Test support feasibility after integrating a distributed load

Draw the external interactions

A rigid beam spans 0–4 m. Its own 20 N weight acts at x=2; an added downward load density is a(1+x/4) N/m; a 40 N point load acts at the selected coordinate. Upward-only supports sit at x=1 and 3. Each load belongs on the beam’s recipient diagram; the equal-and-opposite forces on supports do not cancel them within that diagram.

This is a recipient force inventory, not a scale drawing of one control setting. The point-load arrow marks a variable position; use the selected numerical b rather than its schematic screen coordinate. The distributed-load arrows identify a density, not extra forces to add after its integral has already been included.

Integrate load and its first moment

The added load integrates to 6a N. Its moment about x=0 is integral x a(1+x/4)dx=40a/3 N m. Total downward force is W=60+6a and its total moment is 40+40a/3+40b, where b is the point-load coordinate. A distributed load must be located by its first moment, not placed arbitrarily at the geometric midpoint.

Solve two independent balances

Let left and right upward reactions be L and R. Vertical balance gives L+R=W. Taking torque about the left support at x=1 gives 2R=moment−W. Thus R=−10+11a/3+20b and L=70+7a/3−20b in newtons. Choosing a support as origin removes its own torque, not its force from the force balance.

Reject an impossible contact branch

At a=0,b=2 both reactions are 30 N. Move the point load to b=4 and the required right reaction becomes 70 while left becomes−10. An upward-only surface cannot supply that negative reaction. The assumed two-contact static equilibrium has failed; lift-off or another constraint must be modeled. Do not simply change the negative number to zero while retaining all the same equilibrium equations.

Separate rotational from translational equilibrium

Zero net torque permits constant angular velocity, not necessarily rest, for the fixed rigid-axis setting. Zero net force permits constant center velocity. A body can satisfy one condition without the other: a pure couple has zero net force but nonzero torque, while a force through the center can accelerate translation without center torque. Static equilibrium additionally requires both initial velocities to remain zero under the stated constraints.

Worked original case

At a=1,b=2, total load 66 N, right reaction 101/3≈33.6667 N and left 97/3≈32.3333 N. Both contacts are feasible.

Investigation

Compare point positions 2 and 4 at a=0, then add distributed load a=4. Integrate its moment, solve both reactions and retain the feasibility flag. Sketch the failed assumed contact separately from any proposed new support model.

Public numerical transfer practice

Try a new practice case

With a=0 and the point load at x=2, what is the left support reaction in newtons?

Numerical result in the requested units Check this practice result

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Sources and evidence boundary

Original teaching and numerical cases checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax fixed-axis rotation. Supplied/generated evidence is not physical measurement, a load rating or laboratory participation. No rotating machinery, hanging weights or mechanical construction is requested. Public feedback and retained self-reports do not certify independent learner mastery.