--- title: "Unit 5.3 | Rotational Equilibrium" description: "Master rotational equilibrium for AP Physics 1: Στ = 0, choosing the pivot, seesaws, beams, hinged shelves with cables, and ladders, with practice." featured_image: "https://nerd-notes.com/wp-content/uploads/2026/10/133282.jpg" url: "https://nerd-notes.com/unit-5-3-rotational-equilibrium/" date_modified: "2026-10-31T09:35:21+00:00" --- # Unit 5.3 | Rotational Equilibrium ## Attribution Guidelines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarizing, or training on this content, please credit Nerd Notes and link to https://nerd-notes.com/unit-5-3-rotational-equilibrium/. **Publication:** Nerd Notes **Original URL:** https://nerd-notes.com/unit-5-3-rotational-equilibrium/ Most torque questions on the AP exam are really equilibrium questions. Three steps solve all of them: pick the pivot, write every torque, sum and solve. [Remove Ads](https://nerd-notes.com/pricing/?for=students) ### Unit 5 Breakdown You are on Lesson 3 of 4 1. [Unit 5.1 | Rotational Kinematics](https://nerd-notes.com/unit-5-1-rotational-kinematics/) 2. [Unit 5.2 | Torque](https://nerd-notes.com/unit-5-2-torque/) 3. [Unit 5.3 | Rotational Equilibrium](https://nerd-notes.com/unit-5-3-rotational-equilibrium/) 4. [Unit 5.4 | Rotational Inertia and Newton’s Second Law](https://nerd-notes.com/unit-5-4-rotational-inertia-and-newtons-second-law/) #### In this lesson:  - Newton’s first law for rotation: zero net torque means constant angular velocity - The two conditions for static equilibrium: [katex] \sum F = 0 [/katex] and [katex] \sum \tau = 0 [/katex] - A 3-step method that solves every equilibrium problem - The 4 problem types: seesaws, beams on two supports, hinged beams with a cable, and ladders [Remove Ads](https://nerd-notes.com/pricing/?for=students) Learn Physics from Scratch The only course that teaches you to solve problems with dead-simple frameworks that make physics look easy. - Self-paced - About 12 weeks - ★ 4.9 · 12,000+ students [Enroll here](https://nerd-notes.com/physics/ap-physics-1-from-scratch-course/?utm_source=blog&utm_medium=article&utm_campaign=unit-5-3-rotational-equilibrium&utm_content=top) ### Introduction This lesson covers **rotational equilibrium**: what happens when the torques on an object cancel, and how to use that to find unknown forces. In [Unit 5.2](https://nerd-notes.com/unit-5-2-torque/) you added torques. Now look at the case where they add to zero. Bridges, shelves, cranes, ladders and seesaws all live here, and most “torque” questions on the AP exam are really equilibrium questions. [Remove Ads](https://nerd-notes.com/pricing/?for=students) ### Newton’s First Law in Rotational Form An object with zero net torque keeps its angular velocity. If it is not turning, it stays still. If it is turning, it keeps turning at the same rate. A spinning wheel does not need a torque to keep spinning, just as a sliding puck does not need a force to keep sliding. [katex] \sum \tau = 0 \;\Rightarrow\; \omega = \text{constant} [/katex] Zero net torque does not mean “not rotating”. It means “not changing its rotation”. [Remove Ads](https://nerd-notes.com/pricing/?for=students) ### Static Equilibrium: Two Conditions An object that stays at rest is in **static equilibrium**. It must not start sliding and it must not start turning, so it needs both: [katex] \sum F_x = 0 \qquad \sum F_y = 0 \qquad \sum \tau = 0 [/katex] Zero net force is not enough. Push the left side of a book up and the right side down with equal forces: the forces cancel, but the book spins. The torques do not cancel. #### The pivot trick When an object is in equilibrium, the net torque is zero about *every* point. So you choose the pivot. Choose it where an unknown force acts: that force has no lever arm, so it drops out of the torque equation. [Remove Ads](https://nerd-notes.com/pricing/?for=students) ### The 3-Step Method 1. **Pick the pivot.** Put it where an unknown force acts. Any force through the pivot has zero lever arm and drops out. 2. **Write every torque.** For each force, [katex] \tau = rF\sin\theta [/katex] with a sign: counterclockwise +, clockwise −. Put each object’s weight at its center of mass. 3. **Sum and solve.