AP Physics C: Mechanics · Topic 5.5

Topic 5.5: Rotational Equilibrium and Newton's First Law in Rotational Form

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

Rotational equilibrium means the net torque on a system is zero, so its angular velocity stays constant. A system can be in rotational equilibrium without translational equilibrium, and the reverse. The required content is the same in AP Physics C and AP Physics 1; the torques you sum are not.

AP Physics: Unit 5 (topics 5.5 Rotational Equilibrium and Newton's First Law in Rotational Form). AP Physics C: Mechanics Unit 5, Topic 5.5. One learning objective, 5.5.A: describe the conditions under which a system's angular velocity remains constant. Essential knowledge 5.5.A.1 (a system may exhibit rotational equilibrium, meaning constant angular velocity, without being in translational equilibrium, and vice versa), with 5.5.A.1.i (free-body and force diagrams describe the nature of the forces and torques exerted on an object or rigid system), 5.5.A.1.ii (rotational equilibrium is a configuration of torques such that the net torque exerted on the system is zero, relevant equation sum of tau_i = 0) and 5.5.A.1.iii (the rotational analog of Newton's first law is that a system will have a constant angular velocity only if the net torque exerted on the system is zero); and 5.5.A.2 (a rotational corollary to Newton's second law states that if the torques exerted on a rigid system are not balanced, the system's angular velocity must be changing). One-line boundary statement: 'AP Physics C: Mechanics does not expect students to simultaneously analyze rotation in multiple planes.' All five essential-knowledge statements are printed word for word in the AP Physics 1 CED under the same numbers, and its boundary statement is the same sentence with the course name swapped, so the required content of this topic does not differ between the two courses. The suggested skills do: this course lists 1.A, 2.A, 2.C, 3.B while AP Physics 1 lists 1.C, 2.A, 2.B, 3.B. The equilibrium condition sum of tau_i = 0 is not printed on either equation sheet, and neither is the translational condition; both sheets print Newton's second law in rotational form, of which equilibrium is the zero-angular-acceleration case. None of the fifteen CED sample multiple-choice questions align to Topic 5.5; the two Unit 5 items align to 5.3.B and 5.6.A. Unit 5 sample instructional activity 2, listed under Topic 5.3 as a Create a Plan activity, is a Topic 5.5 problem: design a walkway of given mass suspended from a ceiling and determine the force each of two end supports must provide as a person of given mass walks across.

What Topic 5.5 requires

Topic 5.5 carries one learning objective, 5.5.A: describe the conditions under which a system's angular velocity remains constant. Its suggested skills are 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.C (compare physical quantities between two or more scenarios or at different times and/or locations within a single scenario) and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim).

5.5.A.1 A system may exhibit rotational equilibrium (constant angular velocity) without being in translational equilibrium, and vice versa.

5.5.A.1.i Free-body and force diagrams describe the nature of the forces and torques exerted on an object or rigid system.
5.5.A.1.ii Rotational equilibrium is a configuration of torques such that the net torque exerted on the system is zero. Its relevant equation is

\sum \tau_i = 0

5.5.A.1.iii The rotational analog of Newton's first law is that a system will have a constant angular velocity only if the net torque exerted on the system is zero.

5.5.A.2 A rotational corollary to Newton's second law states that if the torques exerted on a rigid system are not balanced, the system's angular velocity must be changing.

The topic carries a one-line boundary statement: "AP Physics C: Mechanics does not expect students to simultaneously analyze rotation in multiple planes."

Read the parenthesis in 5.5.A.1 carefully, because it is the definition and it is where most of the errors in this topic start. Rotational equilibrium is constant angular velocity, not zero angular velocity. A flywheel spinning steadily at 300 radians per second is in rotational equilibrium. A flywheel at rest that is being spun up is not.

