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3.2 · Model pin, roller, cable, link, and fixed-support reactions

Learn to model pin, roller, cable, link, and fixed-support reactions through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Planar Equilibrium

ENGG 130 Statics study topic 3.2

A support or connector restricts how a body can move, and that restriction appears on a free-body diagram as a reaction. The reaction model depends on the connection: a cable pulls along its length, a roller pushes normal to its surface, and a fixed support can exert both forces and a couple. Correctly modelling these actions is the first step in solving a statics problem. In this lesson, each isolated body is treated as rigid and in planar equilibrium. The examples use only reaction models and the force and moment balance equations needed to find them.

What you will learn

  • Explain which force and moment reactions are allowed at pins, rollers, cables, links, and fixed supports.
  • Draw a free-body diagram that replaces supports and connectors with appropriate unknown reactions.
  • Use planar equilibrium equations to solve for support reactions and check the result.

1. Isolate the body and choose reactions

Define the system before drawing forces. It might be a single beam, a bracket, or one component separated from its supports. Imagine removing the surrounding supports and replacing each connection with the force or moment it can exert on the isolated body. Include known applied loads as well as unknown reactions.
Use a consistent coordinate system, usually +x+x to the right and +y+y upward. A force may be represented by horizontal and vertical components. If you assume a reaction direction and later calculate a negative value, the actual direction is opposite to the one assumed. A moment is a turning effect; in this lesson, counterclockwise moments are positive.
A free-body diagram (FBD) is not a sketch of every part of the surrounding structure. It shows just the chosen body, the applied loads, the reactions acting on it, and useful dimensions. Label unknowns clearly. This makes it possible to write equilibrium equations for that body alone.
  • Choose and name the isolated body before drawing its FBD.
  • Show the forces and moments that the removed supports exert on that body.
  • State axes and a moment sign convention, then use them consistently.

2. Match the reaction model to the connection

An ideal planar pin prevents translation of the connected point in both the horizontal and vertical directions, but it permits rotation. Model its reaction with two components, AxA_x and AyA_y. Do not add a reaction couple at an ideal pin.
An ideal roller or smooth contact can exert a force perpendicular to the contact surface, but not along it. For a horizontal surface, the reaction is vertical. For an inclined surface, draw the reaction normal to that surface. Its sense is commonly assumed to push away from the surface.
A cable can pull but cannot push. Its tension acts along the cable, away from the isolated body at the attachment point. If its direction is known, resolve the tension into components using trigonometry. A link that is pin-connected at both ends and has no other force acting on it is a two-force member: its end forces act along the link’s axis. The link may be in tension or compression, so an assumed direction can be checked by the sign of the answer.
An ideal fixed support prevents both translation and rotation of the body at the support. In a planar FBD, model it with two force components and a reaction couple. The couple is a moment, not a force, and is measured in units such as N·m or kN·m. These ideal reaction models identify what a connection can exert; they do not by themselves determine the reaction values.
  • Pin: two force components; no reaction couple.
  • Roller or smooth contact: one force normal to the contact surface.
  • Cable: tension along the cable, pulling away from the body.
  • Two-force link: force along the member axis; either assumed sense is acceptable if interpreted correctly.
  • Fixed support: two force components and one reaction couple in planar statics.

3. Apply planar equilibrium

For a rigid body at rest in a plane, the sum of horizontal forces, the sum of vertical forces, and the sum of moments must each be zero. These equations govern the examples below. When writing a moment equation, choose a point and use signed perpendicular distances; forces whose lines of action pass through that point have zero moment about it.
Resolve an angled force into components before using the force equations. For a force of magnitude FF at angle θ\theta counterclockwise from the positive horizontal axis, its components are Fx=Fcos⁡θF_x=F\cos\theta and Fy=Fsin⁡θF_y=F\sin\theta. The angle and signs must match the FBD.
An equilibrium calculation should be checked in two ways: add the force components and confirm they balance, then add moments about a point and confirm they balance. A moment equation about a support can often remove that support’s unknown reactions from the first calculation, because their lines of action pass through the chosen point.
∑Fx=0,∑Fy=0,∑M=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M=0
  • Write all three planar equilibrium equations for a general rigid body.
  • Use a moment reference point that simplifies the unknowns where possible.
  • Check both force balance and moment balance after solving.

4. Read signs and units carefully

Reaction components are signed quantities relative to the directions drawn on the FBD. A negative answer does not mean equilibrium failed; it means the actual component acts opposite to the assumed arrow. Report the direction clearly in the final result.
Use force units consistently throughout an example and moment units that include distance. For example, multiplying a force in kN by a distance in m gives a moment in kN·m. Keep the support model visible while solving: adding an unsupported reaction, such as a pin couple, changes the model rather than improving the calculation.
  • A negative reaction component indicates the opposite of its assumed direction.
  • Moment units must combine force and distance.
  • The FBD, equations, and final directions must describe the same model.

