01
Mechanical Advantage
Understand how machines allow a smaller effort to produce a larger useful output force.
The Geometry of Power – Advanced Simple Machines
Understand mechanical advantage, wheel and axle systems, tension, equilibrium and two-mass pulley systems through clear explanations, derivations, worked examples and interactive diagrams.
02
Wheel and Axle
Learn how the ratio of wheel radius to axle radius determines ideal mechanical advantage.
03
Tension
Understand the pulling force in a stretched string, rope or cable and its direction.
04
Equilibrium
Connect balanced forces and zero acceleration with tension and weight.
05
Two-Mass Pulley System
Derive acceleration and tension when unequal masses are connected by a rope over a pulley.
06
Problem-Solving Strategy
Identify forces, choose a direction, write Newton's second-law equations and solve systematically.
Chapter Roadmap
This chapter moves from the idea of machines multiplying force to the mathematics of connected masses.
1
Mechanical Advantage
First understand what a machine changes and what ideal MA tells us.
First understand what a machine changes and what ideal MA tells us.
2
Wheel and Axle
Use geometry of radii to understand force multiplication.
Use geometry of radii to understand force multiplication.
3
Tension
Learn the force transmitted through a stretched rope.
Learn the force transmitted through a stretched rope.
4
Pulley System
Apply Newton's second law to two connected masses.
Apply Newton's second law to two connected masses.
1. Mechanical Advantage
A machine does not create energy. Instead, it can make a task easier by changing the size or direction of the force that we apply.
Mechanical Advantage (MA) is the ratio of the load or useful
output force to the effort or input force.
$$MA=\frac{\text{Load}}{\text{Effort}}=\frac{L}{E}$$
If \(MA>1\), the machine provides force multiplication: the load is larger than the effort.
Important: Mechanical advantage is not the same as efficiency. MA
compares forces, whereas efficiency compares useful output work with input work.
Example
A machine lifts a \(1200\,N\) load using an effort of \(120\,N\).
$$MA=\frac{1200}{120}=\boxed{10}$$
The idealized force advantage is 10.
Quick Check — Mechanical Advantage
- Write the formula for mechanical advantage.
- What does \(MA=1\) mean in terms of load and effort?
- A machine lifts \(600\,N\) using \(100\,N\) effort. Find its MA.
2. Wheel and Axle
A wheel and axle consists of a large wheel fixed to a smaller axle so that both rotate together. The difference in their radii allows the applied effort to produce a larger turning effect at the axle.
Wheel and axle: a simple machine made of a large wheel and a
smaller axle rigidly connected so they rotate together.
Figure 4.1: Wheel and axle machine: Applying effort at larger radius R multiplies force to lift heavy load on axle r.
Why does the wheel need to be larger?
For the same turning effect, applying a force farther from the axis gives a larger torque. The large wheel therefore provides a longer distance from the axis than the small axle.
$$MA=\frac{R_{\text{wheel}}}{R_{\text{axle}}}$$
Derivation from torque balance
For an ideal wheel-and-axle system, input torque equals output torque:
$$E\times R=L\times r$$
Rearranging:
$$\frac{L}{E}=\frac{R}{r}$$
Since \(MA=L/E\),
$$\boxed{MA=\frac{R}{r}}$$
Worked Example
A steering wheel has radius \(30\,cm\). Its axle has radius \(3\,cm\). The resistance force at the axle is \(1200\,N\). Find MA and the effort at the wheel rim.
$$MA=\frac{30}{3}=10$$
$$E=\frac{L}{MA}=\frac{1200}{10}=\boxed{120\,N}$$
Real machines: The ideal calculation ignores friction. In an
actual machine, some input work is lost, so efficiency is less than 100%.
3D Interactive Simulation: Wheel & Axle Machine ($\text{MA} = R/r$)
Three.js 3D Engine
Drag to orbit 3D machine
Wheel Radius: R = 30 cm | Axle Radius: r = 6 cm | Load: L = 600 N
Mechanical Advantage: MA = R/r = 5.0 | Required Effort: E = L/MA = 120 N (Force Multiplied 5.0×)
Mechanical Advantage: MA = R/r = 5.0 | Required Effort: E = L/MA = 120 N (Force Multiplied 5.0×)
Wheel Radius (R):
Axle Radius (r):
Quick Check — Wheel and Axle
- Why is the wheel radius usually greater than the axle radius?
- Write the ideal MA formula for a wheel and axle.
- A wheel has radius \(40\,cm\) and axle radius \(5\,cm\). Find MA.
- If the load is \(800\,N\) for that machine, find the ideal effort.
3. Tension
When a rope, thread or cable is stretched and connected to an object, it can exert a pulling force on that object.
