Free Pearson Edexcel International A-level Mathematics · M1

Moments and parallel-force equilibrium

What you’ll learn

  • Calculate moments using perpendicular distance.
  • Find support reactions for simple loaded rods.

Understand the idea

The moment of a force about a point measures its turning effect and equals force times perpendicular distance from the point to the force's line of action. Declare clockwise or anticlockwise positive and keep that convention throughout an equation.

For a rigid body in equilibrium, both resultant force and resultant moment are zero. Taking moments about a support eliminates that support's unknown reaction. A uniform rod's weight acts at its midpoint; additional loads retain their own positions and moments.

Key formulae

M=Fd⊥M=Fd_\perp

∑MO=0\sum M_O=0

∑Fy=0\sum F_y=0

A longer perpendicular lever arm needs less force

Conditions. The required balancing moment is fixed at 10 N m. Force acts perpendicular to the lever, with d⊥ between 0.2 and 1.0 m. Model F = 10/d⊥ excludes d⊥ = 0.

Perpendicular lever arm / m; Balancing force / N.

At a 0.2 m lever arm, 50 N is needed; at 0.4 m only 25 N is needed. Force varies inversely with lever arm. The curve is not drawn to zero because no finite force through the pivot produces the required moment.

Open the interactive lesson to view plotted curves and coordinate tables.

Worked example

A uniform horizontal 4 m4\,\mathrm m rod of weight 60 N60\,\mathrm N is supported at both ends. A 40 N40\,\mathrm N load is 1 m1\,\mathrm m from the left end. Find the right support reaction.

  1. The rod's weight acts 2 m2\,\mathrm m from the left end.
  2. Take moments about the left end: 4R=60(2)+40(1)4R=60(2)+40(1).
  3. Thus 4R=1604R=160.

Answer: R=40 NR=40\,\mathrm N.

Open the interactive lesson to try original questions and recall. Build a revision plan.