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3.3 · Solve planar particle-equilibrium problems

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

University of Alberta ENGG 130: Engineering Mechanics: Statics

Planar Equilibrium

Free-body diagrams, force components, and equilibrium in two dimensions

A particle-equilibrium problem asks which forces keep a small body or connection point in balance. A particle is a model: its size and shape do not matter for the calculation, so each force is treated as acting at one point. In planar statics, the forces lie in one plane. The central rule is that their vector sum must be zero. A reliable solution starts by isolating the particle, drawing every force on its free-body diagram, and applying the rule one component at a time. This lesson develops that method with original examples and consistent SI units.

What you will learn

  • Isolate a particle and draw a labelled free-body diagram showing every force acting on it.
  • Resolve forces into horizontal and vertical components using a consistent axis and sign convention.
  • Apply the planar particle-equilibrium equations to find unknown force magnitudes.
  • Verify force balance and explain why moments about the particle point provide no additional equation.

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

First name the system: the single particle or connection point whose equilibrium you are solving. Mentally separate it from its surroundings. A free-body diagram (FBD) shows the forces exerted on that isolated particle by cables, contacts, or other connected bodies. It shows forces on the chosen system, not a picture of the surrounding structure in place of those forces.
Represent the particle as a point and draw each force as an arrow starting there. Label each arrow with its known magnitude or an unknown symbol, and show its direction or angle. A cable pulls along its own direction, away from the particle. If a force direction is not known, assume one; a negative result means the actual direction is opposite to the assumed one.
Before calculating, check that every force on the isolated particle appears in the diagram. The number of unknowns also matters: planar particle equilibrium supplies two independent component equations. The diagram is where you identify directions and angle references before writing those equations.
  • Choose one particle as the system and include every force acting on it.
  • A cable pulls along its length and away from the particle.
  • Label force directions and angle references before resolving components.

2. Choose axes and resolve each force

Choose horizontal xx and vertical yy axes through the particle. In this lesson, right and up are positive. For a force of magnitude FF at angle θ\theta measured counterclockwise from positive xx, the horizontal component is Fcos⁡θF\cos\theta and the vertical component is Fsin⁡θF\sin\theta. The signs come from the arrow's direction: a leftward or downward component is negative.
If an angle is measured from the vertical instead, identify which side of the right triangle is adjacent to that angle before choosing sine or cosine. The component along the reference direction uses cosine; the perpendicular component uses sine. Resolve every force using the same axes, then add signed components.
A force is a vector, so balancing magnitudes alone is not enough. Two forces of equal magnitude cancel only if their directions are opposite. In general, the horizontal components must balance, and the vertical components must balance separately.
Fx=Fcos⁡θ,Fy=Fsin⁡θF_x=F\cos\theta,\quad F_y=F\sin\theta
  • State the positive axis directions and how angles are measured.
  • Keep leftward and downward components negative when right and up are positive.
  • Use consistent force units, such as newtons, throughout a calculation.

3. Apply equilibrium and check the answer

For a particle in planar equilibrium, the vector sum of all forces is zero. Separating that vector statement into the chosen axes gives two scalar equations. These are the equations used to find unknown force magnitudes. Keep exact trigonometric expressions as long as practical, and round numerical results after solving.
Because the particle model treats all forces as acting at one point, each force has zero moment arm about that point. Thus the total moment about the particle point is zero automatically; it does not provide another independent equation. Do not use a moment equation as a substitute for either force-component equation.
After solving, substitute the values into both component equations. Each sum should be zero apart from rounding. If a computed force is negative, report its magnitude and reverse its assumed direction. A check using rounded values should still leave only a small residual.
∑Fx=0,∑Fy=0,∑F=0\sum F_x=0,\quad \sum F_y=0,\quad \sum \mathbf{F}=\mathbf{0}
  • Use one equilibrium equation for each in-plane force component.
  • Moments about the particle point are zero because all forces act there.
  • Substitute the result into both force equations to verify balance.

