Equilibrium was the special case where the resultant force is zero. Now let it be anything: the resultant force in the direction of motion equals mass times acceleration, and the suvat equations take it from there.
Newton's second law: resultant force \(= ma\), resolved in the direction of the acceleration. Perpendicular to the motion the forces still balance. Weight is \(W = mg\) with \(g = 10\ \text{m s}^{-2}\), and a sliding particle has friction \(F = \mu R\).
Fig. 1: an 8 kg block pulled along rough horizontal ground by a 50 N force at 30° to the horizontal (angle to scale). The pull reduces R to 55 N, so friction is 13.75 N
Section 01
Three laws, one equation.
In the lab, set the friction coefficient to zero and double the applied force: the acceleration doubles. Then double the mass instead: it halves. Finally add friction and find the force at which the cart just moves at constant speed (zero resultant, zero acceleration).
First law: with zero resultant force, a particle stays at rest or moves with constant velocity.
Second law: resultant force \(= ma\), in the direction of the acceleration. Mass in kg, force in N.
Third law: if A pushes on B, B pushes on A with an equal and opposite force.
Weight: \(W = mg\), \(g = 10\ \text{m s}^{-2}\) (the lab uses 9.81). Air resistance or other resistances act only if the question says so.
Worked example A driving force against resistance
A particle of mass 5 kg on a smooth horizontal surface is pulled by a horizontal force of 20 N against a resistance of 5 N. Find its acceleration.
Resultant force in the direction of motion \(= 20 - 5 = 15\) N.
\(15 = 5a\).
Answer: \(a = 3\ \text{m s}^{-2}\)
Worked example Pulled at an angle on a rough floor (Fig. 1)
A block of mass 8 kg is pulled along rough horizontal ground by a force of 50 N at \(30^\circ\) above the horizontal. The coefficient of friction is 0.25. Find the acceleration.
Vertically (no vertical acceleration): \(R + 50\sin 30^\circ = 80\), so \(R = 55\) N.
The block is moving, so \(F = \mu R = 0.25 \times 55 = 13.75\) N.
Horizontally: \(50\cos 30^\circ - 13.75 = 8a\).
\(43.30 - 13.75 = 29.55 = 8a\), so \(a = 3.69\).
Answer: \(3.69\ \text{m s}^{-2}\)
Examiner trap
\(F = ma\) uses the RESULTANT force. Writing \(50 = 8a\), or \(50\cos 30^\circ = 8a\), ignores friction and is the most common lost mark.
Section 02
Combining with suvat.
Forces give acceleration; acceleration links to motion through the constant acceleration equations. A question may run either way: forces to motion, or motion to forces.
Worked example Motion to force
A block slides across rough horizontal ground. Its speed falls from \(6\ \text{m s}^{-1}\) to zero in a distance of 9 m. Find the coefficient of friction.
The only horizontal force is friction: \(-\mu R = ma\), with \(R = mg = 10m\).
\(-10\mu m = -2m\), so \(\mu = 0.2\).
Answer: \(\mu = 0.2\) (the mass cancels).
Worked example Falling with resistance
A particle of mass 0.5 kg falls from rest against a constant air resistance of 1.5 N. Find its speed after falling 20 m.
Down positive: \(5 - 1.5 = 0.5a\), so \(a = 7\ \text{m s}^{-2}\).
\(v^2 = 0 + 2(7)(20) = 280\).
\(v = 16.7\ \text{m s}^{-1}\).
Answer: \(16.7\ \text{m s}^{-1}\)
Your turn
A car of mass 1200 kg brakes uniformly from \(20\ \text{m s}^{-1}\) to rest in 40 m. Find the braking force, assuming no other horizontal forces.
\(0 = 400 + 80a\), so \(a = -5\ \text{m s}^{-2}\). Braking force \(= 1200 \times 5 = 6000\) N.
Examiner trap
Keep signs consistent. If the particle is slowing down, \(a\) is negative in your positive direction, and so is the resultant force.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
A resultant force of 10 N acts on a 4 kg particle. Find its acceleration.
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What is the weight of a 7 kg mass in Paper 4?
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A particle moves with constant velocity. What can you say about the forces on it?
Quick check+10 XP
A lift accelerates upwards. How does the normal reaction on a passenger compare with their weight?
Section 03
Vertical motion and lifts.
For vertical motion, resolve vertically with the direction of the acceleration as positive. A person in a lift feels the normal reaction from the floor, which differs from their weight whenever the lift accelerates.
Worked example A person in a lift
A woman of mass 60 kg stands in a lift. Find the normal reaction from the floor when the lift (a) accelerates upwards at \(1.5\ \text{m s}^{-2}\), (b) moves up but decelerates at \(2\ \text{m s}^{-2}\).
(a) Up positive: \(R - 600 = 60(1.5)\), so \(R = 690\) N.
(b) Moving up and slowing down means \(a = -2\): \(R - 600 = 60(-2)\).
\(R = 480\) N.
Answer: (a) 690 N (b) 480 N. The woman pushes on the floor with the same force (Newton's third law).
Worked example A crate on a rope
A crate of mass 50 kg is raised from rest by a vertical rope with tension 560 N. Find its speed after 3 s.
Up positive: \(560 - 500 = 50a\), so \(a = 1.2\ \text{m s}^{-2}\).
\(v = 0 + 1.2 \times 3\).
Answer: \(3.6\ \text{m s}^{-1}\)
Examiner trap
"Moving upwards" does not mean "accelerating upwards". A lift moving up while slowing down has a DOWNWARD acceleration, so the reaction is less than the weight.
Stage 2 · Exam
Exam-style questions
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A block of mass 8 kg is pulled along rough horizontal ground (\(\mu = 0.25\)) by a force of 50 N at \(30^\circ\) above the horizontal. Find its acceleration.
Exam style+20 XP
A man of mass 60 kg stands in a lift that is moving upwards and decelerating at \(2\ \text{m s}^{-2}\). Find the normal reaction from the floor.
Exam style+20 XP
A block slides on rough horizontal ground and its speed falls uniformly from \(6\ \text{m s}^{-1}\) to zero in 9 m. Find \(\mu\).
Exam style+20 XP
A particle of mass 0.5 kg falls from rest with constant air resistance 1.5 N. Find its speed after it has fallen 20 m.
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F = ma, applied.
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