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3.1 · Draw complete free-body diagrams

Learn to draw complete free-body diagrams through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Planar Equilibrium

ENGG 130 Engineering Mechanics: Statics — Study topic 3.1

A free-body diagram (FBD) is a drawing of one chosen body separated from its surroundings. The drawing shows the external forces and moments acting on that body, including forces exerted by supports or contacts that have been removed from the picture. It is not merely a sketch of the object: it is a bookkeeping tool for deciding what belongs in the equilibrium equations. A missing force makes the model incomplete; an invented force can make it wrong. This lesson focuses on identifying and representing those interactions clearly. The examples use equilibrium calculations only to show how a complete FBD supports a correct solution.

What you will learn

  • Define the body or system to isolate before drawing its free-body diagram.
  • Replace supports and contacts with the correct external forces and moments.
  • Show applied loads, dimensions, axes, and assumed directions clearly.
  • Use a complete free-body diagram to write and check planar equilibrium equations.

1. Isolate the body and identify every interaction

Start by stating exactly what you are isolating: a particle, a single member, or a larger connected body. Imagine removing everything else. Keep the isolated body’s shape simple, but include enough geometry to locate forces and show their directions. If you isolate a beam, for example, do not leave the wall or ground attached to it; replace their effects with forces or moments at the contact points.
Next, inspect every place where the surroundings touch or act on the isolated body. A pin support in a planar problem can exert horizontal and vertical force components, but it does not prevent rotation by exerting a couple. A roller on a horizontal surface exerts a force perpendicular to that surface, so its reaction is vertical. A cable pulls along its own length and away from the body. A smooth contact can push perpendicular to the contacting surface, but it cannot pull along that surface.
Also include loads applied directly to the body: known forces, applied couples, and distributed loads. A force can be replaced by its horizontal and vertical components if that makes the direction easier to use. Label known magnitudes and directions; for unknown reactions, use clear symbols and choose convenient assumed directions. If a calculated component is negative, its actual direction is opposite the arrow you assumed.
  • Isolate one clearly defined body or system.
  • Replace each removed support or contact with its force or moment effect.
  • Include all applied loads and show where they act.

2. Make the diagram complete and readable

Draw the outline of the isolated body and place forces at their points of application. A force arrow should point in its assumed direction, and its label should identify the force. For an angled force, show the angle or give its components. Keep the original geometry needed for moment calculations, such as distances between a support and a load.
Choose axes before writing component equations. In planar problems, a common choice is positive xx to the right and positive yy upward. State that positive moment is counterclockwise, or state another convention and use it consistently. A force’s moment depends on its perpendicular distance from the chosen point; a force whose line of action passes through that point has zero moment about it.
A complete FBD is separate from the original surroundings drawing. The surroundings drawing helps identify connections; the FBD shows the isolated body and every external action on it. Do not show both sides of an interaction as forces on the same isolated body. For instance, if a support is removed, show the force exerted by the support on the body, not an additional equal-and-opposite force exerted by the body on the support.
  • Show axes, force directions, application points, and useful dimensions.
  • Use a consistent positive direction for force components and moments.
  • Show only forces acting on the isolated body.

3. Use the FBD to state and check equilibrium

For a body at rest in planar statics, the resultant force and resultant moment must each be zero. Resolve angled forces into components when needed. If an angle is measured from the positive horizontal axis, the components are Fx=Fcos⁡θF_x=F\cos\theta and Fy=Fsin⁡θF_y=F\sin\theta. Use signs from the chosen axes rather than treating every component as positive.
After drawing the FBD, write the three planar equilibrium equations. Choose a moment centre that removes as many unknown forces as possible. Solve the equations, then substitute the results back into both force equations and the moment equation. These checks can reveal a missing load, an incorrect lever arm, or a sign error.
A diagram may be complete even before unknown reactions are solved: the key requirement is that every external action is represented. Calculating reactions is a useful test of whether the diagram and its signs are consistent. Keep force units and moment units distinct, such as newtons and newton-metres, or kilonewtons and kilonewton-metres.
∑Fx=0,∑Fy=0,∑MO=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M_O=0
  • Check horizontal force, vertical force, and moment balance.
  • Moments use a force times a perpendicular distance.
  • Use equilibrium calculations to test, not replace, a complete FBD.