The required content is identical to AP Physics 1's

AP Physics 1 has a Topic 5.5 with the same title and the same learning objective. Every essential-knowledge statement above, including all three sub-statements and 5.5.A.2, is printed word for word in the AP Physics 1 course and exam description under the same numbers. The boundary statement is the same sentence with the course name swapped: "AP Physics 1 does not expect students to simultaneously analyze rotation in multiple planes."

That is worth saying plainly rather than dressing up. On the physics of what rotational equilibrium is, the two frameworks are the same document. Topic 5.2 is the other Unit 5 topic where the essential knowledge does not differ, and there even the suggested skills match. Here they do not, which is the one place the CED itself distinguishes the two treatments:

	AP Physics C: Mechanics	AP Physics 1
Practice 1 skill	1.A, create diagrams, tables, charts, or schematics	1.C, create qualitative sketches of graphs
Practice 2 skills	2.A derive, 2.C compare	2.A derive, 2.B calculate
Practice 3 skill	3.B apply a law or model to make a claim	3.B apply a law or model to make a claim

So the algebra-based course pairs deriving with calculating and asks for graph sketches; this course pairs deriving with comparing and asks for diagrams and schematics. That fits the exam behind it. On the AP Physics C: Mechanics multiple-choice section, skill 2.A is weighted at 25 to 30%, the highest band the CED gives any single skill there, and 2.C at 10 to 15%. Practice 1 is not assessed on the multiple-choice section at all; it carries 20 to 35% of the free-response section.

The practical translation: expect to be asked for the support force as an expression, and then to be asked how it changes when something moves. Worked example 2 is exactly that question.

What actually changes once you have Topic 5.3 and Topic 5.4

The condition $\sum \tau_i = 0$ does not change between the courses. What you are allowed to put into the sum does, in three ways.

The torques are cross products. Topic 5.3 in this course defines torque as $\vec{\tau} = \vec{r} \times \vec{F}$ and prints no boundary statement against its direction, while AP Physics 1 prints one saying the direction of torque is beyond the scope of that course. So here $\sum \vec{\tau}_i = \vec{0}$ is a vector equation, and in a planar problem it becomes the single component equation $\sum(x_iF_{y,i} - y_iF_{x,i}) = 0$ . In practice that is the same signed sum the algebra-based course writes, but you get it from a definition rather than from a convention, and you can defend the sign with a right-hand rule.

The load can be distributed. Nothing stops a Physics C problem from putting a load on a beam whose weight per unit length varies with position. Then the total load and its torque are both integrals, $\int w(x)\, dx$ and $\int x\,w(x)\, dx$ , and the ratio of the second to the first is the position at which a single equivalent force would act. Worked example 3 does this. AP Physics 1 has no way to pose it.

The bodies can be nonuniform. A rod whose density varies has its center of mass somewhere other than its midpoint, and finding it needs the center-of-mass integral that only the C sheet prints. Its weight still acts as a single force at that point, which is a result you can derive rather than assume.

What does not change is the method. Draw the force diagram, declare a sign convention and an axis, write $\sum F_x = 0$ , $\sum F_y = 0$ and $\sum \tau = 0$ , and solve. Three equations, at most three unknowns.

The two equilibrium conditions are independent

Statement 5.5.A.1 says a system may exhibit rotational equilibrium without being in translational equilibrium, and vice versa. Both halves of that sentence are testable, and the four combinations are all physically realizable.

Net force	Net torque	Example
zero	zero	a beam resting on two supports, static
zero	not zero	two equal and opposite forces applied at different points, a couple
not zero	zero	a ball dropped in free fall, or a spinning wheel released in midair
not zero	not zero	a rod struck off center and set both moving and turning

The couple in the second row is the cleanest demonstration. Push the top of a steering wheel left and the bottom right with equal forces: $\sum \vec{F} = \vec{0}$ , the wheel's center does not accelerate, and yet both torques have the same sign about the center and the wheel spins up. Force and torque are answering different questions, and $\sum \vec{F} = \vec{0}$ does not imply $\sum \tau = 0$ .