Worked example

Two cables support a hanging connection

A small ring is held in equilibrium by two cables and a vertical load of 300 N300\,\mathrm{N}. The left cable points up and left at 30∘30^\circ above the horizontal, and the right cable points up and right at 45∘45^\circ. Find both cable tensions.
  1. Define and isolate
    The system is the ring, modelled as a particle. Its FBD includes both cable tensions and the downward load.
  2. Choose axes and write equilibrium
    Take right and up as positive. The left cable has a negative horizontal component, while both cable vertical components are positive.
    −TLcos⁡30∘+TRcos⁡45∘=0,TLsin⁡30∘+TRsin⁡45∘−300=0-T_L\cos30^\circ+T_R\cos45^\circ=0,\quad T_L\sin30^\circ+T_R\sin45^\circ-300=0
  3. Solve the equations
    The horizontal equation gives TR=TLcos⁡30∘/cos⁡45∘T_R=T_L\cos30^\circ/\cos45^\circ. Substitution into the vertical equation gives TL(sin⁡30∘+cos⁡30∘)=300 NT_L(\sin30^\circ+\cos30^\circ)=300\,\mathrm{N}. Solving gives the left tension, then the horizontal relation gives the right tension.
    TL=219.6 N,TR=269.3 NT_L=219.6\,\mathrm{N},\quad T_R=269.3\,\mathrm{N}
  4. Verify force and moment balance
    Using the unrounded values, the horizontal terms cancel because TRcos⁡45∘=TLcos⁡30∘T_R\cos45^\circ=T_L\cos30^\circ. The upward components sum to 300 N300\,\mathrm{N}. About the particle point, all moment arms are zero, so the moment sum is zero.
    ∑Fx≈0,∑Fy≈0,∑MO=0\sum F_x\approx0,\quad \sum F_y\approx0,\quad \sum M_O=0
Answer: The left cable tension is 219.6 N219.6\,\mathrm{N} and the right cable tension is 269.3 N269.3\,\mathrm{N}.
Check: Substitution gives opposing horizontal components and a total upward component of 300 N300\,\mathrm{N}, balancing the load.

Worked example

Two unknown forces balance an angled force and a load

A particle is acted on by a known 400 N400\,\mathrm{N} downward force, cable A directed at 120∘120^\circ from positive xx, and cable B directed at 20∘20^\circ from positive xx. Find the cable tensions.
  1. Define the system and axes
    Isolate the particle. Use right and up as positive; both cable angles are measured counterclockwise from positive xx.
  2. Write component equations
    Cable A has a negative horizontal component and a positive vertical component. Cable B has positive components in both directions.
    −Acos⁡60∘+Bcos⁡20∘=0,Asin⁡60∘+Bsin⁡20∘−400=0-A\cos60^\circ+B\cos20^\circ=0,\quad A\sin60^\circ+B\sin20^\circ-400=0
  3. Find both magnitudes
    From the horizontal equation, B=Acos⁡60∘/cos⁡20∘B=A\cos60^\circ/\cos20^\circ. Substitute this relation into the vertical equation, then use the horizontal relation to find BB.
    A=381.7 N,B=203.1 NA=381.7\,\mathrm{N},\quad B=203.1\,\mathrm{N}
  4. Check equilibrium
    The horizontal components are approximately −190.9 N-190.9\,\mathrm{N} and +190.9 N+190.9\,\mathrm{N}. The upward components total approximately 400 N400\,\mathrm{N}, balancing the downward force. The moment sum about the particle point is zero.
    ∑Fx≈0,∑Fy≈0,∑MO=0\sum F_x\approx0,\quad \sum F_y\approx0,\quad \sum M_O=0
Answer: Cable A has tension 381.7 N381.7\,\mathrm{N}, and cable B has tension 203.1 N203.1\,\mathrm{N}.
Check: Both tensions are positive, consistent with the assumed cable directions. Substitution confirms force balance within rounding.