Worked example

Pin-and-roller beam with a horizontal load

A 4.0 m horizontal beam is supported by a pin at A and a roller at B. A 6.0 kN downward force acts 1.0 m from A. A 2.0 kN horizontal force acts to the right, 3.0 m from A. Draw a complete FBD and determine the support reactions.
  1. Define and isolate
    Isolate the beam. The pin supplies two unknown force components, AxA_x and AyA_y. The roller on a horizontal surface supplies one vertical reaction, ByB_y. These are the only support actions needed in this planar model.
  2. Choose signs and write equilibrium
    Take right and upward as positive, and counterclockwise moments as positive. Take moments about A so the two pin reaction components have zero moment arm.
    ∑Fx=0,∑Fy=0,∑MA=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M_A=0
  3. Solve reactions
    The horizontal balance gives the pin reaction opposite the applied horizontal force. The moment balance determines the roller reaction; then the vertical balance gives the pin’s vertical component.
    Ax+2.0=0,Ay+By−6.0=0,4.0By−6.0(1.0)=0A_x+2.0=0,\quad A_y+B_y-6.0=0,\quad 4.0B_y-6.0(1.0)=0
  4. Substitute and verify
    The results are Ax=−2.0 kNA_x=-2.0\,\mathrm{kN}, By=1.5 kNB_y=1.5\,\mathrm{kN}, and Ay=4.5 kNA_y=4.5\,\mathrm{kN}. The negative sign means the actual horizontal pin reaction points left. Check force and moment balance using the indicated units.
    ∑Fx=−2.0+2.0=0,∑Fy=4.5+1.5−6.0=0,∑MA=4.0(1.5)−6.0(1.0)=0\sum F_x=-2.0+2.0=0,\quad \sum F_y=4.5+1.5-6.0=0,\quad \sum M_A=4.0(1.5)-6.0(1.0)=0
Answer: The complete FBD has reactions Ax=2.0 kNA_x=2.0\,\mathrm{kN} left, Ay=4.5 kNA_y=4.5\,\mathrm{kN} upward, and By=1.5 kNB_y=1.5\,\mathrm{kN} upward, along with both applied forces.
Check: The horizontal and vertical force sums are zero, and the moment sum about A is zero. Force values are in kN and moments are in kN·m.

Worked example

Particle held by a cable and horizontal tie

A small ring is held in equilibrium by a horizontal tie and a cable that rises to the left at 30° above the horizontal. A 100 N weight acts downward on the ring. Draw the particle FBD and find the two cable/tie forces.
  1. Define the body and draw forces
    Isolate the ring as a particle, so the forces are represented as acting at one point. The cable pulls along its direction, the tie pulls horizontally, and the weight acts downward. Let TT be the cable tension and HH the tie force.
  2. Resolve forces
    Use positive right and positive up. The cable direction is 150° measured counterclockwise from positive xx, so its horizontal component is leftward and its vertical component is upward.
    ∑Fx=H−Tcos⁡30∘=0,∑Fy=Tsin⁡30∘−100=0\sum F_x=H-T\cos 30^\circ=0,\quad \sum F_y=T\sin 30^\circ-100=0
  3. Solve and check
    The vertical equation gives T=200 NT=200\,\mathrm{N}. Substitution into the horizontal equation gives H=173.2 NH=173.2\,\mathrm{N} to the right. The force components cancel; for a particle, the concurrent forces create no net moment about the ring.
    T=100sin⁡30∘=200 N,H=200cos⁡30∘=173.2 NT=\frac{100}{\sin 30^\circ}=200\,\mathrm{N},\quad H=200\cos 30^\circ=173.2\,\mathrm{N}
Answer: The cable tension is 200 N200\,\mathrm{N}, and the horizontal tie force is 173.2 N173.2\,\mathrm{N} to the right.
Check: Vertically, 200sin⁡30∘−100=0 N200\sin30^\circ-100=0\,\mathrm{N}. Horizontally, 173.2−200cos⁡30∘≈0 N173.2-200\cos30^\circ\approx0\,\mathrm{N}.