The third row matters just as much. A wheel spinning steadily as it falls has a net force equal to its weight and no net torque about its center of mass, so it is in rotational equilibrium while accelerating downward at $g$ .

Statement 5.5.A.1.iii is the first law in rotational form: a system will have a constant angular velocity only if the net torque on it is zero. Statement 5.5.A.2 is its contrapositive stated as its own essential-knowledge item: if the torques are not balanced, the angular velocity must be changing. Two statements, one fact, and the CED prints both in both courses because students reliably use only one direction of it.

Static equilibrium, the case where the system is at rest and stays at rest, needs both conditions at once. That is what makes a beam problem three simultaneous equations rather than one.

Choosing the axis, which is the whole strategy

Topic 5.3's statement 5.3.B.2 says the torque is exerted "about a chosen pivot point". In an equilibrium problem the net torque is zero about every axis, not just about the real hinge, and that freedom is the main technique in this topic.

The move is: put your axis where an unknown force acts. That force's position vector is then zero, its torque is zero, and it drops out of the torque equation entirely, leaving you one equation in one unknown instead of three in three.

Two further consequences worth having in hand:

A second axis gives you a free check, not new information. Once $\sum F_x = 0$ , $\sum F_y = 0$ and $\sum \tau = 0$ about one axis all hold, the torque equation about any other axis is automatically satisfied. Use it to verify a solution, and be suspicious if it does not close.
The axis you choose need not be a physical pivot. It can be a point in empty space. Nothing about $\vec{r} \times \vec{F}$ requires anything to be located there.

The boundary statement puts one limit on all of this: "AP Physics C: Mechanics does not expect students to simultaneously analyze rotation in multiple planes." So expect every force in your diagram to lie in one plane and every torque to point along the one axis perpendicular to it. That is why a single signed equation, rather than three, closes these problems.

One convention to declare and hold: pick $\hat{k}$ out of the page so that positive torque means counterclockwise, and use the same axis for every term in the sum. Mixing two axes inside one torque equation produces an answer that is wrong for reasons no individual line will show.

What the sheets print for Topic 5.5, which is nothing

The equilibrium condition $\sum \tau_i = 0$ , essential knowledge 5.5.A.1.ii, is not printed on the AP Physics C: Mechanics equation sheet, and not printed on the AP Physics 1 sheet either. Neither is a translational $\sum \vec{F} = \vec{0}$ .

That is not an oversight. Both sheets print Newton's second law in rotational form, $\alpha_{\text{sys}} = \frac{\sum\tau}{I_{\text{sys}}} = \frac{\tau_{\text{net}}}{I_{\text{sys}}}$ , and equilibrium is the $\alpha = 0$ case of it. Both sheets likewise print $\vec{a}_{\text{sys}} = \frac{\sum\vec{F}}{m_{\text{sys}}} = \frac{\vec{F}_{\text{net}}}{m_{\text{sys}}}$ , and translational equilibrium is the $a = 0$ case. Set the left side to zero and the numerator must vanish.

So the two conditions you will use most in this topic are the two you have to supply from memory. They are one line each and they follow from a printed equation, which is presumably why College Board leaves them off.

What the C: Mechanics sheet does give you for this topic, and the AP Physics 1 sheet does not:

$\vec{\tau} = \vec{r} \times \vec{F}$ , so each term in the sum is defined as a vector.
The VECTORS box, with $\lvert\vec{A} \times \vec{B}\rvert = AB\sin\theta$ and component addition, so you never have to recall a torque magnitude.
$\vec{r}_{\text{cm}} = \frac{\int \vec{r}\, dm}{\int dm}$ and $\lambda = \frac{d}{d\ell}m(\ell)$ , so a nonuniform body's weight can be located.
$\lvert\vec{F}_f\rvert \le \lvert\mu\vec{F}_N\rvert$ , printed as an inequality on both sheets. Worked example 1 turns on that inequality sign.

Traps

Rotational equilibrium is constant angular velocity, not zero. Statement 5.5.A.1 says so in a parenthesis. A grindstone turning steadily is in rotational equilibrium.

