mathrm{m/s^2}$ , find the net torque about the hinge as a vector, and state which way the beam would begin to turn.

Declare the geometry and the convention before any arithmetic. Put the hinge at the origin, $+x$ along the beam away from the wall, $+y$ up, and $\hat{k}$ out of the page. Every point on the beam then has $y = 0$ , so $\tau_z = xF_y - yF_x$ reduces to $\tau_z = xF_y$ for every force here. A positive $\tau_z$ is counterclockwise.
That reduction is worth noticing: the cable's horizontal component contributes exactly nothing to the torque about the hinge, because its line of action passes through the hinge's height. Statement 5.3.A.1 says the same thing.
Beam's own weight. A uniform beam's weight acts at its center of mass, $x = L/2 = 1.20$ m, with $F_y = -(12.0)(9.8) = -117.6$ N. So $\tau_z = (1.20)(-117.6) = -141.12\ \mathrm{N \cdot m}$ .
Lamp. At $x = 2.00$ m with $F_y = -(6.0)(9.8) = -58.8$ N, so $\tau_z = (2.00)(-58.8) = -117.6\ \mathrm{N \cdot m}$ .
Cable. Its components are $F_x = 250\cos 35^\circ = 204.79$ N and $F_y = 250\sin 35^\circ = 143.39$ N, applied at $x = 2.40$ m. So $\tau_z = (2.40)(143.39) = +344.15\ \mathrm{N \cdot m}$ .
Check the cable torque the other way, using the lever arm from 5.3.A.2. The perpendicular distance from the hinge to the cable's line of action is $r_{\perp} = L\sin 35^\circ = (2.40)(0.57358) = 1.3766$ m, and $r_{\perp}F = (1.3766)(250) = 344.15\ \mathrm{N \cdot m}$ . The two routes agree exactly, as $rF\sin\theta$ says they must.
Sum the signed components: $\tau_{z,\text{net}} = 344.15 - 141.12 - 117.60 = +85.43\ \mathrm{N \cdot m}$ .
Write the answer as the vector 5.3.B.2 asks for: $\vec{\tau}_{\text{net}} = (85.4\ \mathrm{N \cdot m})\hat{k}$ , pointing out of the page. By the right-hand rule of 5.3.B.2.iii, that is a counterclockwise turning tendency as drawn, so the far end of the beam would start to rise.
Sanity check the size. The cable torque alone is 344 N times metres and the two downward loads together contribute 259 N times metres, so a small positive remainder is what should come out. If your net torque exceeds the largest single contribution, you have a sign error.

$\vec{\tau}_{\text{net}} = (85.4\ \mathrm{N \cdot m})\hat{k}$ , out of the page. The beam's far end would begin to rise, turning counterclockwise as drawn. The cable's horizontal component contributes zero torque about the hinge.

A cross product in three dimensions, with the perpendicularity checked

A force $\vec{F} = (12\ \mathrm{N})\hat{i} - (8.0\ \mathrm{N})\hat{j} + (15\ \mathrm{N})\hat{k}$ acts at a point whose position relative to the chosen pivot is $\vec{r} = (0.20\ \mathrm{m})\hat{i} + (0.30\ \mathrm{m})\hat{j} - (0.10\ \mathrm{m})\hat{k}$ . Find (a) the torque as a vector, (b) its magnitude, checked against $rF\sin\theta$ , and (c) verify statement 5.3.B.2.ii directly.