Worked example

Pin and roller under two loads

A 4.0 m horizontal beam is supported by a pin at A and a roller at B. Downward loads of 12 kN and 8 kN act 1.5 m and 3.0 m from A, respectively. Find the support reactions.
  1. Isolate and model
    Isolate the beam. The pin at A supplies AxA_x and AyA_y; the roller on a horizontal surface supplies only the vertical reaction ByB_y. Assume these unknown arrows point right or upward.
  2. Take moments about A
    Counterclockwise is positive. The pin reactions have zero moment about A. The downward loads turn clockwise, while ByB_y turns counterclockwise.
    4.0By−(12)(1.5)−(8)(3.0)=04.0B_y-(12)(1.5)-(8)(3.0)=0
  3. Solve reactions
    The moment equation gives By=10.5 kNB_y=10.5\,\text{kN}. Vertical force balance then gives Ay=9.5 kNA_y=9.5\,\text{kN}. Since there are no horizontal applied loads, horizontal balance gives Ax=0A_x=0.
    By=10.5 kN,Ay=9.5 kN,Ax=0B_y=10.5\,\text{kN},\quad A_y=9.5\,\text{kN},\quad A_x=0
  4. Verify equilibrium
    The upward reactions total 20.0 kN20.0\,\text{kN}, matching the downward loads. About A, the reaction moment is 42 kN⋅m42\,\text{kN}\cdot\text{m} counterclockwise and the loads produce 42 kN⋅m42\,\text{kN}\cdot\text{m} clockwise.
    ∑Fy=9.5+10.5−12−8=0\sum F_y=9.5+10.5-12-8=0
Answer: The reactions are Ax=0A_x=0, Ay=9.5 kNA_y=9.5\,\text{kN} upward, and By=10.5 kNB_y=10.5\,\text{kN} upward.
Check: Moment check about A: 10.5(4.0)−12(1.5)−8(3.0)=0 kN⋅m10.5(4.0)-12(1.5)-8(3.0)=0\,\text{kN}\cdot\text{m}.

Worked example

Cable attached to a fixed-supported beam

A 3.0 m beam is fixed at A. A cable attached at its right end runs up and left, with direction components proportional to (−3,4)(-3,4). Downward loads of 4 kN and 2 kN act 1.0 m and 2.0 m from A. Find the cable tension and the fixed-support reactions.
  1. Resolve the cable force
    The direction ratios form a 3-4-5 triangle. Thus the cable components on the beam are −3T/5-3T/5 horizontally and 4T/54T/5 vertically. The fixed support contributes AxA_x, AyA_y, and MAM_A.
    Tx=−35T,Ty=45TT_x=-\frac{3}{5}T,\quad T_y=\frac{4}{5}T
  2. Find the tension from moments
    Take moments about A, counterclockwise positive. The cable’s vertical component acts 3.0 m from A; its horizontal component has zero moment arm about A because it lies along the beam. The downward loads create clockwise moments.
    3(45T)−4(1.0)−2(2.0)=03\left(\frac{4}{5}T\right)-4(1.0)-2(2.0)=0
  3. Solve the reactions
    The equation gives T=10/3 kNT=10/3\,\text{kN}. Horizontal balance gives Ax=2.0 kNA_x=2.0\,\text{kN} to the right. Vertical balance gives Ay=10/3 kNA_y=10/3\,\text{kN} upward. The fixed couple is zero for this particular loading because the cable tension balances the loads’ moment.
    T=103 kN,Ax=2.0 kN,Ay=103 kN,MA=0T=\frac{10}{3}\,\text{kN},\quad A_x=2.0\,\text{kN},\quad A_y=\frac{10}{3}\,\text{kN},\quad M_A=0
  4. Verify balance
    The cable’s horizontal component is 2.0 kN left, balanced by AxA_x. Upward force totals 10/3+8/3=6 kN10/3+8/3=6\,\text{kN}, matching the loads. The cable’s counterclockwise moment is 12 kN·m, matching the loads’ clockwise moment.
    ∑Fx=2−35(103)=0\sum F_x=2-\frac{3}{5}\left(\frac{10}{3}\right)=0
Answer: The cable tension is 10/3 kN10/3\,\text{kN}. The fixed-support reactions are Ax=2.0 kNA_x=2.0\,\text{kN} right, Ay=10/3 kNA_y=10/3\,\text{kN} up, and MA=0 kN⋅mM_A=0\,\text{kN}\cdot\text{m}.
Check: The vertical force balance is 10/3+8/3−4−2=0 kN10/3+8/3-4-2=0\,\text{kN}. The moment balance about A is 12−4−4=0 kN⋅m12-4-4=0\,\text{kN}\cdot\text{m}.

