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3.4 · Solve planar rigid-body equilibrium problems

Learn to solve planar rigid-body equilibrium problems through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Planar Equilibrium

A free-body-diagram method for balancing forces and moments

A planar rigid-body equilibrium problem asks which external forces and moments balance a body in a plane. Treat the body as one object whose shape and dimensions do not change during the analysis. Its size matters because a force applied at one point can create a different moment from the same force applied elsewhere. The basic method is to isolate the body, show all external actions on a free-body diagram, and apply force and moment balance. The examples below use support arrangements for which the equilibrium equations are sufficient to determine the unknown reactions.

What you will learn

  • Isolate a planar rigid body and identify its external forces, couples, and support reactions.
  • Choose axes and signs, then apply the three planar equilibrium equations.
  • Solve for unknown reactions and verify both force and moment balance.

1. Define the body and draw its free-body diagram

First state exactly what you are isolating, such as a beam or bracket. Imagine removing its surroundings. Then show every force and couple the surroundings exert on the isolated body. These include applied loads and support reactions. Do not include forces that the isolated body exerts on its surroundings.
Use the support model to decide which reactions belong on the diagram. In a planar model, a pin can exert horizontal and vertical force components but no reaction couple. A roller on a horizontal surface exerts a vertical reaction. A fixed support can exert horizontal and vertical forces and a reaction couple. A cable pulls along its length. These idealized models describe how the support acts on the body.
Show a force where it is applied, because its location affects its moment. An applied couple is shown as a moment with a direction; it is not a force and does not need a lever arm. If a distributed load is replaced by a single resultant, preserve both its total force and its line of action.
  • Isolate one clearly defined body.
  • Show every external force, couple, and support reaction.
  • Retain the application point of each force.

2. Choose signs and write the equilibrium equations

For a planar problem, a convenient choice is positive xx to the right and positive yy upward. Take counterclockwise moments as positive. Mark or state the convention before solving, and keep it throughout. If you assume an unknown reaction points in a direction that turns out to be wrong, its calculated value will be negative.
A force can be split into horizontal and vertical components. For a force of magnitude FF at angle θ\theta counterclockwise from positive xx, its components are Fx=Fcos⁡θF_x=F\cos\theta and Fy=Fsin⁡θF_y=F\sin\theta. A moment about a point describes the force's turning effect. With force components applied at coordinates (x,y)(x,y) relative to that point, the signed moment is M=xFy−yFxM=xF_y-yF_x. For a horizontal beam, a downward force to the right of the moment point creates a clockwise, negative moment.
An applied couple is included directly with its sign. In planar equilibrium, the horizontal force sum, vertical force sum, and moment sum about any selected point must each be zero. Choose a moment point that removes unknown forces where possible: a force whose line of action passes through that point contributes no moment there.
∑Fx=0,∑Fy=0,∑MO=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M_O=0
  • State axes and the positive moment direction before calculating.
  • Use two force-balance equations and one moment-balance equation.
  • Choose a moment point that simplifies the unknowns.

3. Solve and check the result

Write equations from the diagram, keeping track of each force's direction and location. When practical, first write the equations symbolically; then substitute the given values. A force multiplied by a perpendicular distance gives a moment, so the units must be consistent: for example, kilonewtons times metres gives kilonewton-metres.
Solve the equations and report reaction directions as well as magnitudes. A negative component means the actual force acts opposite to its assumed arrow. Avoid rounding intermediate values too early, especially when later equations use those values.
Check the answer independently. Add the horizontal components and the vertical components to confirm their respective sums are zero. Then calculate moments about a convenient point, including applied couples, and confirm the signed moment sum is zero. Small residuals can result from rounding; they should be consistent with the displayed precision.
  • Keep force and moment units distinct and consistent.
  • Interpret negative reaction components as reversed assumed directions.
  • Verify both force balance and moment balance.