Worked example

Beam under a uniform distributed load

A 5.0 m beam is supported by a pin at A and a roller at B. A uniform downward load of 2.0 kN/m covers the first 3.0 m from A. Draw a complete FBD using an equivalent resultant for the distributed load, then find the reactions.
  1. Replace the distributed load
    The isolated body is the whole beam. The uniform load has total magnitude equal to its intensity times its loaded length. Because it is uniform, its single equivalent downward force acts at the midpoint of the 3.0 m loaded region, or 1.5 m from A. Do not include both the distributed load and its equivalent resultant on this FBD.
    W=(2.0 kN/m)(3.0 m)=6.0 kNW=(2.0\,\mathrm{kN/m})(3.0\,\mathrm{m})=6.0\,\mathrm{kN}
  2. Write equilibrium equations
    Take upward force and counterclockwise moment as positive. The pin has vertical reaction AyA_y and horizontal reaction AxA_x; no horizontal load acts, so the horizontal balance will set AxA_x to zero.
    Ax=0,Ay+By−6.0=0,5.0By−6.0(1.5)=0A_x=0,\quad A_y+B_y-6.0=0,\quad 5.0B_y-6.0(1.5)=0
  3. Solve and verify
    Taking moments about A gives By=1.8 kNB_y=1.8\,\mathrm{kN}. Vertical force balance then gives Ay=4.2 kNA_y=4.2\,\mathrm{kN}. Substitution confirms that both force components and the moment sum vanish.
    By=1.8 kN,Ay=4.2 kN,∑MA=5.0(1.8)−6.0(1.5)=0B_y=1.8\,\mathrm{kN},\quad A_y=4.2\,\mathrm{kN},\quad \sum M_A=5.0(1.8)-6.0(1.5)=0
Answer: The complete equivalent-load FBD has Ax=0A_x=0, Ay=4.2 kNA_y=4.2\,\mathrm{kN} upward, By=1.8 kNB_y=1.8\,\mathrm{kN} upward, and the 6.0 kN downward resultant 1.5 m from A.
Check: The vertical balance is 4.2+1.8−6.0=0 kN4.2+1.8-6.0=0\,\mathrm{kN}. The horizontal balance is zero, and the moment balance is 5.0(1.8)−6.0(1.5)=0 kN⋅m5.0(1.8)-6.0(1.5)=0\,\mathrm{kN\cdot m}.

Common mistakes and how to avoid them

Leaving a support on the isolated-body sketch without showing its reaction.
Correction: Remove the surroundings and replace each support with its appropriate force components or moment.
Giving a roller two reaction components, or a cable a force perpendicular to itself.
Correction: Use the physical constraint: a roller reaction is normal to its surface, while a cable pulls along its length.
Adding a distributed load and its equivalent resultant to the same equilibrium FBD.
Correction: Use either the original distribution or its equivalent single resultant, not both.
Treating a negative reaction as proof that the equations failed.
Correction: A negative result means the actual force acts opposite to the direction assumed in the diagram.
Forgetting dimensions or using a slanted distance instead of a perpendicular moment arm.
Correction: Show the geometry needed for moments and use the shortest perpendicular distance from the moment centre to the force line of action.

Lesson summary

  • Define and isolate one body before listing forces.
  • Show every applied load and every force or moment exerted by a removed support or contact.
  • Choose axes and moment sign conventions, then label force directions and useful distances.
  • Use force and moment equilibrium to test whether the FBD and calculated reactions are consistent.

Check your understanding

Question 1

A beam rests on a roller on a level floor. Which reaction direction belongs on its FBD?
  1. Horizontal only
  2. Vertical only, normal to the floor
  3. Both horizontal and vertical
  4. A clockwise couple only
Show answer and explanation
Vertical only, normal to the floor
A roller on a horizontal surface exerts a force perpendicular to that surface, so the reaction is vertical.

Question 2

A uniform load of 4.0 kN/m covers 2.0 m of a beam. What is the magnitude of its equivalent resultant?
  1. 2.0 kN
  2. 4.0 kN
  3. 6.0 kN
  4. 8.0 kN
Show answer and explanation
8.0 kN
The resultant magnitude is load intensity times loaded length: (4.0 kN/m)(2.0 m)=8.0 kN(4.0\,\mathrm{kN/m})(2.0\,\mathrm{m})=8.0\,\mathrm{kN}.

Question 3

You assume a pin’s horizontal reaction points right, and calculation gives −3.0 kN-3.0\,\mathrm{kN}. What does this mean?
  1. The actual reaction is 3.0 kN to the left
  2. The actual reaction is 3.0 kN to the right
  3. The pin exerts no horizontal force
  4. The FBD must show a 3.0 kN couple
Show answer and explanation
The actual reaction is 3.0 kN to the left
The negative sign indicates that the actual force direction is opposite the assumed rightward direction.

Key terms

Free-body diagram
A drawing of an isolated body showing all external forces and moments acting on it.
Reaction
A force or moment exerted on the isolated body by a support or contact.
Resultant
A single force that has the same overall force and moment effect as a distributed load on the chosen body.
Moment arm
The perpendicular distance from a chosen point to a force’s line of action.

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