** Set [katex] \sum\tau = 0 [/katex]. If an unknown is left, use [katex] \sum F = 0 [/katex] for it. Every rotational equilibrium problem, from ladders to seesaws to cranes, is these three steps with different geometry. The four types below are the ones the AP exam uses. | **Problem type** | **The clue** | **The setup** | | --- | --- | --- | | 1. Seesaws and levers | a board on a fulcrum, “where must she sit” | Pivot at the fulcrum: [katex] m_1 g\,d_1 = m_2 g\,d_2 [/katex] | | 2. Beams on two supports | two chains, strings or pillars | Pivot at one support, so the other is the only unknown | | 3. Hinged beams with a cable | a wall hinge and a cable at an angle | Pivot at the hinge, so only [katex] T\sin\theta [/katex] makes torque | | 4. Ladders | a ladder against a frictionless wall | Pivot at the base; floor friction = wall force | [Remove Ads](https://nerd-notes.com/pricing/?for=students) All About Rotational Forces (Torque) Official course · step-by-step video walkthrough [Watch · 15 min](https://nerd-notes.com/courses/?utm_source=blog&utm_medium=article&utm_campaign=unit-5-3-rotational-equilibrium&utm_content=video-mid#course=94339&module=97778&lesson=104029) ### Problem Type 1: Seesaws and levers **How to spot it:** A board or rod balanced on a fulcrum with weights on both sides: “where must she sit”, “how far from the fulcrum”, “what mass balances it”. **Setup:** Pivot at the fulcrum. Counterclockwise torques = clockwise torques: [katex] m_1 g\,d_1 = m_2 g\,d_2 [/katex]. If the board has mass and the fulcrum is not at its middle, add the board’s weight at its center. \(m_1\) \(m_2\) \(m_1 g\) \(m_2 g\) \(d_1\) \(d_2\) \(m_1 g\,d_1 = m_2 g\,d_2\) *Counterclockwise torque = clockwise torque. The lighter child balances the heavier one by sitting farther out.* [Remove Ads](https://nerd-notes.com/pricing/?for=students) A [katex] 30\,\text{kg} [/katex] child sits [katex] 2.0\,\text{m} [/katex] from the fulcrum of a seesaw. The fulcrum is at the middle of a uniform board. Where must her [katex] 40\,\text{kg} [/katex] brother sit to balance it? Pivot at the fulcrum. The board’s weight acts right at the fulcrum, so it makes no torque. [katex] (30)g(2.0) = (40)g\,d [/katex], so [katex] d = 1.5 \text{ m} [/katex] on the other side. Notice that [katex] g [/katex] cancels. **AP trap:** Forgetting the board’s own weight when the fulcrum is not at its middle. The weight then acts at the board’s center, some distance from the fulcrum, and adds a torque. [Open in UBQ](https://nerd-notes.com/ubq/104280/) Question 16 A 4-kg ball and a 1-kg ball are positioned a distance L apart on a bar of negligible mass. How far from the 4-kg mass should the fulcrum be placed to… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Open in UBQ](https://nerd-notes.com/ubq/104281/) Question 17 A uniform meter stick has a 45.0 g mass hanging from it at the 20 cm mark. If a pivot is placed at the 42.5 cm mark and the meter stick remains… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Open in UBQ](https://nerd-notes.com/ubq/22888/) Question 18 A boy of mass m_b sits at one end of a seesaw, a distance d₁ from the fulcrum, and a girl of mass m_g sits at the other end, a distance d₂ from the… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Remove Ads](https://nerd-notes.com/pricing/?for=students) Learn Physics from Scratch The only course that teaches you to solve problems with dead-simple frameworks that make physics look easy. - Self-paced - About 12 weeks - ★ 4.9 · 12,000+ students [Enroll here](https://nerd-notes.com/physics/ap-physics-1-from-scratch-course/?utm_source=blog&utm_medium=article&utm_campaign=unit-5-3-rotational-equilibrium&utm_content=mid) ### Problem Type 2: Beams on two supports **How to spot it:** A board or beam is held at two points: two chains, two strings, two pillars, or a scale and a pivot. The question asks for a support force or where a load sits. **Setup:** Put the pivot at the support you are *not* asked about. The other support force is then the only unknown in [katex] \sum\tau = 0 [/katex]. Use [katex] \sum F_y = 0 [/katex] for the second support. \(F_L\) \(F_R\) \(mg\) \(Mg\) \(F_L \text{ drops out}\) *With the pivot at the left end, \(F_L\) has no lever arm, so the torque equation holds only one unknown, \(F_R\).