Zero net force does not give you zero net torque. A couple has one and not the other. Statement 5.5.A.1 makes the independence explicit, and it is the most commonly assumed-away sentence in the topic.

Zero net torque does not mean nothing is happening. A spinning wheel in free fall has no net torque about its center of mass and a net force equal to its weight.

Do not sum torques about two different axes in one equation. Every term in a $\sum \tau = 0$ must be measured from the same point.

Use a force diagram, not a free-body diagram. Statement 5.5.A.1.i names both, and statement 5.3.B.1.ii says the difference: a force diagram also depicts the location at which each force is exerted relative to the axis of rotation. Collapsing the body to a point destroys the information a torque equation needs.

Friction at a surface is an inequality, not an equation. Both sheets print $\lvert\vec{F}_f\rvert \le \lvert\mu\vec{F}_N\rvert$ . In a ladder problem the equilibrium condition fixes the friction force that is required, and the inequality then tells you whether the surface can supply it. Solving for a minimum coefficient means setting the required value equal to the maximum available one.

Do not assume a distributed load acts at the midpoint. It acts at the load's own centroid, which for a triangular load is two thirds of the way along, as worked example 3 shows.

Count your equations. A planar equilibrium problem gives you exactly three: two force components and one torque. If you have four unknowns, the problem is statically indeterminate and something else must be given.

How Topic 5.5 is assessed

None of the fifteen sample multiple-choice questions in the AP Physics C: Mechanics course and exam description align to Topic 5.5. The two Unit 5 questions in that set align to 5.3.B and 5.6.A. That does not make the topic optional: Unit 5 carries 10 to 15% of the multiple-choice section overall, and the required content and skills page applies to every topic.

The CED's own Unit 5 sample instructional activity 2, listed under Topic 5.3 and typed as a Create a Plan activity, is a Topic 5.5 problem in everything but label. It asks students to design a walkway of given mass to be suspended from a ceiling, and to determine the amount of force the two supports, one on each end, must be able to provide as a person of given mass walks across it. Worked example 2 works that scenario, and the answer is a pair of expressions rather than a pair of numbers, which is what skill 2.C is looking for.

On the free-response section, the question type most likely to carry a Topic 5.5 setup is Question 1, Mathematical Routines, worth 10 points with a suggested time of 20 to 25 minutes. Its skill list is 1.A, 1.C, 2.A, 2.B, 3.B and 3.C, and the CED describes it as assessing students' ability to use mathematics to analyze a scenario and make predictions, with students expected to symbolically derive relationships between variables as well as calculate numerical values, and to create and use representations that describe the scenario, giving a free-body diagram as an example.

The task verbs are worth knowing exactly, since they set the standard for an answer. The CED defines Derive as "Starting with a fundamental law or relationship, perform a series of mathematical steps to arrive at a final answer", and Justify as "Provide qualitative reasoning beyond mathematical derivations or expressions to support, qualify, or defend a claim". A justification that only restates the algebra does not meet the definition.

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

For this topic in particular, the honest statement is that your framework and this one print the same essential knowledge and the same boundary statement. If you are in AP Physics 1, read the [AP Physics 1 Topic 5.5 page](/ap-physics-1/unit-5-torque-and-rotational-dynamics/5-5-rotational-equilibrium-and-newtons-first-law-in-rotational-form) instead. It covers the same five statements, and it is written to your exam, whose suggested skills for this topic are 1.C and 2.B where this course lists 1.A and 2.C. This page is for AP Physics C: Mechanics students.

What this page adds is what a calculus-based course can put into the sum: torques defined as cross products with directions you can defend, distributed loads that need an integral, and nonuniform bodies whose center of mass you locate rather than assume. Worked example 3 is the one that has no AP Physics 1 equivalent.

AP Physics C: Mechanics is a calculus-based introductory college-level course equivalent to the first course in a college calculus-based physics sequence, and its only stated prerequisite is that students have taken or are concurrently taking calculus.

