Why machines make work easier but never reduce it: work, power, moments, levers, pulleys, ramps and gears, explained with diagrams, animations and worked examples.
Archimedes is said to have boasted that, with a long enough lever and a place to stand, he could move the Earth. The remark is attributed to him by later writers, and it captures a real idea: machines let a small force move a big load. But machines do not give you something for nothing. Every tool from a pair of scissors to a crane obeys the same rule: what you gain in force you lose in distance. To understand this, we need three linked ideas: work (energy transferred by a force), power (how fast work is done) and machines (devices that change the size or direction of a force).
In everyday speech, "work" means any effort, such as studying or carrying a heavy bag while standing still. In physics, work has a precise meaning: work is done when a force moves an object in the direction of the force. Work is a transfer of energy.
W = F × d (work done = force × distance moved in the direction of the force)
W in joules (J), F in newtons (N), d in metres (m). One joule is the work done when a force of 1 N moves an object 1 m. If the force is at an angle θ to the movement, W = F d cos θ: only the part of the force along the motion does work.
It follows that no work is done in these cases, even though you might feel tired:
(a) A person pushes a trolley with a constant force of 50 N through 8.0 m. (b) A student lifts a 10 kg box slowly from the floor to a shelf 2.0 m high. Find the work done in each case. (g = 9.8 N/kg)
(a) W = Fd = 50 × 8.0 = 400 J.
(b) The lifting force equals the weight: F = mg = 10 × 9.8 = 98 N. W = Fd = 98 × 2.0 = 196 J (this equals the gain in gravitational potential energy, mgh).
A child pulls a sledge with a rope that makes an angle of 30° with the horizontal ground. The tension is 100 N and the sledge moves 5.0 m. How much work does the rope do?
W = F d cos θ = 100 × 5.0 × cos 30° = 500 × 0.866 = 433 J.
If the rope were horizontal (θ = 0°), the work would be the full 500 J.
The area under a force–distance graph equals the work done. For a constant force the graph is a horizontal line and the area is a rectangle (F × d). For a force that changes, such as stretching a spring, the area can be found by splitting it into simple shapes.
Doing work transfers energy, so the two are measured in the same unit. When the resultant force on an object does work, its kinetic energy changes by the same amount: net work done = change in kinetic energy (the work–energy principle). Work done lifting an object at constant speed equals its gain in potential energy. Work done against friction becomes thermal energy.
A resultant force of 2000 N acts on a 1000 kg car, initially at rest, over a distance of 50 m. Find its final speed.
Step 1. Work done = Fd = 2000 × 50 = 100 000 J.
Step 2. This becomes kinetic energy: ½mv² = 100 000, so v² = 2 × 100 000 ÷ 1000 = 200.
Step 3. v = √200 ≈ 14 m/s (about 51 km/h).
Power is the rate at which work is done or energy is transferred. Two people can lift the same box to the same shelf (the same work), but the one who does it in half the time has twice the power.
P = W ÷ t or P = E ÷ t
P = F × v (for a constant force moving an object at constant speed v)
The unit is the watt (W): 1 W = 1 J/s. 1 kW = 1000 W; 1 MW = 10⁶ W.
The unit is named after the Scottish engineer James Watt, whose improvements to the steam engine in the later 18th century helped drive the Industrial Revolution. Watt also introduced horsepower to compare engines with working horses: 1 horsepower (mechanical) is about 746 W.
A crane lifts a 500 kg load through 12 m in 20 s at constant speed. What power does it deliver to the load?
Step 1. Work = mgh = 500 × 9.8 × 12 = 58 800 J.
Step 2. P = W ÷ t = 58 800 ÷ 20 = 2940 W ≈ 2.9 kW.
A car travels at a steady 25 m/s against total resistive forces of 800 N. What power must the engine deliver to the wheels?
At constant speed, the driving force equals the resistance: F = 800 N. P = Fv = 800 × 25 = 20 000 W = 20 kW.