Tension is the pulling force transmitted through a stretched
string, rope, thread or cable. It acts along the length of the string and pulls the attached object.
$$\boxed{\text{SI unit of tension}=N}$$
Direction rule: Tension always acts along the string and away
from the object being considered.
Tension in a stationary hanging mass
Consider a mass \(m\) hanging at rest from a vertical rope.
Because the mass is stationary, its acceleration is zero. Newton's second law gives:
$$T-mg=0$$
$$\boxed{T=mg}$$
Worked Example
A \(5\,kg\) object hangs stationary from a rope. Taking \(g=9.8\,m/s^2\):
$$T=mg=5(9.8)=\boxed{49\,N}$$
Quick Check — Tension
- What is tension?
- In which direction does tension act?
- A \(6\,kg\) mass hangs at rest. Find the tension using \(g=9.8\,m/s^2\).
4. Tension When a Mass Accelerates Upward
Suppose a mass \(m\) is being lifted upward with acceleration \(a\). The upward tension must now be greater than the downward weight.
$$T-mg=ma$$
Therefore:
$$\boxed{T=m(g+a)}$$
Worked Example
A \(4\,kg\) mass is lifted upward with acceleration \(2\,m/s^2\). Take \(g=9.8\,m/s^2\).
$$T=4(9.8+2)=4(11.8)=\boxed{47.2\,N}$$
Compare carefully: stationary mass → \(T=mg\); upward
acceleration → \(T=m(g+a)\). Tension is larger than weight when the mass accelerates upward.
Quick Check — Accelerating Mass
- Why is tension greater than weight when a mass accelerates upward?
- Write the formula for upward acceleration.
- Find the tension in a \(3\,kg\) mass accelerating upward at \(2\,m/s^2\), taking \(g=9.8\,m/s^2\).
5. Two-Mass Pulley System
When two unequal masses are connected by a light string over an ideal pulley, the heavier mass moves downward and the lighter mass moves upward.
Step 1 — Draw forces on the heavier mass
Let \(m_1>m_2\), so \(m_1\) moves downward. On \(m_1\), weight acts downward and tension acts upward.
$$m_1g-T=m_1a$$
Step 2 — Draw forces on the lighter mass
The lighter mass \(m_2\) moves upward. Tension acts upward and weight acts downward.
$$T-m_2g=m_2a$$
Step 3 — Add the equations
$$m_1g-T+T-m_2g=m_1a+m_2a$$
$$(m_1-m_2)g=(m_1+m_2)a$$
Step 4 — Solve for acceleration
$$\boxed{a=\frac{(m_1-m_2)g}{m_1+m_2}}$$
Step 5 — Find tension
Use either mass equation. For the heavier mass:
$$T=m_1(g-a)$$
or for the lighter mass:
$$T=m_2(g+a)$$
Worked Example
Two masses \(6\,kg\) and \(2\,kg\) are connected over an ideal pulley. Find the acceleration and tension. Take \(g=9.8\,m/s^2\).
$$a=\frac{(6-2)9.8}{6+2}=\frac{39.2}{8}=\boxed{4.9\,m/s^2}$$
$$T=2(9.8+4.9)=\boxed{29.4\,N}$$
One system, one acceleration: If the string is taut and the
masses remain connected, both masses have the same magnitude of acceleration, but in opposite
directions.
3D Interactive Simulation: Two-Mass Pulley System (Atwood Machine)
Three.js 3D Engine
Drag to orbit 3D pulley
Mass 1 (Left): m₁ = 6.0 kg | Mass 2 (Right): m₂ = 2.0 kg
System Acceleration: a = (m₁−m₂)g/(m₁+m₂) = 4.90 m/s² | String Tension: T = 29.40 N
System Acceleration: a = (m₁−m₂)g/(m₁+m₂) = 4.90 m/s² | String Tension: T = 29.40 N
Mass 1 (Left, m₁):
Mass 2 (Right, m₂):
Quick Check — Two-Mass Pulley
- Which mass moves downward when two unequal masses are connected?
- Why is the acceleration the same in magnitude for both masses?
- Write the two Newton's second-law equations for \(m_1>m_2\).
- Find the acceleration for \(m_1=5\,kg,\;m_2=3\,kg,\;g=9.8\,m/s^2\).
6. How to Solve Pulley Problems Reliably
| Step | What to do |
|---|---|
| 1 | Identify the heavier mass and predict the direction of motion. |
| 2 | Draw the forces acting on each mass. |
| 3 | Choose positive direction separately for each mass. |
| 4 | Apply Newton's second law, \(F_{\text{net}}=ma\). |
| 5 | Add equations to eliminate tension when finding acceleration. |
| 6 | Substitute acceleration into either equation to find tension. |
| 7 | Check units and whether the result makes physical sense. |
Quick Check — Strategy
- What should you identify before writing the equations?
- Why is adding the two equations useful?
- What should you check after obtaining the numerical answer?
