DoAssignment.ca

5.4 · Analyze forces in a simple machine

Learn to analyze forces in a simple machine through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Frames and Machines

Use free-body diagrams and equilibrium to find effort, load, and support forces

A simple machine changes how an applied force is used. A lever can exchange a long effort arm for a larger force on a load; a pulley can share a load between rope segments; and a wheel-and-axle can trade a larger turning radius for a smaller one. In statics, the central task is to isolate a machine part and find the forces that keep it in equilibrium. The analysis uses familiar vector components, force balance, and moment balance. It does not require motion analysis. This lesson uses idealized machines: contacts are modelled as smooth where appropriate, and rope weight and friction are neglected only when stated. These assumptions make the force relationships clear; they should not be mistaken for a description of every real machine.

What you will learn

  • Identify the body to isolate when analyzing a simple machine.
  • Represent machine contacts and applied forces on a labelled free-body diagram.
  • Use planar force and moment equilibrium to solve for unknown forces.
  • Interpret an ideal machine’s force advantage while checking the assumptions behind it.

1. Set up the machine as a statics problem

First define the body being analyzed. It may be a lever, a pulley and its attached load, or a wheel-and-axle. Isolate that body from its surroundings and show every external force on it: applied effort, load or weight, rope tensions, and support reactions. Do not draw forces that the isolated body applies to something else.
Choose axes that make the force directions easy to resolve. For a planar problem, take xx horizontal to the right and yy vertical upward unless the geometry suggests another convenient choice. State the moment sign convention; here, counterclockwise moments are positive. A force component is found from its direction, for example Fx=Fcos⁡θF_x=F\cos\theta and Fy=Fsin⁡θF_y=F\sin\theta when θ\theta is measured counterclockwise from positive xx.
For a body at rest, the sum of external forces in each direction and the sum of external moments about any point must be zero. Choose a moment point that removes as many unknown reactions as possible. For a lever, the pivot is often convenient. A force whose line of action passes through the moment point creates no moment about that point.
A moment’s magnitude is force times perpendicular distance to its line of action. For a perpendicular force on a lever, this becomes force times lever-arm length. Preserve units: a force in newtons multiplied by a distance in metres gives a moment in newton-metres.
∑Fx=0,∑Fy=0,∑MO=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M_O=0
  • Isolate one machine part and include all forces acting on it.
  • Use force components when a force is not aligned with an axis.
  • A moment arm is a perpendicular distance, not necessarily the distance along the body.

2. Relate effort and load without skipping the free-body diagram

A lever’s force advantage follows from moment balance about its pivot. The effort and load may act on opposite sides or on the same side; their moment directions must be read from the diagram rather than memorized. If each force is perpendicular to the lever, the moment terms are simply force multiplied by distance from the pivot.
A rope in an ideal, light, continuous rope passing over smooth pulleys has the same tension throughout each uninterrupted segment. A movable load supported by two vertical segments therefore has two upward tension forces on the isolated load-and-pulley assembly. This is a force-counting result from its free-body diagram, not a rule to apply without checking how many segments actually support the moving assembly.
For a wheel-and-axle, effort and load act at different radii. If their forces are tangential, each makes a moment equal to its magnitude times its radius. The axle support can exert a force on the wheel, but if that force acts through the axle centre it creates no moment about that centre.
The ideal force advantage is a ratio of load force to effort force. It describes the force relation under the stated ideal assumptions; it does not replace equilibrium analysis. The support reactions still need to be found when the problem asks for them, and every result should be checked against both force and moment balance.
ideal force advantage=load forceeffort force\text{ideal force advantage}=\frac{\text{load force}}{\text{effort force}}
  • Use the actual perpendicular moment arms in lever and wheel-and-axle problems.
  • Count the supporting rope segments on the isolated moving assembly.
  • Ideal-machine force ratios depend on stated modelling assumptions.