Worked example

Pin and roller with an angled force and couple

A horizontal beam ABAB is 4.0 m4.0\,\mathrm{m} long. A pin at AA and a roller at BB support it. A 6.0 kN6.0\,\mathrm{kN} downward force acts 1.5 m1.5\,\mathrm{m} from AA. A 4.0 kN4.0\,\mathrm{kN} force acts 3.0 m3.0\,\mathrm{m} from AA, directed 60∘60^\circ below the positive horizontal axis. A clockwise couple of 2.0 kN⋅m2.0\,\mathrm{kN\cdot m} also acts. Find the support reactions.
  1. Resolve the angled force
    The force points right and down. Its horizontal component is positive and its vertical component is negative with the chosen axes.
    Fx=4cos⁡60∘=2.00 kN,Fy=−4sin⁡60∘=−3.464 kNF_x=4\cos60^\circ=2.00\,\mathrm{kN},\quad F_y=-4\sin60^\circ=-3.464\,\mathrm{kN}
  2. Take moments about A
    This choice removes both pin reactions from the moment equation. The roller reaction is counterclockwise about AA; the downward forces and clockwise couple contribute negative moments.
    4By−6(1.5)−3.464(3.0)−2.0=04B_y-6(1.5)-3.464(3.0)-2.0=0
  3. Solve force balance
    The moment equation gives the roller reaction. Horizontal and vertical force balance then determine the pin reaction components.
    By=5.348 kN,Ax=−2.00 kN,Ay=4.116 kNB_y=5.348\,\mathrm{kN},\quad A_x=-2.00\,\mathrm{kN},\quad A_y=4.116\,\mathrm{kN}
  4. Verify equilibrium
    The horizontal reactions oppose the applied horizontal component. The vertical reactions balance the downward forces, and the moments about AA cancel. The negative AxA_x means the actual pin force is leftward.
    ∑Fx=−2.00+2.00=0,∑Fy=4.116+5.348−6−3.464=0,∑MA≈0 kN⋅m\sum F_x=-2.00+2.00=0,\quad \sum F_y=4.116+5.348-6-3.464=0,\quad \sum M_A\approx0\,\mathrm{kN\cdot m}
Answer: The reactions are 2.00 kN2.00\,\mathrm{kN} left and 4.116 kN4.116\,\mathrm{kN} upward at AA, and 5.348 kN5.348\,\mathrm{kN} upward at BB.
Check: The force components and the moments about AA balance, within rounding.

Worked example

Cantilever with a force and an applied couple

A 3.0 m3.0\,\mathrm{m} horizontal cantilever is fixed at AA. A 5.0 kN5.0\,\mathrm{kN} downward force acts 1.0 m1.0\,\mathrm{m} from AA. At 2.0 m2.0\,\mathrm{m} from AA, a 4.0 kN4.0\,\mathrm{kN} force acts up and to the left at 150∘150^\circ from the positive horizontal axis. A clockwise couple of 2.0 kN⋅m2.0\,\mathrm{kN\cdot m} acts on the beam. Find the fixed-support reactions.
  1. Resolve the applied force
    At 150∘150^\circ, the force points left and upward, so its horizontal component is negative and its vertical component is positive.
    Fx=4cos⁡150∘=−3.464 kN,Fy=4sin⁡150∘=2.00 kNF_x=4\cos150^\circ=-3.464\,\mathrm{kN},\quad F_y=4\sin150^\circ=2.00\,\mathrm{kN}
  2. Balance horizontal and vertical forces
    The fixed support supplies force components that cancel the applied force components in each direction.
    Ax=3.464 kN,Ay=3.00 kNA_x=3.464\,\mathrm{kN},\quad A_y=3.00\,\mathrm{kN}
  3. Balance moments about A
    The downward force contributes a clockwise moment, while the upward component at 2.0 m2.0\,\mathrm{m} contributes a counterclockwise moment. Include the applied clockwise couple directly.
    MA−5.0(1.0)+2.0(2.0)−2.0=0,MA=3.0 kN⋅mM_A-5.0(1.0)+2.0(2.0)-2.0=0,\quad M_A=3.0\,\mathrm{kN\cdot m}
  4. Check force and moment balance
    The support forces cancel the applied force components. The support couple balances the net applied moment.
    ∑Fx=3.464−3.464=0,∑Fy=3−5+2=0,∑MA=3−5+4−2=0\sum F_x=3.464-3.464=0,\quad \sum F_y=3-5+2=0,\quad \sum M_A=3-5+4-2=0
Answer: The fixed support exerts 3.464 kN3.464\,\mathrm{kN} to the right, 3.00 kN3.00\,\mathrm{kN} upward, and a 3.00 kN⋅m3.00\,\mathrm{kN\cdot m} counterclockwise couple.
Check: The force sums and moment sum are each zero.