Air resistance grows rapidly with speed, so at higher speeds the power needed rises much faster than in proportion to speed.
For a sense of scale, a person climbing stairs briefly develops a few hundred watts; well-trained cyclists can sustain roughly 300–400 W for an hour, while elite sprinters can exceed 1000 W for a few seconds. A typical household kettle is 2–3 kW.
A machine is a device that makes work easier by changing the size, direction or point of application of a force. Machines can:
The effort is the force you apply; the load is the force the machine overcomes. Three numbers describe a machine:
Mechanical advantage (MA) = load ÷ effort (a ratio of forces)
Velocity ratio (VR) = distance moved by effort ÷ distance moved by load
Efficiency = (MA ÷ VR) × 100% = (work out ÷ work in) × 100%
For an ideal machine with no friction, work out equals work in, so MA = VR. Real machines lose some work to friction, so their efficiency is always below 100% and MA is smaller than VR. No machine can output more work than it receives; that would break the law of conservation of energy.
A force can turn an object about a pivot (a fixed point). The turning effect is called the moment of the force:
moment = force × perpendicular distance from the pivot to the line of action of the force
The unit is the newton-metre (N·m). Pushing a door far from its hinges (a large distance) needs much less force than pushing near the hinges.
The principle of moments says that when an object is in equilibrium (balanced and not turning), the sum of the clockwise moments equals the sum of the anticlockwise moments about any pivot.
A child weighing 300 N sits 1.5 m from the pivot on one side of a seesaw. How far from the pivot must a 450 N adult sit on the other side to balance?
Step 1. Clockwise moment = anticlockwise moment: 300 × 1.5 = 450 × d.
Step 2. 450 = 450d, so d = 1.0 m.
A lever is a rigid bar that turns about a fixed point called the fulcrum. By the principle of moments, effort × effort arm = load × load arm, so a long effort arm gives a mechanical advantage greater than 1: MA = effort arm ÷ load arm. Levers are classified by the order of the three points. A handy way to remember: F-L-E 1-2-3: the class number tells you which of Fulcrum, Load, Effort is in the middle.
A crowbar has an effort arm of 1.20 m and a load arm of 0.15 m. What is its ideal mechanical advantage? What effort lifts a 1600 N load?
Step 1. MA = effort arm ÷ load arm = 1.20 ÷ 0.15 = 8.
Step 2. Effort = load ÷ MA = 1600 ÷ 8 = 200 N.
Step 3. Check the trade-off: the effort moves 8 times as far as the load. To raise the load by 1 cm you push the end through 8 cm.
Your own body uses levers. Your forearm is a class 3 lever: the biceps pulls on a point only a few centimetres from the elbow (the fulcrum), while the load sits in your hand much farther away. The muscle must therefore pull with a force much larger than the load, but the hand moves quickly through a large distance.
A pulley is a grooved wheel with a rope running over it. A fixed pulley only changes the direction of the force (MA = 1); that is how a flag is raised. A movable pulley moves with the load, and the rope's tension supports it on two sides, halving the effort. A block and tackle combines several pulleys, and for an ideal system MA = number of rope strands supporting the load.
A block and tackle has 4 strands supporting a load of 800 N and is 80% efficient. (a) Find the ideal effort. (b) Find the actual effort. (c) How much rope must be pulled to raise the load by 2.0 m?
(a) Ideal MA = 4, so ideal effort = 800 ÷ 4 = 200 N.
(b) Actual MA = efficiency × VR = 0.80 × 4 = 3.2, so actual effort = 800 ÷ 3.2 = 250 N.
(c) VR = 4, so the rope moves 4 × 2.0 = 8.0 m.
An inclined plane (ramp) lets you raise a load by pushing it a longer distance with a smaller force. For an ideal frictionless ramp of length ℓ and height h, MA = VR = ℓ ÷ h. The work is the same as lifting vertically: effort × ℓ = load × h.
A 500 N crate is pushed up a smooth ramp 6.0 m long onto a platform 1.5 m high. Find the ideal effort and the work done.
