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Physics · Force and motion I

Applying Newton's laws in context

You know F = ma, but in a lift question you cannot decide which forces to put in.

Newton’s laws connect force to motion. Draw the forces on one object, find the resultant, then apply F = ma.

This lesson is part of SPM Physics force and motion I. It needs the diagrams from drawing force diagrams.

What routine works for any situation?

  1. Choose one object.
  2. Draw the forces on it.
  3. Take one direction as positive and find the resultant.
  4. Write F = ma and solve.

Worked example: a person in a lift

A 60 kg person stands on a scale in a lift that accelerates upward at 1.5 m s⁻². Use g = 9.8 m s⁻².

The forces on the person are the weight down and the normal reaction R up. Take up as positive: R − mg = ma.

R = m(g + a) = 60 × (9.8 + 1.5) = 60 × 11.3 = 678 N. The scale reads more than the weight, 588 N, because the lift is accelerating upward.

At constant speed, a = 0, so R = mg = 60 × 9.8 = 588 N.

Worked example: a car

A 1200 kg car has an engine force of 3000 N and a total resistance of 600 N. The resultant is 3000 − 600 = 2400 N forward, so a = 2400 ÷ 1200 = 2.0 m s⁻².

Worked example: a third-law pair

A rocket pushes hot gas downward with 5000 N, and the gas pushes the rocket upward with 5000 N. These two forces are a third-law pair.

They act on different objects, so they never cancel each other. Whether the rocket accelerates depends only on the forces on the rocket: the push from the gas and its own weight.

Worked example: a third-law pair

A rocket pushes hot gas downward with 5000 N, and the gas pushes the rocket upward with 5000 N. These two forces are a third-law pair.

They act on different objects, so they never cancel each other. Whether the rocket accelerates depends only on the forces on the rocket: the push from the gas and its own weight.

The mistake that loses marks

A student writes R = mg for the accelerating lift and gets 588 N. That is correct only when the acceleration is zero.

Use the second law with both forces. If the resultant is not zero, the forces are not equal.

Check yourself

The same person is in a lift that moves upward at a steady 2.0 m s⁻¹. Find the scale reading, and say which law explains it.

Answer

The velocity is constant, so the resultant force is zero. R = mg = 60 × 9.8 = 588 N.

This follows from Newton’s first law: no resultant force means no change in velocity.

What to study next

Continue with distinguishing work, energy and power. Then test the routine in the practice set.

The units and significant figure checker checks the last line. If you want a teacher to go through new situations with you, see online one-to-one Physics tuition.

Common questions

What do Newton's three laws say?

First: an object stays at rest or at constant velocity when the resultant force is zero. Second: the resultant force equals mass times acceleration. Third: when one object pushes another, the second pushes back with an equal and opposite force on the first.

Why does a person in a lift feel heavier when it accelerates up?

The floor must push up with more than the person's weight to give an upward acceleration. The scale reading is the normal reaction, so it is larger than mg. In steady motion at constant speed, the reading equals the weight.

What is the difference between mass and weight?

Mass is the amount of matter and is measured in kilograms. Weight is the gravitational force on the object, measured in newtons, and equals mass times g. Mass stays the same everywhere, but weight changes with g.

Do the two forces in a third-law pair cancel?

No. They act on different objects, so they cannot be added to get a resultant on one object. Each force affects only the object it acts on.

A teacher working one-to-one can give you a new situation each time and ask you to name the object, the forces and the law before you calculate, so the routine becomes automatic.

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