From here, Topic 5.6 is what happens when the sum is not zero, and Topic 5.4 supplies the rotational inertia that then divides it. The how to calculate torque guide and the how to draw a free-body diagram guide own the step-by-step routines, and the Unit 5 hub lists all six topics.

A ladder against a frictionless wall, solved symbolically

A uniform ladder of mass $m$ and length $L$ leans against a frictionless vertical wall, making an angle $\theta$ with the horizontal floor. The floor exerts friction on the ladder's base. Derive the minimum coefficient of static friction that keeps the ladder from slipping, then evaluate it at $\theta = 60^\circ$ and at $\theta = 75^\circ$ .

Set up the geometry and the convention. Put the base at the origin with the ladder rising to the right, so its top touches the wall at $(L\cos\theta,\ L\sin\theta)$ . Take $+x$ toward the wall, $+y$ up, and $\hat{k}$ out of the page, so positive torque is counterclockwise.
Draw the force diagram, per 5.5.A.1.i. Four forces: the ladder's weight $mg$ downward at its midpoint $(L\cos\theta/2,\ L\sin\theta/2)$ ; the wall's normal force $N_w$ horizontal at the top, pointing in $-x$ because the wall is frictionless and can only push; the floor's normal force $N_f$ upward at the base; and static friction $f$ at the base, pointing in $+x$ to oppose the slipping tendency.
Translational equilibrium, both components. $\sum F_y = 0$ gives $N_f = mg$ . $\sum F_x = 0$ gives $f = N_w$ .
Rotational equilibrium about the base, which kills both unknown base forces at once because their position vectors are zero. Using $\tau_z = xF_y - yF_x$ : the weight contributes $(L\cos\theta/2)(-mg) = -\frac{1}{2}mgL\cos\theta$ , and the wall's force contributes $-(L\sin\theta)(-N_w) = +N_wL\sin\theta$ .
Set the sum to zero, per 5.5.A.1.ii: $N_wL\sin\theta - \frac{1}{2}mgL\cos\theta = 0$ , so $N_w = \dfrac{mg\cos\theta}{2\sin\theta} = \dfrac{mg}{2\tan\theta}$ . The length $L$ cancels.
Combine with the force equations. The friction actually required is $f = N_w = mg/(2\tan\theta)$ , and the sheet prints the limit as an inequality, $\lvert\vec{F}_f\rvert \le \lvert\mu\vec{F}_N\rvert$ , so the ladder holds provided $mg/(2\tan\theta) \le \mu mg$ .
The mass cancels too, leaving $\mu \ge \dfrac{1}{2\tan\theta}$ . Neither the ladder's mass nor its length appears. That is the kind of result skill 2.A is weighted for, and it is more informative than any single number.
Evaluate. At $\theta = 60^\circ$ , $\tan 60^\circ = 1.7321$ , so $\mu_{\min} = 1/3.4641 = 0.289$ . At $\theta = 75^\circ$ , $\tan 75^\circ = 3.7321$ , so $\mu_{\min} = 1/7.4641 = 0.134$ .
Read the functional dependence, which is skill 2.C. As $\theta$ increases the ladder becomes more vertical and needs less friction; as $\theta$ approaches zero, $\tan\theta$ approaches zero and $\mu_{\min}$ grows without bound, which is why a ladder laid at a shallow angle always slips no matter how rough the floor.
Check the whole solution by taking torques about the top instead. About that point the wall force drops out and you get $N_fL\cos\theta - fL\sin\theta - \frac{1}{2}mgL\cos\theta = 0$ . Substituting $N_f = mg$ and $f = mg/(2\tan\theta)$ gives $mgL\cos\theta - \frac{mgL\sin\theta}{2\tan\theta} - \frac{1}{2}mgL\cos\theta = mgL\cos\theta - \frac{1}{2}mgL\cos\theta - \frac{1}{2}mgL\cos\theta = 0$ . It closes.