(a) Apply 5.3.B.2 in components. For $\vec{\tau} = \vec{r} \times \vec{F}$ the three components are $\tau_x = r_yF_z - r_zF_y$ , $\tau_y = r_zF_x - r_xF_z$ and $\tau_z = r_xF_y - r_yF_x$ . Keep the cyclic order; a swapped pair flips a sign.
$\tau_x = (0.30)(15) - (-0.10)(-8.0) = 4.50 - 0.80 = 3.70\ \mathrm{N \cdot m}$ .
$\tau_y = (-0.10)(12) - (0.20)(15) = -1.20 - 3.00 = -4.20\ \mathrm{N \cdot m}$ .
$\tau_z = (0.20)(-8.0) - (0.30)(12) = -1.60 - 3.60 = -5.20\ \mathrm{N \cdot m}$ . So $\vec{\tau} = (3.70\hat{i} - 4.20\hat{j} - 5.20\hat{k})\ \mathrm{N \cdot m}$ .
(b) $\lvert\vec{\tau}\rvert = \sqrt{(3.70)^2 + (4.20)^2 + (5.20)^2} = \sqrt{13.69 + 17.64 + 27.04} = \sqrt{58.37} = 7.64\ \mathrm{N \cdot m}$ .
Now the check against 5.3.B.2.i. $r = \sqrt{0.04 + 0.09 + 0.01} = \sqrt{0.14} = 0.37417$ m and $F = \sqrt{144 + 64 + 225} = \sqrt{433} = 20.809$ N, so $rF = 7.7859\ \mathrm{N \cdot m}$ and $\sin\theta = 7.6400/7.7859 = 0.98126$ .
Do not stop at $\arcsin(0.98126) = 78.9^\circ$ . The sine cannot distinguish an angle from its supplement. Use the dot product, which the sheet's VECTORS box prints: $\vec{r} \cdot \vec{F} = (0.20)(12) + (0.30)(-8.0) + (-0.10)(15) = 2.40 - 2.40 - 1.50 = -1.50$ , so $\cos\theta = -1.50/7.7859 = -0.19266$ and $\theta = 101.1^\circ$ .
The two are consistent: $\sin 101.1^\circ = 0.98126$ , matching the cross product exactly. The torque magnitude does not care which of the two angles it is, but a direction question does.
(c) Statement 5.3.B.2.ii says the result is perpendicular to both inputs. Test it with dot products. $\vec{\tau} \cdot \vec{r} = (3.70)(0.20) + (-4.20)(0.30) + (-5.20)(-0.10) = 0.740 - 1.260 + 0.520 = 0$ .
$\vec{\tau} \cdot \vec{F} = (3.70)(12) + (-4.20)(-8.0) + (-5.20)(15) = 44.4 + 33.6 - 78.0 = 0$ . Both are exactly zero, so $\vec{\tau}$ is normal to the plane containing $\vec{r}$ and $\vec{F}$ , exactly as 5.3.B.2.ii states.
One caution about scope. Computing a single torque vector in three dimensions is fine. Topic 5.5's boundary statement rules out analyzing rotation in more than one plane at the same time, which is a different thing. On the exam, expect both vectors to lie in one plane and the torque to have one nonzero component.

(a) $\vec{\tau} = (3.70\hat{i} - 4.20\hat{j} - 5.20\hat{k})\ \mathrm{N \cdot m}$ . (b) $\lvert\vec{\tau}\rvert = 7.64\ \mathrm{N \cdot m}$ , matching $rF\sin\theta$ with $\theta = 101.1^\circ$ as fixed by the dot product. (c) $\vec{\tau} \cdot \vec{r} = 0$ and $\vec{\tau} \cdot \vec{F} = 0$ exactly.

The torque of gravity on a rod whose density is not uniform

A thin rod of length $L = 1.50$ m is hinged at its left end and held horizontal. Its linear mass density is $\lambda(x) = \lambda_0(1 + x/L)$ with $\lambda_0 = 2.4\ \mathrm{kg/m}$ , where $x$ is measured from the hinge. Find (a) the rod's mass, (b) the torque that gravity exerts about the hinge, by direct integration, (c) the rod's center of mass, and (d) how the answer compares with a uniform rod of the same mass and length.