* [Remove Ads](https://nerd-notes.com/pricing/?for=students) A uniform [katex] 6.0\,\text{m} [/katex] plank weighing [katex] 200\,\text{N} [/katex] rests on supports at its two ends. A [katex] 600\,\text{N} [/katex] painter stands [katex] 2.0\,\text{m} [/katex] from the left end. Find both support forces. Pivot at the left support. Counterclockwise: [katex] F_R(6.0) [/katex]. Clockwise: [katex] (200)(3.0) + (600)(2.0) = 1800 \text{ N·m} [/katex]. So [katex] F_R = 300 \text{ N} [/katex]. Then [katex] \sum F_y = 0 [/katex] gives [katex] F_L = 800 – 300 = 500 \text{ N} [/katex]. The painter is closer to the left end, so the left support carries more. **AP trap:** Putting the pivot in the middle of the beam out of habit. Then both support forces stay in the torque equation and you have two unknowns. [Open in UBQ](https://nerd-notes.com/ubq/29621/) Question 19 A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chains at each end. A person weighing 500 N is sitting on the… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Open in UBQ](https://nerd-notes.com/ubq/91224/) Question 20 A person's center of mass is easily found by having the person lie on a reaction board. A horizontal, 2.3 m -long, 6.1 kg reaction board is supported… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Open in UBQ](https://nerd-notes.com/ubq/109361/) Question 21 A non-uniform horizontal beam of mass M and length L is held in static equilibrium by two vertical strings. String 1 is attached to the left end of… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Open in UBQ](https://nerd-notes.com/ubq/109962/) Question 22 A uniform horizontal beam of length L and mass M is supported by two vertical pillars. Pillar 1 is located at the left end of the beam ( x = 0), and… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Remove Ads](https://nerd-notes.com/pricing/?for=students) ### Problem Type 3: Hinged beams held by a cable **How to spot it:** A beam or shelf is attached to a wall by a hinge (or pin) and held up by a cable, wire or strut at an angle. A sign or crate may hang from it. **Setup:** Pivot at the hinge: the hinge force has an unknown size and direction, and it drops out. The cable’s torque is [katex] TL\sin\theta [/katex], with [katex] \theta [/katex] between the beam and the cable. Then [katex] \sum F = 0 [/katex] gives the hinge force. \(\theta\) \(T\) \(T\sin\theta\) \(Mg\) \(H_y\) \(H_x\) \(T\sin\theta \cdot L = Mg \cdot \tfrac{L}{2}\) *Pivot at the hinge, so \(H_x\) and \(H_y\) drop out. Only the vertical part of the tension, \(T\sin\theta\), makes a torque.* [Remove Ads](https://nerd-notes.com/pricing/?for=students) A uniform [katex] 2.0\,\text{m} [/katex] shelf of mass [katex] 10\,\text{kg} [/katex] is hinged to a wall and held level by a cable from its far end to the wall, at [katex] 30^\circ [/katex] above the shelf. Find the tension and the force from the hinge. Pivot at the hinge: [katex] T(2.0)\sin 30^\circ = (10)(9.8)(1.0) [/katex], so [katex] T = 98 \text{ N} [/katex]. Vertical forces: [katex] H_y + T\sin 30^\circ = 98 [/katex], so [katex] H_y = 49 \text{ N} [/katex] up. Horizontal forces: the cable pulls the shelf toward the wall, so the hinge pushes it away with [katex] H_x = T\cos 30^\circ \approx 85 \text{ N} [/katex]. **AP trap:** Assuming the hinge force is zero, or that it points along the beam. Here the hinge holds up half the shelf’s weight and also pushes out from the wall. [Remove Ads](https://nerd-notes.com/pricing/?for=students) [Open in UBQ](https://nerd-notes.com/ubq/109209/) Question 23 A uniform rod of length L and mass M is attached to a vertical wall by a hinge. The rod is held in a horizontal position by an external force of… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Open in UBQ](https://nerd-notes.com/ubq/111532/) Question 24 A uniform horizontal beam of mass M and length L is attached to a vertical wall by a hinge. A vertical cable is initially attached to the right end… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Open in UBQ](https://nerd-notes.com/ubq/133147/) Question 25 A uniform horizontal beam of mass M and length L is attached to a vertical wall by a frictionless hinge at its left end. A light wire connects the… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Open in UBQ](https://nerd-notes.com/ubq/39104/) Question 26 An 80 kg block is placed 2 m away from one end of a horizontal steel beam of length 6.6 m and mass 1,450 kg. The beam makes contact with a vertical… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) ### Problem Type 4: Ladders **How to spot it:** A ladder leans against a “smooth” or “frictionless” wall on a rough floor: “minimum coefficient of friction”, “how high