Balance a beam with force and torque
You will be able to: Solve a static beam using a convenient pivot and both equilibrium conditions.
Why does the support nearest a person carry more load?
A person stands near one end of a supported platform. Both supports push upward, but the nearer one usually carries more load. Torque balance determines how the support forces share the total weight.
A useful starting point: Balanced torque can mean steady rotation →
Words and symbols before equations
- Support reaction R
- Force exerted by a support on the beam.
- Uniform beam
- Mass evenly distributed, so its weight acts through the midpoint in uniform gravity.
- Chosen calculation pivot
- Point used for torque accounting; it need not be a physical hinge.
- Static model
- The beam is at rest with zero net force and zero net torque.
What this picture assumes
Static uniform 4 m beam, weight 40 N, supported upward at both ends. Person weight 80 N. The person remains between supports. Forces are vertical; lengths and force arrows have labeled scales.
Connect the picture to the physics
Draw both support forces upward, the beam’s weight downward at its center, and the person’s weight downward at the person’s position. Use distances measured from one chosen point.
Choose the left support for torque accounting. Its own reaction has zero lever arm, so the torque equation contains only the right reaction and the two weights. Once the right reaction is known, vertical force balance gives the left one.
A support that can only push cannot supply a negative upward reaction. If an assumed load position predicts such a value, the original contact model may fail and tipping must be reconsidered. Our investigation keeps the person between the supports, where both reactions stay positive.
A worked example, step by step
A 4 m uniform beam weighs 40 N and supports an 80 N person 3 m from its left end. Find the upward end reactions.
- About the left end, CCW positive: 4R_R−40(2)−80(3)=0.
- Thus R_R=(80+240)/4=80 N.
- Vertical force balance: R_L+80−40−80=0, giving R_L=40 N.
- Check torque about the right end: −40(4)+40(2)+80(1)=0. The same reactions satisfy either pivot choice.
Include the beam’s own weight when it is not negligible. Torque balance alone does not give every support force.
Why choose a pivot at a support?
Compare with an explanation
That support force has zero lever arm there, removing one unknown from the torque equation; force balance still includes it.
Predict. Change one thing. Explain.
Move the person along the 4 m beam. Track both support arrows and check their sum remains 120 N. Predict the equal-reaction position before setting it.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Left reaction=40 N; right reaction=80 N. Sum=120 N. About the left: 320−80−240=0 N·m.
Static uniform 4 m beam, weight 40 N, supported upward at both ends. Person weight 80 N. The person remains between supports. Forces are vertical; lengths and force arrows have labeled scales.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use an angular-motion or torque relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionUse the 4 m, 40 N beam with an 80 N person at x=1 m. (a) Write torques about the left. (b) Find R_R. (c) Find R_L. (d) Check the total upward force.
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Compare with the answer and four-point rubric
- 1 point: 4R_R−40(2)−80(1)=0.
- 1 point: R_R=40 N.
- 1 point: R_L=120−40=80 N.
- 1 point: R_L+R_R=120 N equals the combined weights.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Why keep the beam’s weight?
It contributes force and generally torque.
RECALL 2Best pivot choice?
Often a point through an unknown force, simplifying that torque equation.
RECALL 3What if a push-only support requires negative reaction?
Reconsider contact and tipping; the assumed static model may be invalid.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Balance a beam with force and torque
- About left support: LR_R−W_beam(L/2)−W_person x=0.
- Vertical balance: R_L+R_R=W_beam+W_person.
Remember: Include the beam’s own weight when it is not negligible. Torque balance alone does not give every support force.
Conditions: Static uniform 4 m beam, weight 40 N, supported upward at both ends. Person weight 80 N. The person remains between supports. Forces are vertical; lengths and force arrows have labeled scales.
Refresh Kid · Unit 5 · Objectives 5.5.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.5, objectives 5.5.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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