Worked example

Pin-connected link and beam

A horizontal beam has a pin at A and is connected at a point 2.0 m from A to a link that rises to the right at 60∘60^\circ above horizontal. A 6.0 kN downward load acts 3.0 m from A. Find the link force and pin reactions. The link has pins at both ends and no other force acting on it.
  1. Model the link
    Isolate the beam. Since the link has forces only at its two pinned ends, its force acts along the link. Assume it pushes the beam up and right with magnitude FF. The pin at A has components AxA_x and AyA_y.
    Fx=Fcos⁡60∘,Fy=Fsin⁡60∘F_x=F\cos60^\circ,\quad F_y=F\sin60^\circ
  2. Use moment balance
    Take moments about A. Only the link’s vertical component and the downward load create moments about A. The link force acts 2.0 m from A; the load acts 3.0 m from A.
    2.0Fsin⁡60∘−6.0(3.0)=02.0F\sin60^\circ-6.0(3.0)=0
  3. Find link and pin forces
    Solving gives F=10.4 kNF=10.4\,\text{kN}, rounded to three significant figures. Horizontal balance makes AxA_x oppose the link’s rightward component, while vertical balance gives a downward AyA_y because the link’s upward component exceeds the applied load.
    F=10.4 kN,Ax=−5.20 kN,Ay=−3.00 kNF=10.4\,\text{kN},\quad A_x=-5.20\,\text{kN},\quad A_y=-3.00\,\text{kN}
  4. Interpret and verify
    The negative pin components mean AxA_x acts left and AyA_y acts down. The link force is positive under the assumed compressive direction, so the link pushes along its axis as drawn. Component sums and the moment about A are zero, allowing for rounding.
    ∑Fx=−5.20+10.4cos⁡60∘=0\sum F_x=-5.20+10.4\cos60^\circ=0
Answer: The link pushes on the beam with approximately 10.4 kN10.4\,\text{kN} along the link. The pin reaction at A is 5.20 kN5.20\,\text{kN} left and 3.00 kN3.00\,\text{kN} down.
Check: Vertical balance: −3.00+10.4sin⁡60∘−6.0≈0 kN-3.00+10.4\sin60^\circ-6.0\approx0\,\text{kN}. Moment balance: 2.0(10.4sin⁡60∘)−18.0≈0 kN⋅m2.0(10.4\sin60^\circ)-18.0\approx0\,\text{kN}\cdot\text{m}.

Common mistakes and how to avoid them

Drawing a moment reaction at a pin or allowing a roller to exert a force along its surface.
Correction: Use the ideal connection model: a pin has two force components and no couple; a roller or smooth contact has one normal force.
Drawing a cable force toward the body or not along the cable.
Correction: A cable pulls along its own axis and away from the isolated body at its attachment.
Treating a link force as necessarily vertical or horizontal.
Correction: For a two-force member, the force acts along the member’s axis. Resolve it into components if needed.
Keeping a negative reaction pointed in its assumed direction.
Correction: A negative value means the actual component acts in the opposite direction; report that direction.
Checking only force balance.
Correction: Also check moment balance about a stated point, using consistent moment units and signs.

Lesson summary

  • Draw an FBD of the isolated body and replace each connection with its correct ideal reaction model.
  • Pins have two force components; rollers and smooth contacts have a normal force; cables pull along their length; two-force links act along their axis; fixed supports have two force components and a couple.
  • Use horizontal-force, vertical-force, and moment equilibrium, then verify both force and moment balance.
  • Interpret negative unknowns as directions opposite to the assumed arrows.

Check your understanding

Question 1

A beam rests on a smooth horizontal roller. Which reaction model is appropriate at that contact?
  1. A horizontal force and a vertical force
  2. A vertical force only
  3. A vertical force and a reaction couple
  4. A force along the roller surface
Show answer and explanation
A vertical force only
A smooth roller supplies one force normal to its contact surface. For a horizontal surface, that force is vertical.

Question 2

A pin-connected link has forces only at its two ends. In which direction does its force act?
  1. Along the link’s axis
  2. Perpendicular to the link’s axis
  3. Always vertically
  4. Always away from the connected beam
Show answer and explanation
Along the link’s axis
A member with only two end forces has those forces along the line joining its ends. The sign of a solved force identifies its actual sense.

Question 3

A calculated horizontal pin reaction is −2.5 kN-2.5\,\text{kN} when the FBD arrow was drawn rightward. What should be reported?
  1. A 2.5 kN reaction to the right
  2. A 2.5 kN reaction to the left
  3. No horizontal reaction
  4. A 2.5 kN counterclockwise couple
Show answer and explanation
A 2.5 kN reaction to the left
The negative sign indicates the reaction acts opposite to the assumed rightward direction, so it acts left.

Key terms

Reaction
A force or moment exerted on an isolated body by a support or connection.
Free-body diagram
A sketch of an isolated body showing the external forces, moments, and dimensions used in its equilibrium analysis.
Two-force member
A member with forces acting only at two ends; those forces act along the member’s axis.
Reaction couple
A moment exerted by a support, represented separately from the support’s force components.

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