A ramp, a lever, a pulley: simple machines let you lift heavy things with a smaller force. They seem to give something for nothing. They don't: the work stays the same, and you pay for the smaller force with a longer distance.
Key idea
An ideal (frictionless) machine puts out the same work it takes in: $$F_{\text{in}}\,d_{\text{in}} = F_{\text{out}}\,d_{\text{out}}$$ So a smaller force must move a longer distance. Mechanical advantage = output force ÷ input force.
Why it works
To raise a 600 N box by 1 m, you must give it $600 \times 1 = 600$ J, however you do it. Lift it straight up: 600 N for 1 m. Push it up a 3 m ramp: the same 600 J spread over 3 m needs only $600 \div 3 = 200$ N. The ramp trades force for distance. Real machines have friction, so you need a bit more force than the ideal value.
Worked example: A ramp
Find the push needed to slide a 600 N box up a frictionless 3 m ramp onto a 1 m platform, and the mechanical advantage.
- Work out: $600\,\mathrm{N} \times 1\,\mathrm{m} = 600$ J.
- Work in must match: $F \times 3\,\mathrm{m} = 600$ J, so $F = 200$ N.
- Mechanical advantage: $600 \div 200 = 3$.
Answer: 200 N; mechanical advantage 3
Worked example: A pulley block
A block and tackle with 4 rope strands holding the load lifts a 400 N engine by 0.5 m. Ignoring friction, find the pull and how much rope you pull.
- Four strands share the load: $400 \div 4 = 100$ N.
- Each strand must shorten by 0.5 m, so you pull $4 \times 0.5 = 2$ m of rope.
- Check the work: $100 \times 2 = 200$ J in, $400 \times 0.5 = 200$ J out.
Answer: 100 N, pulling 2 m of rope
Common mistake
Thinking a machine reduces the work needed. It reduces the force; the work is the same (or a bit more, because of friction). Check: input force × input distance should match output force × output distance.