Worked example

Simply supported beam with a triangular distributed load

A 5.0 m5.0\,\mathrm{m} horizontal beam has a pin at AA and a roller at BB. A downward triangular distributed load increases linearly from zero at AA to 4.0 kN/m4.0\,\mathrm{kN/m} at BB. A separate 3.0 kN3.0\,\mathrm{kN} downward force acts 2.0 m2.0\,\mathrm{m} from AA. Find the support reactions.
  1. Replace the distributed load by a resultant
    The resultant magnitude is the area under the load-intensity diagram. For a triangular load that grows from zero at AA, its line of action is two-thirds of the beam length from AA.
    R=12(5.0)(4.0)=10.0 kN,xR=23(5.0)=3.333 mR=\tfrac12(5.0)(4.0)=10.0\,\mathrm{kN},\quad x_R=\tfrac23(5.0)=3.333\,\mathrm{m}
  2. Take moments about A
    The pin reactions create no moment about AA. Both downward loads create clockwise moments, balanced by the roller reaction.
    5By−10(3.333)−3(2.0)=0,By=7.867 kN5B_y-10(3.333)-3(2.0)=0,\quad B_y=7.867\,\mathrm{kN}
  3. Apply force balance
    There are no horizontal applied loads, so the pin's horizontal reaction is zero. The upward reactions must sum to the combined downward load.
    Ax=0,Ay+By−10−3=0,Ay=5.133 kNA_x=0,\quad A_y+B_y-10-3=0,\quad A_y=5.133\,\mathrm{kN}
  4. Verify the reactions
    The reactions sum to the total downward load. Their moments about AA also cancel the moments of both downward loads.
    ∑Fy=5.133+7.867−10−3=0,∑MA=7.867(5)−10(3.333)−3(2)≈0 kN⋅m\sum F_y=5.133+7.867-10-3=0,\quad \sum M_A=7.867(5)-10(3.333)-3(2)\approx0\,\mathrm{kN\cdot m}
Answer: The pin reaction is 5.133 kN5.133\,\mathrm{kN} upward with zero horizontal component; the roller reaction is 7.867 kN7.867\,\mathrm{kN} upward.
Check: The upward reactions total 13.0 kN13.0\,\mathrm{kN}, matching the total downward load. Their moment balance about AA is satisfied within rounding.

Common mistakes and how to avoid them

Leaving out a support reaction or assigning an ideal roller extra reaction components.
Correction: Use the stated support model to show only the reactions it can exert in the planar setup.
Calculating a force's moment from its magnitude without accounting for its direction and lever arm.
Correction: Use signed force components and their distances from the selected moment point.
Treating an applied couple like a force that needs a distance from the moment point.
Correction: Include a couple directly as a signed moment.
Assuming a negative reaction means the equilibrium equations are wrong.
Correction: A negative result means the actual reaction points opposite to its assumed direction.
Checking force balance but not moment balance.
Correction: Verify both force-component sums and a moment sum before accepting the solution.

Lesson summary

  • Define and isolate one rigid body, then draw a complete free-body diagram.
  • Choose axes and a sign convention before writing equations.
  • Apply the two planar force equations and one moment equation.
  • Report reaction directions and units, then verify force and moment balance.

Check your understanding

Question 1

A downward force acts to the right of a moment point. With counterclockwise positive, what is the sign of its moment?
  1. Positive
  2. Negative
  3. Zero
  4. It depends only on the force magnitude.
Show answer and explanation
Negative
The force tends to turn the body clockwise about the point, so its signed moment is negative.

Question 2

A horizontal reaction is calculated as −1.5 kN-1.5\,\mathrm{kN} when right was assumed positive. What does this mean?
  1. The reaction is 1.5 kN1.5\,\mathrm{kN} to the left.
  2. The reaction is 1.5 kN1.5\,\mathrm{kN} upward.
  3. The body cannot be in equilibrium.
  4. The reaction is zero after rounding.
Show answer and explanation
The reaction is 1.5 kN1.5\,\mathrm{kN} to the left.
The negative sign means the actual horizontal reaction is opposite to the assumed rightward direction.

Key terms

Rigid body
An idealized body whose shape and dimensions are treated as unchanged during an equilibrium analysis.
Free-body diagram
A diagram of an isolated body showing its external forces, couples, and support reactions.
Reaction
A force or couple exerted by a support on the isolated body.
Couple
A pure turning effect specified by a moment and direction, with no net force.
Resultant
A single force representing the combined force effect of a distributed load, applied along its correct line of action.

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