Set the convention: $+x$ along the rod away from the hinge, $+y$ up, $\hat{k}$ out of the page, so gravity's torque will come out negative and therefore clockwise.
(a) With $dm = \lambda(x)\, dx$ , the mass is $M = \int_0^L \lambda_0(1 + x/L)\, dx = \lambda_0\left[L + \frac{L}{2}\right] = \frac{3}{2}\lambda_0 L$ . The equation sheet prints $\lambda = \frac{d}{d\ell}m(\ell)$ , which is this relation read backwards.
$M = 1.5(2.4)(1.50) = 5.40$ kg.
(b) Every mass element $dm$ sits at position $x$ and has weight $g\, dm$ pointing in $-\hat{j}$ , so its torque about the hinge is $d\tau_z = -x g\, dm$ . Add them up: $\lvert\tau\rvert = g\int_0^L x\lambda(x)\, dx = g\lambda_0\int_0^L \left(x + \frac{x^2}{L}\right) dx = g\lambda_0\left[\frac{L^2}{2} + \frac{L^2}{3}\right] = \frac{5}{6}g\lambda_0 L^2$ .
$\lvert\tau\rvert = \frac{5}{6}(9.8)(2.4)(1.50)^2 = \frac{5}{6}(9.8)(2.4)(2.25) = (9.8)(2.4)(1.875) = 44.1\ \mathrm{N \cdot m}$ , so $\vec{\tau} = -(44.1\ \mathrm{N \cdot m})\hat{k}$ , clockwise as set up.
(c) The sheet prints $\vec{r}_{\text{cm}} = \frac{\int \vec{r}\, dm}{\int dm}$ . Here $x_{\text{cm}} = \frac{\lambda_0(L^2/2 + L^2/3)}{\frac{3}{2}\lambda_0 L} = \frac{(5/6)L^2}{(3/2)L} = \frac{5L}{9} = 0.833$ m, which is past the midpoint toward the heavy end, as it must be.
Now notice what parts (b) and (c) have in common. The integral in the numerator of $x_{\text{cm}}$ is the same integral that gave the torque, divided by $g$ . So $\lvert\tau\rvert = Mgx_{\text{cm}} = (5.40)(9.8)(0.83333) = 44.1\ \mathrm{N \cdot m}$ , agreeing to every digit.
That agreement is the proof of a rule AP Physics 1 has to assert: for the purpose of computing gravitational torque, a rigid body's whole weight may be treated as acting at its center of mass. Here you derived it in one line instead of accepting it.
(d) A uniform rod of the same mass has $x_{\text{cm}} = L/2 = 0.75$ m, giving $\lvert\tau\rvert = (5.40)(9.8)(0.75) = 39.69\ \mathrm{N \cdot m}$ . The ratio is $(5L/9)/(L/2) = 10/9$ , so assuming uniformity here understates the torque by about 11%.
Sanity check the direction of the error. The density increases toward the far end, so more mass sits at larger $x$ , so the torque should exceed the uniform value. It does.

(a) $M = 5.40$ kg. (b) $\vec{\tau} = -(44.1\ \mathrm{N \cdot m})\hat{k}$ , clockwise. (c) $x_{\text{cm}} = 5L/9 = 0.833$ m. (d) A uniform rod of the same mass would give $39.7\ \mathrm{N \cdot m}$ ; the ratio is exactly 10/9, and the integral is what proves the weight may be placed at the center of mass.

Frequently asked questions

Is the direction of torque tested on AP Physics C Mechanics?

Yes. AP Physics C: Mechanics prints no boundary statement under Topic 5.3, and its essential knowledge 5.3.B.2 defines torque as the cross product of the position vector and the force vector, followed by three sub-statements devoted to getting a direction out of it: 5.3.B.2.i gives the magnitude as the product of the two magnitudes and the sine of the angle between them, 5.3.B.2.ii says the result is perpendicular to both vectors and therefore normal to the plane they define, and 5.3.B.2.iii says the direction can be qualitatively determined by applying the appropriate right-hand rule. AP Physics 1 goes the other way: its Topic 5.3 boundary statement reads 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. The two equation sheets match their frameworks, one printing the cross product with arrows and the other printing a magnitude.

How do I use the right-hand rule for torque?