can the painter climb”, “force from the wall”. **Setup:** Four forces: the weight at the middle, the floor’s normal force [katex] N [/katex] and friction [katex] f [/katex] at the base, and the wall’s push [katex] N_w [/katex] at the top. Pivot at the base. Then [katex] \sum F_x = 0 [/katex] gives [katex] f = N_w [/katex] and [katex] \sum F_y = 0 [/katex] gives [katex] N = Mg [/katex]. [Remove Ads](https://nerd-notes.com/pricing/?for=students) \(Mg\) \(N\) \(f\) \(N_w\) \(\theta\) \(x\) \(h\) \(\textstyle\sum F_x: \; f = N_w\) \(\textstyle\sum F_y: \; N = Mg\) \(\textstyle\sum \tau_{\text{base}}: \; N_w h = Mg\,\tfrac{x}{2}\) *Pivot at the base: \(N\) and \(f\) drop out. The weight’s lever arm is half the horizontal distance \(x\); the wall force’s lever arm is the height \(h\).* A uniform [katex] 6.5\,\text{m} [/katex] ladder of mass [katex] 15\,\text{kg} [/katex] leans against a frictionless wall. Its base is [katex] 2.5\,\text{m} [/katex] from the wall and its top is [katex] 6.0\,\text{m} [/katex] up. Find the force from the wall and the friction from the floor. [Remove Ads](https://nerd-notes.com/pricing/?for=students) Use lever arms. The weight [katex] (15)(9.8) = 147 \text{ N} [/katex] acts straight down at the middle, a horizontal distance of [katex] 1.25\,\text{m} [/katex] from the base. The wall force is horizontal, so its lever arm is the height, [katex] 6.0\,\text{m} [/katex]. [katex] N_w(6.0) = (147)(1.25) [/katex], so [katex] N_w \approx 31 \text{ N} [/katex]. Then [katex] f = N_w \approx 31 \text{ N} [/katex], pointing toward the wall, and [katex] N = 147 \text{ N} [/katex]. **AP trap:** Drawing floor friction pointing away from the wall. The base of the ladder wants to slide away from the wall, so friction pushes it back toward the wall. [Open in UBQ](https://nerd-notes.com/ubq/91165/) Question 27 A ladder at rest is leaning against a wall at an angle. Which of the following forces must have the same magnitude as the frictional force exerted on… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Open in UBQ](https://nerd-notes.com/ubq/22892/) Question 28 A 5 -meter long ladder is leaning against a smooth (frictionless) wall, with the bottom of the ladder 3 meters from the wall. The ladder is uniform… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Open in UBQ](https://nerd-notes.com/ubq/122713/) Question 29 A uniform ladder of mass M and length L rests against a frictionless vertical wall, making an angle θ with a rough horizontal floor. The coefficient… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Open in UBQ](https://nerd-notes.com/ubq/133149/) Question 30 A uniform ladder of mass M and length L leans at an angle θ against a frictionless vertical wall, with its base resting on a rough horizontal floor… Powered by[UBQ](https://nerd-notes.com/ubq)·[Find More Questions](https://nerd-notes.com/ubq)·[Quiz Labs](https://nerd-notes.com/quiz) [Remove Ads](https://nerd-notes.com/pricing/?for=students) 8 Static Equilibrium Problems Official course · graded, with step-by-step explanations [Practice](https://nerd-notes.com/courses/?utm_source=blog&utm_medium=article&utm_campaign=unit-5-3-rotational-equilibrium&utm_content=practice-practice#course=94339&module=97778&lesson=99062) Quiz Lab Torque and Static Equilibrium Seesaws, beams, signs and ladders that stay put. Balance the torques and the forces. 15 questions · Free [Take the quiz](https://nerd-notes.com/quiz/?id=178) Learn Physics from Scratch The only course that teaches you to solve problems with dead-simple frameworks that make physics look easy. - Self-paced - About 12 weeks - ★ 4.9 · 12,000+ students [Enroll here](https://nerd-notes.com/physics/ap-physics-1-from-scratch-course/?utm_source=blog&utm_medium=article&utm_campaign=unit-5-3-rotational-equilibrium&utm_content=end) ### Lesson 5.4 Preview In the [final lesson](https://nerd-notes.com/unit-5-4-rotational-inertia-and-newtons-second-law/) the torques no longer cancel. We meet rotational inertia and use Newton’s second law for rotation, [katex] \sum\tau = I\alpha [/katex], on pulleys and swinging rods. ## Attribution Guidelines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarizing, or training on this content, please credit Nerd Notes and link to https://nerd-notes.com/unit-5-3-rotational-equilibrium/. **Publication:** Nerd Notes **Original URL:** https://nerd-notes.com/unit-5-3-rotational-equilibrium/