Exercise 4.1 — The Geometry of Power
Use \(g=9.8\,m/s^2\) unless another value is specified.
1. Define mechanical advantage and write its formula.
2. A machine lifts a \(600\,N\) load using an effort of
\(100\,N\). Find its mechanical advantage.
Solution: \(MA=L/E=600/100=\boxed{6}\).
3. Explain why a wheel-and-axle system has a larger wheel and a
smaller axle.
Solution: The larger radius provides a greater turning effect
for the applied effort, allowing force multiplication at the smaller axle.
4. A wheel has radius \(40\,cm\) and axle radius \(5\,cm\).
Find its ideal mechanical advantage.
Solution: \(MA=R/r=40/5=\boxed{8}\).
5. An ideal wheel-and-axle machine has MA \(=8\) and lifts a
\(800\,N\) load. Find the effort.
Solution: \(E=L/MA=800/8=\boxed{100\,N}\).
6. A \(5\,kg\) object is suspended at rest from a rope. Find
the tension.
Solution: \(T=mg=5(9.8)=\boxed{49\,N}\).
7. A \(4\,kg\) mass is lifted upward with acceleration
\(2\,m/s^2\). Find the tension.
Solution: \(T=m(g+a)=4(9.8+2)=\boxed{47.2\,N}\).
8. Two masses \(6\,kg\) and \(2\,kg\) are connected over an
ideal pulley. Find acceleration and tension.
Solution:
\(a=(6-2)9.8/(6+2)=\boxed{4.9\,m/s^2}\).
\(T=2(9.8+4.9)=\boxed{29.4\,N}\).
\(a=(6-2)9.8/(6+2)=\boxed{4.9\,m/s^2}\).
\(T=2(9.8+4.9)=\boxed{29.4\,N}\).
9. Two masses \(2\,kg\) and \(6\,kg\) are connected over a
frictionless pulley. Find the acceleration and tension.
Solution:
\(a=(6-2)9.8/(6+2)=\boxed{4.9\,m/s^2}\).
Using the \(2\,kg\) mass: \(T=2(9.8+4.9)=\boxed{29.4\,N}\).
\(a=(6-2)9.8/(6+2)=\boxed{4.9\,m/s^2}\).
Using the \(2\,kg\) mass: \(T=2(9.8+4.9)=\boxed{29.4\,N}\).
10. Show the directions of weight and tension for both objects
in a two-mass pulley system.
Solution: For the heavier downward-moving mass, weight is
downward and tension upward. For the lighter upward-moving mass, tension is upward and weight
downward.
11. Explain why equal masses on an ideal pulley do not
accelerate.
Solution: Their weights are equal, so the driving difference in
force is zero. Hence \(a=0\) for the ideal system.
12. A system has \(m_1>m_2\). Derive the acceleration formula
for the two-mass pulley system.
Solution:
\(m_1g-T=m_1a\), \(T-m_2g=m_2a\). Adding gives \((m_1-m_2)g=(m_1+m_2)a\). Therefore \[ \boxed{a=\frac{(m_1-m_2)g}{m_1+m_2}}. \]
\(m_1g-T=m_1a\), \(T-m_2g=m_2a\). Adding gives \((m_1-m_2)g=(m_1+m_2)a\). Therefore \[ \boxed{a=\frac{(m_1-m_2)g}{m_1+m_2}}. \]
13. Why is tension not always equal to \(mg\)?
Solution: \(T=mg\) only for a stationary mass or zero
acceleration in the simple vertical case. With acceleration, Newton's second law changes the
relation; for upward acceleration \(T=m(g+a)\).
14. A machine has ideal MA \(=10\). If the load is \(1200\,N\),
calculate the effort.
Solution: \(E=L/MA=1200/10=\boxed{120\,N}\).
Worksheets
Printable worksheets for mechanical advantage, wheel and axle, tension and two-mass pulley systems will be added here.
Coming SoonChapter 4 — Quick Revision
| Concept | Remember |
|---|---|
| Mechanical Advantage | \(MA=L/E\) |
| Wheel and Axle | \(MA=R/r\) for an ideal system. |
| Efficiency | \(\eta=(\text{useful output work}/\text{input work})\times100\%\). |
| Tension | Pulling force transmitted through a stretched string; acts along the string. |
| Stationary Mass | \(T=mg\). |
| Upward Acceleration | \(T=m(g+a)\). |
| Two-Mass Pulley | \(a=(m_1-m_2)g/(m_1+m_2)\) when \(m_1>m_2\). |
| Heavier Mass | Moves downward in the ideal unequal-mass system. |
| Connected Masses | Same magnitude of acceleration if connected by a taut inextensible string. |
Chapter Test
A full chapter test covering mechanical advantage, wheel and axle, tension, equilibrium and pulley-system numericals will be added here.
Coming Soon