Add turning effects about one axis
You will be able to: Construct a force diagram and sum signed torques about one axis.
Can two downward forces turn a bar in opposite senses?
Push downward on the left end of a center-pivoted bar and it tends to turn counterclockwise. Push downward on its right end and it tends to turn clockwise. The force directions match, but their locations give opposite turning effects.
A useful starting point: Torque: push where it makes a difference →
Words and symbols before equations
- Net torque Στ
- Sum of all signed torques about the same axis.
- Turning sense
- Clockwise or counterclockwise tendency in the specified view.
- Pivot reaction
- Force exerted by the support; its torque about that pivot is zero when it acts through the axis.
What this picture assumes
Horizontal rigid bar, center pivot. A 10 N downward force acts at x=−0.5 m, the selected downward force at x=+1 m. Bar weight acts through pivot, giving zero torque. Pivot reaction is omitted from the torque bars because its lever arm is zero.
Connect the picture to the physics
Draw the whole bar and mark the pivot, force directions and application distances. Choose CCW positive. A downward force left of the pivot contributes positive torque; the same direction right of the pivot contributes negative torque.
Calculate each perpendicular lever arm separately, then add signed torques. Do not merely subtract force magnitudes unless their lever arms are equal. A smaller force farther away can balance a larger nearby force.
A force through the chosen pivot contributes zero torque about that pivot, but may still matter in the translational force balance. Net torque and net force are separate questions. Changing the calculation axis changes individual torque contributions.
A worked example, step by step
A 10 N force acts downward 0.5 m left of a pivot. A 4 N force acts downward 1 m right of it. Find net torque; then find the right-hand force that balances torques.
- Left torque τ_L=+(10)(0.5)=+5 N·m.
- Right torque τ_R=−(4)(1)=−4 N·m.
- Στ=5−4=+1 N·m, a counterclockwise tendency.
- For balance: 5−F_R(1)=0, so F_R=5 N. The balancing forces need not be equal because their lever arms differ.
Determine turning sense from the force location, not from whether its arrow points up or down alone.
If net torque is zero, must the two applied forces be equal?
Compare with an explanation
No. It is their torques that balance; different lever arms can compensate for different force magnitudes.
Predict. Change one thing. Explain.
Vary the downward force 1 m to the right while keeping 10 N at 0.5 m to the left. Find torque balance, then predict the sign after increasing the right-hand force.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Left torque=+5 N·m; right torque=-5 N·m; net=0 N·m. Torques balance.
Horizontal rigid bar, center pivot. A 10 N downward force acts at x=−0.5 m, the selected downward force at x=+1 m. Bar weight acts through pivot, giving zero torque. Pivot reaction is omitted from the torque bars because its lever arm is zero.
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 questionA bar has downward forces of 12 N at 0.5 m left and 3 N at 1 m right of its pivot. (a) Find each torque, CCW positive. (b) Find net torque. (c) Find a replacement right-hand force giving balance. (d) Explain why the pivot force cannot simply be omitted from a force balance.
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Compare with the answer and four-point rubric
- 1 point: +6 N·m and −3 N·m.
- 1 point: +3 N·m.
- 1 point: 6 N downward at the same right-hand location.
- 1 point: It can have nonzero force even though its torque about the pivot is zero.
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 1What must be common when summing torques?
The chosen axis and sign convention.
RECALL 2Downward force left of a central pivot?
Counterclockwise tendency in this view.
RECALL 3Zero torque means zero force?
No. Force and torque are different balances.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Add turning effects about one axis
- Στ=Σ(signed force × perpendicular lever arm).
- A force through the selected axis contributes zero torque about that axis.
Remember: Determine turning sense from the force location, not from whether its arrow points up or down alone.
Conditions: Horizontal rigid bar, center pivot. A 10 N downward force acts at x=−0.5 m, the selected downward force at x=+1 m. Bar weight acts through pivot, giving zero torque. Pivot reaction is omitted from the torque bars because its lever arm is zero.
Refresh Kid · Unit 5 · Objectives 5.3.A, 5.3.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.3, objectives 5.3.A, 5.3.B. 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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