Moments, Supports & Distributed Loads

A moment measures a force’s tendency to rotate a body about a point or axis. In planar work its signed magnitude is

where \(d_\perp\) is the shortest distance from point \(O\) to the force’s line of action. Sliding a force along that line does not change its moment; moving it sideways does.

Replace a load distribution without losing its effect

A distributed load \(w(x)\) in N/m becomes an equivalent resultant

acting at

Graphically, \(R\) is the area under the load diagram and \(\bar{x}\) is its centroid. A uniform 800 N/m load over 1.5 m therefore becomes 1200 N acting at the segment midpoint, 0.75 m from either end. A triangular load peaks at one end; its resultant acts one-third of the base length from the high-intensity end.

Worked simply supported beam

A 2 m beam has a pin at A, a roller at B, and a uniform downward load of 600 N/m over its full length.

• Resultant: \(R=600(2)=1200\) N at midspan.
• Moments about A: \(2B_y-1200(1)=0\), so \(B_y=600\) N.
• Vertical equilibrium: \(A_y+B_y-1200=0\), so \(A_y=600\) N.

Symmetry predicts the same result, but the equations prove it. If the load covered only the left metre, its 600 N resultant would act 0.5 m from A; then \(B_y=150\) N and \(A_y=450\) N.

Couples and equivalent systems

Two equal, opposite, parallel forces separated by distance \(d\) form a couple \(M=Fd\). Their net force is zero, but their moment remains. A pure couple is a free vector in rigid-body statics: its rotational effect is the same about every point. This is why a motor torque, wrench input, or fastener couple must not be discarded because the force arrows cancel.

Rule-of-thumb checks

• The sum of vertical reactions must equal the total vertical load.
• A resultant must lie within a positive load distribution’s footprint.
• Reactions shift toward the side where the load’s centroid shifts.
• Treat a fixed support moment as an unknown reaction, not a decorative arrow.

Design exercise

Create a load table for a small conveyor section: belt and frame weight, product weight, tension, motor torque, and any jam load. Separate normal, startup, fault, and service load cases. Draw one FBD per case and solve reactions. Statics is not one timeless diagram; machines occupy states, and each state can create a different critical load.