3. Solve, interpret, and verify

Write the equilibrium equations before substituting numbers. Solve the moment equation for the effort or load when possible, then use the force equations to find reactions or confirm the remaining unknowns. This order is efficient because taking moments about a pin removes the pin reaction from that equation.
Keep assumed force directions consistent through the algebra. If an unknown solves as negative, its actual direction is opposite the arrow you assumed. This is a useful sign interpretation, not a reason to discard the calculation.
Finally, substitute the answers back into independent checks. The horizontal and vertical force sums should each be zero, and the moment sum about a convenient point should also be zero. Round only after the balance checks, so small rounding differences do not obscure an otherwise exact equilibrium.
∑F=0,∑MO=0\sum \mathbf{F}=\mathbf{0},\quad \sum M_O=0
  • Use moment balance to find the main machine force relation, then force balance for reactions.
  • A negative answer means the actual force direction is opposite the assumed arrow.
  • Check force balance and moment balance independently.

Worked example

Lever with a pivot between effort and load

A horizontal, light lever is held at a pin pivot. A 200 N load acts downward 0.20 m to the right of the pivot. An effort acts vertically downward 0.40 m to its left. Find the effort and the pin reaction. Assume the lever is in equilibrium.
  1. Choose the system and signs
    Isolate the lever. Let upward forces be positive and counterclockwise moments be positive. The pin can exert horizontal and vertical reactions; no horizontal loads act, so its horizontal reaction is zero.
    ∑Fx=0\sum F_x=0
  2. Balance moments about the pivot
    The pin reaction passes through the pivot and has zero moment arm. The downward effort on the left creates a counterclockwise moment, while the downward load on the right creates a clockwise moment.
    E(0.40 m)−(200 N)(0.20 m)=0E(0.40\,\mathrm{m})-(200\,\mathrm{N})(0.20\,\mathrm{m})=0
  3. Find effort and vertical reaction
    Solving the moment equation gives the effort. Then vertical force balance requires the pin to support the combined downward forces.
    E=100 N,Ay−E−200 N=0E=100\,\mathrm{N},\quad A_y-E-200\,\mathrm{N}=0
  4. Verify equilibrium
    The calculated pin force is upward. The vertical forces sum to zero, and the clockwise and counterclockwise moments about the pivot are equal.
    300−100−200=0 N,(100)(0.40)−(200)(0.20)=0 N⋅m300-100-200=0\,\mathrm{N},\quad (100)(0.40)-(200)(0.20)=0\,\mathrm{N\cdot m}
Answer: The required effort is 100 N downward, and the pin reaction is 300 N upward. The horizontal pin reaction is zero.
Check: The effort and load moments both have magnitude 40 N·m and opposite signs. The upward 300 N reaction balances the total 300 N downward force.

Worked example

Load supported by a movable pulley

A light movable pulley and its load have a total weight of 360 N. Two vertical segments of an ideal rope support the pulley assembly. Find the rope tension and the downward effort at the free end. Neglect rope weight and assume smooth contact.
  1. Isolate the moving assembly
    The pulley and load are treated as one body. The two rope segments each pull upward with tension TT. The assembly’s weight acts downward through its centre; its precise line of action does not affect vertical force balance.
    ∑Fy=0\sum F_y=0
  2. Solve vertical equilibrium
    Because the rope is ideal and continuous over a smooth pulley, both supporting segments have the same tension. Their upward forces balance the assembly’s weight.
    T+T−360 N=0T+T-360\,\mathrm{N}=0
  3. Relate effort to tension
    The free end of the rope must be pulled downward with a force equal in magnitude to the rope tension in this ideal model.
    T=180 N,E=TT=180\,\mathrm{N},\quad E=T
  4. Verify forces and moments
    The two upward tensions total 360 N, balancing the weight. For the symmetric assembly shown, the tension moments about its centre cancel; the weight has no moment about that point.
    180+180−360=0 N180+180-360=0\,\mathrm{N}
Answer: The rope tension is 180 N, so the required downward effort is 180 N.
Check: Vertical force balance is satisfied. About the assembly centre, equal tensions at equal offsets create opposite moments, so the net moment is zero.