Worked example

Use component balance to find two cable tensions

A particle is acted on by a 250 N250\,\mathrm{N} force to the right and a 200 N200\,\mathrm{N} force at 120∘120^\circ from positive xx. Two cables pull at 210∘210^\circ and 270∘270^\circ. Find their tensions.
  1. Isolate and resolve known forces
    The system is the particle. The 200 N200\,\mathrm{N} force contributes −100 N-100\,\mathrm{N} horizontally and approximately 173.2 N173.2\,\mathrm{N} vertically.
    200cos⁡120∘=−100 N,200sin⁡120∘=173.2 N200\cos120^\circ=-100\,\mathrm{N},\quad 200\sin120^\circ=173.2\,\mathrm{N}
  2. Write equilibrium equations
    The cable at 210∘210^\circ has negative horizontal and vertical components. Cable 2 acts vertically downward.
    250−100−T1cos⁡30∘=0,173.2−T1sin⁡30∘−T2=0250-100-T_1\cos30^\circ=0,\quad 173.2-T_1\sin30^\circ-T_2=0
  3. Solve for cable tensions
    The horizontal equation determines T1T_1. Substituting that value into the vertical equation determines T2T_2.
    T1=173.2 N,T2=86.6 NT_1=173.2\,\mathrm{N},\quad T_2=86.6\,\mathrm{N}
  4. Verify both balances
    Horizontally, 250−100−173.2cos⁡30∘250-100-173.2\cos30^\circ is approximately zero. Vertically, 173.2−173.2sin⁡30∘−86.6173.2-173.2\sin30^\circ-86.6 is approximately zero. All forces act at the particle, so their moment sum about it is zero.
    ∑Fx≈0,∑Fy≈0,∑MO=0\sum F_x\approx0,\quad \sum F_y\approx0,\quad \sum M_O=0
Answer: The cable tensions are T1=173.2 NT_1=173.2\,\mathrm{N} and T2=86.6 NT_2=86.6\,\mathrm{N}.
Check: Both force-component sums are zero within rounding, and the positive values agree with the assumed cable directions.

Common mistakes and how to avoid them

Leaving out a force because it is a load or appears to belong to the surroundings.
Correction: Include every force exerted on the isolated particle, including known loads and cable pulls.
Giving every force component a positive sign.
Correction: With right and up positive, components pointing left or down are negative.
Using sine for the component adjacent to the stated angle.
Correction: Identify the adjacent and opposite sides relative to the angle before choosing cosine or sine.
Treating particle equilibrium as requiring a separate useful moment equation.
Correction: Moments about the particle point are automatically zero. Find unknowns using the two force-component equations.
Rounding forces too early and reporting an unbalanced result.
Correction: Keep extra digits during calculation, then check both component sums before rounding the final answer.

Lesson summary

  • Define the particle system and draw an FBD containing every force acting on it.
  • Choose axes, state the angle convention, and resolve forces into signed components.
  • Apply ∑Fx=0\sum F_x=0 and ∑Fy=0\sum F_y=0 to solve for unknown force magnitudes.
  • Check both force sums after substitution; moments about the particle point are zero because the forces act through it.

Check your understanding

Question 1

A 50 N50\,\mathrm{N} force acts at 60∘60^\circ counterclockwise from positive xx. What is its vertical component?
  1. +43.3 N+43.3\,\mathrm{N}
  2. +25.0 N+25.0\,\mathrm{N}
  3. −43.3 N-43.3\,\mathrm{N}
  4. correctIndex":0,"explanation":"The vertical component is 50sin⁡60∘=43.3 N50\sin60^\circ=43.3\,\mathrm{N}. It is positive because the force points upward."}]}]}
Show answer and explanation
+43.3 N+43.3\,\mathrm{N}
The vertical component is 50sin⁡60∘=43.3 N50\sin60^\circ=43.3\,\mathrm{N}. It is positive because the force points upward.

Question 2

A particle has a 90 N90\,\mathrm{N} force to the right and a 90 N90\,\mathrm{N} force to the left. What is their resultant horizontal force?
  1. 0 N0\,\mathrm{N}
  2. 180 N180\,\mathrm{N} to the right
  3. 90 N90\,\mathrm{N} to the right
  4. correctIndex":0,"explanation":"The signed sum is +90−90=0 N+90-90=0\,\mathrm{N}, so the horizontal forces cancel."}]}]}
Show answer and explanation
0 N0\,\mathrm{N}
The signed sum is +90−90=0 N+90-90=0\,\mathrm{N}, so the horizontal forces cancel.

Key terms

Particle
A model of a body whose size and shape are ignored so all forces are treated as acting at one point.
Free-body diagram
A sketch of an isolated system showing all external forces acting on it.
Equilibrium
A condition in which the vector sum of the forces on the particle is zero.
Component
The part of a vector along a chosen axis; horizontal and vertical components are used for planar force balance.

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