$\mu_{\min} = \dfrac{1}{2\tan\theta}$ , independent of both the ladder's mass and its length. At $60^\circ$ that is 0.289, and at $75^\circ$ it is 0.134. A shallower ladder needs more friction, without limit as the angle approaches zero.

The CED's suspended walkway, as a function of where the person stands

This is sample instructional activity 2 from the AP Physics C: Mechanics Unit 5 guide. A uniform walkway of mass $M = 240$ kg and length $L = 4.0$ m hangs from a ceiling by a support at each end. A person of mass $m = 65$ kg stands a distance $x$ from the left support. Derive expressions for the tension in each support as a function of $x$ , then evaluate them at $x = 0$ , $x = L/2$ and $x = L$ , and state the largest force either support ever carries. Use $g = 9.8\ \mathrm{m/s^2}$ .

Force diagram, per 5.5.A.1.i. Four forces: $F_L$ up at the left end, $F_R$ up at the right end, the walkway's weight $Mg$ down at its midpoint $L/2$ because it is uniform, and the person's weight $mg$ down at $x$ .
The system is static, so both equilibrium conditions hold at once. Take $\hat{k}$ out of the page and put the axis at the left support, which removes $F_L$ from the torque equation because its position vector is zero.
Rotational equilibrium about the left support: $F_R L - Mg\dfrac{L}{2} - mgx = 0$ , so $F_R(x) = \dfrac{Mg}{2} + mg\dfrac{x}{L}$ .
Translational equilibrium: $F_L + F_R = (M + m)g$ , so $F_L(x) = (M+m)g - F_R = \dfrac{Mg}{2} + mg\left(1 - \dfrac{x}{L}\right)$ .
Read the structure before substituting. Each support carries half the walkway's weight no matter what, plus a share of the person's weight that is linear in position: the far support gets the fraction $x/L$ and the near one gets $1 - x/L$ . The two shares always add to 1.
Numbers. $Mg/2 = (240)(9.8)/2 = 1176$ N and $mg = (65)(9.8) = 637$ N. Total weight is $(240 + 65)(9.8) = 2989$ N.
At $x = 0$ , standing on the left support: $F_L = 1176 + 637 = 1813$ N and $F_R = 1176$ N. Sum $= 2989$ N.
At $x = L/2 = 2.0$ m: $F_L = F_R = 1176 + 318.5 = 1494.5$ N. Sum $= 2989$ N.
At $x = L = 4.0$ m: $F_L = 1176$ N and $F_R = 1813$ N. Sum $= 2989$ N. The answer is symmetric under reflection, as it must be for a uniform walkway with identical end supports.
The largest force either support ever carries is 1813 N, reached when the person stands directly over that support. That is the number the activity is actually asking for, since a support must be rated for the worst case rather than the average.
Sanity check the extremes. Standing directly over a support puts all of the person's weight on it, and the walkway's weight is still split evenly, which is exactly $1176 + 637$ . And at every position the two support forces sum to 2989 N, because the total weight does not depend on where anyone stands.

$F_R(x) = \dfrac{Mg}{2} + mg\dfrac{x}{L}$ and $F_L(x) = \dfrac{Mg}{2} + mg\left(1 - \dfrac{x}{L}\right)$ . At $x = 0$ : 1813 N and 1176 N. At the midpoint: 1494.5 N each. At $x = L$ : 1176 N and 1813 N. Each support must be rated for 1813 N.

A beam under a load that is not uniform

A light beam of length $L = 3.0$ m rests on a support at each end. It carries a load whose weight per unit length increases linearly from zero at the left end, $w(x) = w_0 x/L$ with $w_0 = 900\ \mathrm{N/m}$ . Neglect the beam's own weight. Find (a) the total load, (b) the position at which a single equivalent force would act, and (c) the force each support provides.

