Moment & Equilibrium

\sum F_{x} = 0, \sum F_{y} = 0, \sum M_{A} = 0

A rigid body in 2-D equilibrium has zero net force and zero net moment about any point. Moment of a force about A is M = Fd, with d the perpendicular lever arm, or M = r × F.

$F$ (N or lb): Force
$d$ (m or ft): Perpendicular lever arm
$M$ (N·m or ft·lb): Moment

Step-by-step solved example

A 4 m simply supported beam carries 12 kN at midspan. Find the left reaction.

Moment about a support finds the far reaction, then ΣFy.

1. Moment about right support
ΣM_B = 0: A_y(4) − 12(2) = 0 → A_y = 6 kN.
$A_{y} = 6 kN$
2. Check ΣF_y
A_y + B_y = 12 → B_y = 6 kN. Symmetric as expected.

Answer: A_y = 6 kN

10 practice questions

0/10 correct

1.Moment of 50 N about a point 0.4 m perpendicular away is

2.A 10 kN load 3 m from A on a 6 m simple beam: R_A is

3.If ΣF = 0 but ΣM ≠ 0 the body

4.Varignon's theorem states that

5.A couple of 40 N with 0.5 m arm has moment

6.Cantilever 2 m with 8 kN at the free end: fixed moment is

7.Three 2-D equilibrium equations exist for a

8.A 30° force of 100 N has x-component

9.Internal hinge in a beam adds

10.For a 9 kN load 2 m from A on a 6 m beam, R_B is