Point the fingers of your right hand along the position vector, the one running from the axis of rotation to the point where the force is applied. Curl them through the smaller angle toward the force vector. Your extended thumb then points along the torque vector. Essential knowledge 5.3.B.2.iii in the AP Physics C: Mechanics course description is the statement that licenses this, and it says the direction can be qualitatively determined this way. Order matters, because the cross product is not commutative: position first, then force. If both vectors lie in the plane of the page, the torque points either straight out of the page, which corresponds to a counterclockwise turning tendency, or straight into it, which corresponds to clockwise. Reversing the order of the two vectors reverses the direction of the result.

What is the difference between torque in AP Physics 1 and AP Physics C?

One equation, three extra sub-statements, and one boundary statement. AP Physics C: Mechanics prints nine numbered items under Topic 5.3 and AP Physics 1 prints six. Five of those six are word for word identical in the two course descriptions, including the rule that torque comes only from the force component perpendicular to the position vector, the definition of the lever arm as the perpendicular distance from the axis to the line of action, and the entire treatment of force diagrams. The two frameworks separate at 5.3.B.2. AP Physics 1 gives the magnitude, the position times the perpendicular force, equal to the position times the force times the sine of the angle between them, and then prints a boundary statement putting the direction of torque beyond the scope of the course. AP Physics C: Mechanics gives torque as the cross product of the position vector and the force vector, adds three sub-statements on the cross product, perpendicularity and the right-hand rule, and prints no boundary statement at all.

Is the cross product formula on the AP Physics C Mechanics equation sheet?

Yes, twice over. The mechanics equation table prints torque as the cross product of the position vector and the force vector, with arrows over all three symbols. Separately, the Table of Information appendix carries a VECTORS box that prints the magnitude of a cross product as the product of the two magnitudes and the sine of the angle between them, written with absolute-value bars, along with the dot product as the product of the magnitudes and the cosine, unit-vector notation, and vector addition by components. The AP Physics 1 Table of Information has no VECTORS box on either of its two pages, so none of that is available on that exam. If you are checking a transcription of the sheet that lists only the physics equations, the VECTORS box may be missing from it even though it is printed in the real booklet.

Why does only the perpendicular component of a force produce torque?

Because the component along the position vector has no turning effect about that axis. Essential knowledge 5.3.A.1 states it directly: torque results only from the force component perpendicular to the position vector from the axis of rotation to the point of application. The cross product encodes the same fact, since the magnitude carries a factor of the sine of the angle between the two vectors, and that sine is zero when the force points straight along or straight against the position vector. Physically, pushing a door directly toward or directly away from its hinge does not turn it, however hard you push, because the force line passes through the axis and the lever arm is zero. Essential knowledge 5.3.A.2 gives the equivalent reading: the lever arm is the perpendicular distance from the axis to the line of action, and a line of action through the axis has a lever arm of zero.

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

One piece of information: where the forces act. Essential knowledge 5.3.B.1.ii, which is printed identically in both AP Physics C: Mechanics and AP Physics 1, 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. A free-body diagram treats the object as a point and shows only the vectors, which is all translational dynamics needs. A torque depends on where each force is applied, so collapsing the object to a point destroys the very quantity you are trying to compute. Statement 5.3.B.1.i adds that force diagrams are used to analyze the torques exerted on a rigid system, and 5.5.A.1.i says both kinds of diagram describe the nature of the forces and torques on an object or rigid system.

Does the axis I choose change the torque?

Yes, and that is why essential knowledge 5.3.B.2 specifies the torque exerted about a chosen pivot point. The position vector in the cross product is measured from whatever axis you pick, so the same force generally produces a different torque about different axes, and it produces zero torque about any axis lying on its own line of action. A torque quoted without an axis is not a well-defined quantity. Practically, this freedom is useful rather than annoying: in a static problem you choose the axis at the point where an unknown force acts, which makes that force's position vector zero and removes it from the torque equation entirely. Whatever you choose, state it in the first line and use the same axis for every torque in the sum.