This is the case AP Physics 1 has no way to pose: the load is not a point weight and not a uniform one, so both its total and its torque are integrals. First check the given: at $x = L$ the load density is $w_0 L/L = 900\ \mathrm{N/m}$ , and at $x = 0$ it is zero.
(a) The load on a slice of width $dx$ is $w(x)\, dx$ , so the total is $W = \int_0^L \dfrac{w_0x}{L}\, dx = \dfrac{w_0}{L}\cdot\dfrac{L^2}{2} = \dfrac{w_0L}{2}$ .
$W = (900)(3.0)/2 = 1350$ N. Sanity check: the density rises linearly from 0 to $900\ \mathrm{N/m}$ , so its average is $450\ \mathrm{N/m}$ over 3.0 m, which is 1350 N.
(b) Take torques about the left support, with $\hat{k}$ out of the page and $+x$ to the right, so each slice contributes $-x\,w(x)\, dx$ . The magnitude of the load's total torque about that point is $\int_0^L x\,\dfrac{w_0x}{L}\, dx = \dfrac{w_0}{L}\cdot\dfrac{L^3}{3} = \dfrac{w_0L^2}{3}$ .
$= (900)(3.0)^2/3 = (900)(9.0)/3 = 2700\ \mathrm{N \cdot m}$ .
A single force $W$ placed at position $\bar{x}$ would give the same torque, so $\bar{x} = \dfrac{\text{torque}}{W} = \dfrac{w_0L^2/3}{w_0L/2} = \dfrac{2L}{3}$ .
$\bar{x} = 2(3.0)/3 = 2.0$ m from the left support, not the midpoint at 1.5 m. Assuming the midpoint here would understate the right support by a third of the load's contribution.
(c) Now the equilibrium conditions. Rotational equilibrium about the left support, per 5.5.A.1.ii: $F_R L = 2700\ \mathrm{N \cdot m}$ , so $F_R = 2700/3.0 = 900$ N.
Translational equilibrium: $F_L = W - F_R = 1350 - 900 = 450$ N.
Check by taking torques about the right support instead, which should close independently. The load's torque about that point is $\int_0^L (L - x)\dfrac{w_0x}{L}\, dx = \dfrac{w_0}{L}\left[\dfrac{L^3}{2} - \dfrac{L^3}{3}\right] = \dfrac{w_0L^2}{6} = (900)(9.0)/6 = 1350\ \mathrm{N \cdot m}$ , so $F_L = 1350/3.0 = 450$ N. It agrees.
Read the result. The right support carries twice what the left one does, $900$ N against $450$ N, in the ratio 2 to 1, because the load's centroid sits two thirds of the way toward it. For a uniform load the two would be equal.

(a) $W = w_0L/2 = 1350$ N. (b) The equivalent single force acts at $\bar{x} = 2L/3 = 2.0$ m from the left, not at the midpoint. (c) $F_L = 450$ N and $F_R = 900$ N, in the ratio 1 to 2, confirmed by taking torques about each support in turn.

Frequently asked questions

What is rotational equilibrium in AP Physics?

Rotational equilibrium is the condition in which the net torque exerted on a system is zero, and the CED defines it that way in essential knowledge 5.5.A.1.ii. Its consequence, stated in 5.5.A.1.iii as the rotational analog of Newton's first law, is that a system will have a constant angular velocity only if the net torque on it is zero. The critical word is constant, not zero: essential knowledge 5.5.A.1 puts constant angular velocity in parentheses as the definition, so a wheel spinning steadily at any rate is in rotational equilibrium just as much as a wheel at rest. Essential knowledge 5.5.A.2 states the same fact from the other side: if the torques exerted on a rigid system are not balanced, the system's angular velocity must be changing. All of this is printed identically in AP Physics C: Mechanics and AP Physics 1.

Can something be in rotational equilibrium but not translational equilibrium?

