Force Systems and Equilibrium

Key Takeaways

  • A force is a vector quantity with magnitude, direction, and point of application.
  • The resultant of concurrent forces: R = ΣFx i + ΣFy j — resolve each force into components.
  • For equilibrium: ΣFx = 0, ΣFy = 0, and ΣM = 0 (sum of forces and moments equals zero).
  • Free-body diagrams (FBDs) are essential — isolate the body and show ALL external forces and reactions.
  • Support reactions: pin supports provide two force components (Rx, Ry); rollers provide one (perpendicular to surface).
  • A couple is two equal, opposite, parallel forces that create a pure moment M = F × d.
Last updated: March 2026

Force Systems and Equilibrium

FE Exam Weight: Statics accounts for 9-14 questions (~10% of the exam). Statics is fundamental to Dynamics, Mechanics of Materials, and Structural Analysis.

Vectors and Force Components

A force vector F can be resolved into rectangular components: Fx=FcosθFy=FsinθF_x = F \cos \theta \qquad F_y = F \sin \theta

Magnitude and direction from components: F=Fx2+Fy2θ=arctanFyFxF = \sqrt{F_x^2 + F_y^2} \qquad \theta = \arctan\frac{F_y}{F_x}

3D Force Components

Fx=FcosαFy=FcosβFz=FcosγF_x = F \cos \alpha \qquad F_y = F \cos \beta \qquad F_z = F \cos \gamma

where α, β, γ are the angles between the force and the x, y, z axes respectively.

Direction cosines: cos²α + cos²β + cos²γ = 1

Moment of a Force

The moment of a force about a point is the tendency of the force to cause rotation:

MO=F×dM_O = F \times d

where d is the perpendicular distance from the point to the line of action of the force.

Vector cross product form: MO=r×F\vec{M}_O = \vec{r} \times \vec{F}

where r is the position vector from O to any point on the line of action of F.

Sign Convention: Counterclockwise (CCW) is typically positive; clockwise (CW) is negative.

Varignon's Theorem

The moment of a force about a point equals the sum of the moments of its components about the same point: MO=Fxdy+FydxM_O = F_x \cdot d_y + F_y \cdot d_x

This is extremely useful — instead of finding the perpendicular distance, resolve the force into components and sum moments of each component.

Couples

A couple consists of two equal, opposite, parallel forces separated by distance d: M=F×dM = F \times d

Properties of couples:

  • A couple produces a pure moment (rotation only, no translation)
  • The moment of a couple is the same about ANY point
  • A couple can be moved anywhere on the body without changing its effect
  • Couples can only be balanced by other couples

Equilibrium Conditions

For a body in static equilibrium:

2D Equilibrium: Fx=0Fy=0MO=0\sum F_x = 0 \qquad \sum F_y = 0 \qquad \sum M_O = 0

These three equations can solve for up to three unknowns in 2D.

3D Equilibrium: Fx=0Fy=0Fz=0\sum F_x = 0 \quad \sum F_y = 0 \quad \sum F_z = 0 Mx=0My=0Mz=0\sum M_x = 0 \quad \sum M_y = 0 \quad \sum M_z = 0

Six equations → up to six unknowns in 3D.

Free-Body Diagrams (FBDs)

Steps to draw an FBD:

  1. Isolate the body (or portion of a body)
  2. Show all external forces (applied loads, weights)
  3. Replace supports with their reaction forces
  4. Include dimensions and angles
  5. Label all forces and moments

Common Support Reactions

Support TypeReactions ProvidedUnknowns
Roller1 force (⊥ to surface)1
Pin (hinge)2 force components (Rx, Ry)2
Fixed (cantilever)2 forces + 1 moment (Rx, Ry, M)3
Cable/rope1 force (tension along cable)1
Smooth surface1 force (⊥ to surface, push only)1

Exam Strategy: When solving for unknown reactions, choose your moment point wisely. Taking moments about a point where two unknowns intersect eliminates both from the equation, giving you the third unknown directly.

Test Your Knowledge

A 100 N force acts at 30° above the horizontal. What is its horizontal component?

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Test Your Knowledge

A simply supported beam has a pin support at A and a roller at B, 6 m apart. A 12 kN load is applied 2 m from A. What is the reaction at B?

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B
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D
Test Your Knowledge

How many independent equilibrium equations are available for a 2D rigid body?

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B
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D