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SCIENCE · INVESTIGATE AND EXPLAIN

Use two equilibrium conditions for an extended load

Static equilibrium means both zero translational acceleration and zero angular acceleration, with the object initially at rest. For a planar rigid body, solve force components and signed torques about a convenient axis. A torque calculation alone cannot determine whether the body is also translating.

Learning outcome: With W=80 N compare x=1, 2 and 3 m. Draw all forces at their true locations. Compute both reactions and check net torque about both supports. Then change the load while keeping its location fixed and separate the beam-weight contribution from the added-load contribution.

Starting recap

Review signed graph slopes and areas, extended-body force diagrams, Newton’s laws and center-of-mass coordinates. State the rotation axis.

All lessons are open directly. Your saved experiments remain when you leave and return.

Try

Put an 80 N load at x=3 m. Right reaction R=(40×2+80×3)/4=80 N, and left reaction L=120−80=40 N. About the right support, −4×40+40×2+80×1=0 N m, confirming the same equilibrium using another axis.

Open the experiment

Predict

As the load moves right while its weight stays fixed, what remains constant: each reaction, their sum, or their ratio?

Experiment

With W=80 N compare x=1, 2 and 3 m. Draw all forces at their true locations. Compute both reactions and check net torque about both supports. Then change the load while keeping its location fixed and separate the beam-weight contribution from the added-load contribution.

Solve force and torque balance independently

Worked teaching model · synthetic values · A 4 m uniform beam weighs 40 N and rests on ideal vertical supports at both ends. One downward point load is placed within the span. Static rigid-body model only; no bending, material strength or safe construction claim.

Reset keeps your prediction, captures and practice history.

Extended beam force locations

Left support at x=0 m; beam weight 40 N at x=2 m; downward load 80 N at x=3 m; right support at x=4 m. Arrow lengths are schematic, not a force scale.

leftrightL upR up40 N80 N load

Application locations and directions are meaningful; force arrows are schematic unless explicitly labeled otherwise. Exact quantities and units follow in the ledger.

Calculated evidence ledger — synthetic teaching model
QuantityValueUnit
Upward left support reaction40N
Upward right support reaction80N
Net vertical force0N
Right support torque about left320N m
Beam weight torque about left-80N m
Load torque about left-240N m
Net torque about left0N m
Accessible plot coordinates
Primary plotted points
time (s)angular velocity (rad/s)

Force balance fixes the sum of reactions; torque balance fixes their distribution.

About the left support, 4Rright−40×2−Wx=0. Then Rleft+Rright=40+W. A support reaction of zero is a limiting case, not a negative pulling force available from an ordinary resting contact.

Force balance fixes the sum of reactions; torque balance fixes their distribution.

About the left support, 4Rright−40×2−Wx=0. Then Rleft+Rright=40+W. A support reaction of zero is a limiting case, not a negative pulling force available from an ordinary resting contact.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

1. Frame the question

Static equilibrium means both zero translational acceleration and zero angular acceleration, with the object initially at rest. For a planar rigid body, solve force components and signed torques about a convenient axis. A torque calculation alone cannot determine whether the body is also translating.

2. Build the mechanism

Our original virtual beam has length 4 m, weight 40 N at its midpoint, ideal upward supports at both ends, and a downward point load W at position x from the left. Vertical force balance gives L+R=40+W. Taking moments about the left support eliminates L from that torque equation, yielding 4R−40×2−Wx=0. Then calculate L from force balance.

The weight acts at the center because the beam is uniform and gravity is treated as uniform. A nonuniform beam would need its actual center of mass. The support chosen as torque origin has not disappeared physically; its lever arm about that origin is zero. Checking moments about the other support provides an independent bookkeeping test.

Resting supports push but do not ordinarily pull the beam downward. A negative solved support reaction would therefore indicate that the assumed two-contact equilibrium is not feasible without a different constraint. The present bounded control range keeps the load inside the span. This is not a structural-strength model: rigidity ignores bending and the equations provide no construction or load-rating guarantee.

3. Work an example

Put an 80 N load at x=3 m. Right reaction R=(40×2+80×3)/4=80 N, and left reaction L=120−80=40 N. About the right support, −4×40+40×2+80×1=0 N m, confirming the same equilibrium using another axis.

4. Predict and investigate

Predict first: As the load moves right while its weight stays fixed, what remains constant: each reaction, their sum, or their ratio?

With W=80 N compare x=1, 2 and 3 m. Draw all forces at their true locations. Compute both reactions and check net torque about both supports. Then change the load while keeping its location fixed and separate the beam-weight contribution from the added-load contribution. Retain settings and raw evidence in the lesson notebook.

5. Transfer and explain

Try a new practice case

For the same beam, place a 60 N load 1 m from the left. Find the right reaction.

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Reflect and retain evidence

What additional evidence would be needed before turning an ideal equilibrium calculation into a real claim about a structure’s safe capacity?

Retain your first response, an extended-body diagram or labeled graph, units, assumptions and a changed-case prediction. Public worked checks are practice, not independent mastery evidence. Synthetic model captures are not a performed physical experiment.

Framework and source notes

Original teaching prose and numerical cases. College Board AP Physics 1 CED, Unit 5 topic mapping: 5.3; 5.5. Checked 30 October 2026. Planar clockwise/counterclockwise analysis; supplied formulas for extended bodies, no required calculus or three-dimensional torque-vector direction. Framework cross-reference, not AP authorization.

Apply

A 4 m uniform beam weighs 40 N. An 80 N load sits 3 m from the left support. What is the upward right reaction?

Public practice checks a specific claim. It does not establish independent mastery.

What additional evidence would be needed before turning an ideal equilibrium calculation into a real claim about a structure’s safe capacity?

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

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