Yes, and essential knowledge 5.5.A.1 says so directly: a system may exhibit rotational equilibrium, meaning constant angular velocity, without being in translational equilibrium, and vice versa. A spinning wheel released in midair is the clearest case of the first kind. It has a net force equal to its weight and accelerates downward, while the net torque about its center of mass is zero, so its angular velocity holds steady. The reverse case is a couple: push the top of a steering wheel one way and the bottom the other with equal forces, and the net force is zero while both torques add, so the center does not accelerate but the wheel spins up. Static equilibrium, at rest and staying at rest, requires both conditions at once, which is why a beam problem needs three equations rather than one.

Is Topic 5.5 different in AP Physics C and AP Physics 1?

The required course content is not. Essential knowledge 5.5.A.1, its three sub-statements and 5.5.A.2 are printed word for word in both course and exam descriptions under the same numbers, and the boundary statement is the same sentence with the course name swapped: neither course expects students to simultaneously analyze rotation in multiple planes. The suggested skills do differ. AP Physics C: Mechanics lists 1.A, create diagrams and schematics, and 2.C, compare physical quantities between scenarios, while AP Physics 1 lists 1.C, create qualitative sketches of graphs, and 2.B, calculate an unknown quantity. Both list 2.A and 3.B. What also differs is what can go into the sum of torques: in the calculus-based course each torque is a cross product with a direction, the load on a beam can be distributed and need an integral, and a body's density can vary so its center of mass has to be located rather than assumed.

Which axis should I choose for a torque equation?

In an equilibrium problem the net torque is zero about every axis, so choose the one that makes the algebra shortest: put it at the point where an unknown force acts. That force's position vector is then zero, its torque vanishes, and the torque equation contains one fewer unknown. On a beam supported at two points, taking torques about one support solves directly for the other support's force. Essential knowledge 5.3.B.2 phrases torque as being exerted about a chosen pivot point, and the choice is genuinely yours; the axis does not have to be a physical hinge and can sit in empty space. Two rules go with the freedom. Every term in one torque equation must be measured from the same axis. And once you have used two force equations and one torque equation, a torque equation about a second axis gives no new information, which makes it a useful check on the answer.

Is the sum of torques equals zero on the AP Physics equation sheet?

No, on neither the AP Physics C: Mechanics sheet nor the AP Physics 1 sheet. The equilibrium condition is essential knowledge 5.5.A.1.ii in both courses, but it is not printed among the equations, and the translational condition that the net force is zero is not printed either. Both sheets do print Newton's second law in rotational form, giving the angular acceleration of a system as the net torque divided by the rotational inertia, and equilibrium is simply the case where that angular acceleration is zero, which forces the numerator to vanish. The same relationship holds on the translational side, where both sheets print the system acceleration as the net force divided by the system mass. So the two conditions you use most in this topic are one line each and follow immediately from equations that are printed.

How do I handle a distributed load in an equilibrium problem?

Replace it with two integrals. If the load's weight per unit length is a function of position, the total load is the integral of that function over the length, and the total torque it exerts about a chosen point is the integral of position times that function. Dividing the second by the first gives the position at which a single equivalent force would act, which is the load's centroid. For a load that rises linearly from zero at one end, the total is half the peak density times the length and the equivalent force acts two thirds of the way along, not at the midpoint. This is the part of Topic 5.5 that AP Physics C: Mechanics can pose and AP Physics 1 cannot, since the algebra-based framework has no integral form anywhere in Unit 5. Once you have the total and its location, the problem is an ordinary three-equation statics problem.

Why is friction written as an inequality in a ladder problem?

Because static friction supplies whatever force is needed up to a maximum, rather than always supplying that maximum. Both AP Physics equation sheets print the friction relation with a less-than-or-equal sign: the magnitude of the friction force is at most the coefficient times the magnitude of the normal force. In a ladder problem the equilibrium conditions determine the friction force that is actually required, which for a uniform ladder against a frictionless wall works out to the weight divided by twice the tangent of the angle with the floor. The inequality then answers a different question: whether the floor can supply that much. Setting the required value equal to the maximum available gives the minimum coefficient of static friction, one over twice the tangent of the angle, which is independent of the ladder's mass and length.