Maths › Mechanics › Forces and Newton's laws
Forces and Newton's laws
Draw the forces, add them up, divide by the mass. Newton's second law is three instructions long, and almost every mechanics mark hangs off the first one, the diagram.
Builds on Kinematics with constant acceleration.
Where it earns its keep: Newton's laws and the resultant force on InkPhysics.
IN THIS TOPIC
- Draw free body diagrams with weight, normal reaction, tension, thrust and friction correctly placed.
- Apply F = ma along a chosen direction, and the equilibrium condition when a = 0.
- Resolve a force at an angle into components, and see what that does to the normal reaction.
- Work with forces written in i and j notation, including the magnitude and direction of a resultant.
- Combine F = ma with suvat when a constant force produces constant acceleration.
COMMON MISCONCEPTION
A moving object always has a force pushing it along in the direction of motion.
The diagram is the physics
Draw every force on the object and no force that is not on the object. Weight mg straight down. Normal reaction R perpendicular to the surface. Tension pulling along a string, thrust pushing along a rod, friction opposing the sliding. Forces the object exerts on other things belong on their diagrams, not on this one.
Newton's laws at work
The first law says that no resultant force means no change of velocity, so rest and steady speed are the same condition, equilibrium. The second law quantifies everything else. Resultant force equals mass times acceleration, applied along a direction you have chosen, with the signs kept consistent. The third law says forces come in equal and opposite pairs acting on different objects, and that is the reason a free body diagram only ever shows one member of each pair.
WORKED EXAMPLE
Resultant first, then divide
A 5 kg crate is dragged by a horizontal rope with tension 30 N against a constant resistance of 10 N. Find the acceleration, and the speed after 4 s from rest.
Along the motion: F = 30 − 10 = 20 N, so a = 20/5 = 4 m s⁻².
A constant force gives constant acceleration, so suvat applies: v = 0 + 4 × 4 = 16 m s⁻¹.
Vertically nothing moves, so R = mg = 49 N, and that quiet second equation confirms the crate stays on the floor.
Forces at an angle
Real ropes are rarely horizontal. A force T at angle θ to the horizontal splits into T cos θ along the ground and T sin θ up, and both halves do work in the equations. The vertical component is the one candidates drop.
WORKED EXAMPLE
An angled tow
A 10 kg sledge is pulled by a rope at 30° above the horizontal with tension 50 N, against a horizontal resistance of 15 N. Find the acceleration and the normal reaction.
Horizontally: 50 cos 30° − 15 = 43.3 − 15 = 28.3 N, so a = 28.3/10 = 2.83 m s⁻².
Vertically there is no motion, so R + 50 sin 30° = 10g, giving R = 98 − 25 = 73 N.
R has dropped well below mg because the rope is taking part of the sledge's weight. Assuming R = mg here would be wrong by 25 N, and in a friction question that error propagates into everything.
Forces as vectors
Forces can also arrive written in components, as (3i + 5j) N. Adding them is then component by component, and F = ma becomes a vector equation with the acceleration pointing along the resultant.
Take forces (3i + 5j) N and (−i + 2j) N acting on a particle of mass 2 kg. The resultant is 2i + 7j N, so a = i + 3.5j m s⁻², with magnitude √(1² + 3.5²) ≈ 3.64 m s⁻². The direction comes from tan⁻¹(3.5/1) ≈ 74.1° from the i direction. Quote the angle from a stated reference direction, because an unlabelled angle earns nothing.
Equilibrium in this language is short. The resultant vector is zero, which means the i components sum to zero and the j components sum to zero, so one vector statement yields two scalar equations.
ASSESSMENT FOCUS
- Draw the free body diagram first. Examiners award it, and everything downstream leans on it.
- Write F = ma along a declared direction and keep every sign consistent with that declaration.
- When a pull or push is at an angle, resolve it in both directions before writing anything, and check what has happened to R.
- Equilibrium means the resultant is zero in each direction. Resolve twice, then solve.
- F = ma followed by suvat is the standard two-step. Say when the force, and therefore the acceleration, is constant.
CHECK YOURSELF
A 1200 kg car accelerates at 2.5 m s⁻² against a total resistance of 600 N. Find the driving force.
Show a hint
F in F = ma is the resultant.
Show the answer
Resultant needed: ma = 1200 × 2.5 = 3000 N.
Driving force = 3000 + 600 = 3600 N.
Diagram first. Every force on the object, nothing it exerts on anything else.
Resolve, sum with signs, apply F = ma. Equilibrium is the a = 0 special case.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the forces and newton's laws questions page.
CHECK YOUR PROGRESS
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- Draw free body diagrams with weight, normal reaction, tension, thrust and friction correctly placed.
- Apply F = ma along a chosen direction, and the equilibrium condition when a = 0.
- Resolve a force at an angle into components, and see what that does to the normal reaction.
- Work with forces written in i and j notation, including the magnitude and direction of a resultant.
- Combine F = ma with suvat when a constant force produces constant acceleration.
Open the full revision checklist to see every objective in the course in one place.