Worked example

Tangential effort on a wheel-and-axle

A wheel-and-axle is held at its centre by a pin. A 60 N load acts tangentially downward at an axle radius of 0.08 m. An upward tangential effort acts at a wheel radius of 0.32 m on the same side of the axle centre as the load. Find the effort and the pin reaction. Treat the wheel-and-axle as light and neglect friction.
  1. Take moments about the axle centre
    The pin reaction acts through the centre and therefore has no moment about it. On the same side of the centre, the upward effort and downward load create opposing moments.
    E(0.32 m)−(60 N)(0.08 m)=0E(0.32\,\mathrm{m})-(60\,\mathrm{N})(0.08\,\mathrm{m})=0
  2. Find effort and support force
    Solve the moment equation for the effort. Then use vertical force balance to determine the pin reaction. No horizontal force is applied, so the horizontal pin reaction is zero.
    E=15 N,Oy+15 N−60 N=0E=15\,\mathrm{N},\quad O_y+15\,\mathrm{N}-60\,\mathrm{N}=0
  3. Verify the balances
    The pin reaction must act upward. Substitution shows equal and opposite moments about the centre and zero resultant vertical force.
    Oy=45 N,45+15−60=0 N,(15)(0.32)−(60)(0.08)=0 N⋅mO_y=45\,\mathrm{N},\quad 45+15-60=0\,\mathrm{N},\quad (15)(0.32)-(60)(0.08)=0\,\mathrm{N\cdot m}
Answer: The required effort is 15 N upward, and the pin reaction is 45 N upward. The horizontal pin reaction is zero.
Check: The effort and load each create a 4.8 N·m moment about the centre in opposite directions. All vertical forces sum to zero.

Common mistakes and how to avoid them

Using the distance from the body’s end instead of the perpendicular distance from the pivot or axle.
Correction: Measure each moment arm from the chosen moment point to the force’s line of action, perpendicular to that line.
Assuming a movable pulley always has a particular number of supporting rope segments.
Correction: Isolate the moving assembly and count the rope segments that pull directly on it.
Leaving out a pin reaction because it does not appear in the moment equation about the pin.
Correction: The pin reaction has zero moment about the pin, but it still belongs in the force balance.
Treating ideal force advantage as a substitute for equilibrium.
Correction: State the assumptions, draw the free-body diagram, and check both force and moment balance.

Lesson summary

  • Define and isolate the machine part whose forces are being found.
  • Draw the applied forces, loads, rope tensions, and support reactions with clear directions.
  • Apply planar force and moment equilibrium using consistent axes, signs, and units.
  • For ideal levers and wheel-and-axles, compare opposing moments; for a supported pulley assembly, count each supporting tension.
  • Verify the final forces and moments independently.

Check your understanding

Question 1

A perpendicular effort acts 0.30 m from a lever pivot and balances a 150 N load acting 0.10 m on the other side. What is the effort magnitude?
  1. 50 N
  2. 150 N
  3. 450 N
  4. 75 N
Show answer and explanation
50 N
Moment balance gives E(0.30 m)=(150 N)(0.10 m)E(0.30\,\mathrm{m})=(150\,\mathrm{N})(0.10\,\mathrm{m}), so E=50 NE=50\,\mathrm{N}.

Question 2

An ideal movable pulley’s assembly weighs 240 N and is supported by three vertical rope segments of equal tension. What is the tension in each segment?
  1. 80 N
  2. 120 N
  3. 240 N
  4. 720 N
Show answer and explanation
80 N
The three upward tensions balance the weight, so 3T=240 N3T=240\,\mathrm{N} and T=80 NT=80\,\mathrm{N}.

Key terms

Effort
The force applied to operate a simple machine.
Load
The force the machine acts against or supports.
Moment arm
The perpendicular distance from a chosen point to a force’s line of action.
Pin reaction
The force exerted by a pin support on an attached body; in a planar problem it may have horizontal and vertical components.
Ideal machine
A simplified model that neglects effects such as friction or rope weight when explicitly assumed.

Continue through ENGG 130

View the complete ENGG 130 University of Alberta ENGG 130: Engineering Mechanics: Statics curriculum and lessons

About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows University of Alberta ENGG 130: Engineering Mechanics: Statics, study topic 5.4. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question