Ideal MA = ℓ ÷ h = 6.0 ÷ 1.5 = 4, so effort = 500 ÷ 4 = 125 N.
Work = effort × ℓ = 125 × 6.0 = 750 J, the same as lifting straight up: 500 × 1.5 = 750 J.
In practice friction means a larger effort is needed, and rollers or a smoother surface reduce it.
A wheel and axle is a lever that turns in a full circle: a large wheel fixed to a smaller axle. Turning the wheel by its rim (large radius R) turns the axle with a much larger force at its small radius r: MA = R ÷ r. Steering wheels, door handles, screwdrivers, taps and windlasses use this. If instead you turn the axle, you get speed but less force, as in a bicycle's wheels driven through the axle.
Gears are toothed wheels that mesh together. The driver gear turns the driven gear in the opposite direction. The speeds depend on the number of teeth:
speed of driven gear ÷ speed of driver = teeth on driver ÷ teeth on driven
A small gear driving a large one turns it more slowly but with greater turning force (torque); a large gear driving a small one gives more speed and less torque.
A driver gear with 20 teeth turns at 60 revolutions per minute and drives a gear with 60 teeth. Find the speed of the second gear.
Speed = 60 × (20 ÷ 60) = 20 revolutions per minute. The second gear turns three times more slowly but with about three times the turning force.
A bicycle shows many machines at once: pedals are levers on a crank, the chain links gear wheels (sprockets), brakes use levers, and the wheel is a wheel and axle. Low gears (a small front sprocket and large rear one) give a high mechanical advantage for hills; high gears give speed on the flat.
In a real machine, friction between moving parts and the weight of moving parts (such as pulleys and ropes) mean that some of the work input is wasted as heat and sound. Efficiency = (useful work output ÷ work input) × 100%.
A student applies a force of 40 N through 0.50 m on a lever to lift a 120 N load through 0.14 m. Find the efficiency.
Work in = 40 × 0.50 = 20 J. Work out = 120 × 0.14 = 16.8 J.
Efficiency = 16.8 ÷ 20 × 100% = 84%. The other 3.2 J went to friction and flexing.
We improve efficiency by lubricating moving parts, using bearings, making parts lighter and smoother, and aligning ropes properly. We can also ask whether a machine is worth using: a machine with a high MA but low efficiency can still be very useful, such as a car jack, which lets one person raise a heavy car.
Try each question before opening the answer.
W = Fd = 75 × 4.0 = 300 J.
Zero. The force (upward) is perpendicular to the movement (horizontal).
P = W ÷ t = 3600 ÷ 12 = 300 W.
Work = mgh = 70 × 9.8 × 4.0 = 2744 J. P = 2744 ÷ 5.0 ≈ 549 W.
Moment = 20 × 0.30 = 6.0 N·m.
200 × 0.60 = W × 0.40, so W = 120 ÷ 0.40 = 300 N.
Actual MA = 0.75 × 3 = 2.25. Effort = 600 ÷ 2.25 ≈ 267 N.
The effort is closer to the fulcrum than the load, so the effort must be larger than the load. It is useful because the load moves faster and farther than the effort, as with a fishing rod or tweezers.
Practical ideas: use a metre rule balanced on a pivot with slotted masses to verify the principle of moments; build a pulley system with two or three string pulleys and measure effort, load and distances to calculate MA, VR and efficiency; use a plank as a ramp at different angles and a newton meter to measure the pull for a trolley; compare the speed of a bicycle in different gears; time students climbing a flight of stairs to estimate their power (with a safe, flat, uncrowded staircase). Check the setup of any hanging masses so they cannot fall on feet. Numerical values are rounded; please check that they match the data booklet used by your examination board.
This article is original EDUSAMBAM writing. Definitions, constants and historical facts were checked against the sources below. Diagrams are original and were not copied from any source.
Note on the numbers: human power outputs and horsepower conversions are approximate. Please check numerical values against your own syllabus or exam board specification before relying on them.
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