Mini-Lesson

Forces and Newton's laws

2 missions. Guess, learn, then check. Your place is saved.

Mission 1On scales in a lift, what do they read as it speeds up going up?
Mission 2Tilt a slope until a box slides. Why doesn't its weight matter?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

On scales in a lift, what do they read as it speeds up going up?

Make a guess: On scales in a lift, what do they read as it speeds up going up?

Mechanics · Newton's laws

Newton's laws and the forces you meet

First law: a body stays at rest or moves with constant velocity unless a resultant force acts.
Second law: resultant force $= ma$ = ma, in the direction of the acceleration. Forces add as vectors, so $F = m a$ F = ma too.
Third law: if A pushes B, B pushes A with an equal and opposite force.
Forces: weight $W = m g$ W = mg, normal reaction $R$ R, tension in a string, thrust in a rod, resistance. $R = 0$ R = 0 means contact is lost.

Worked example Mechanics · Newton's laws

A car accelerating

Question

A car of mass 900 kg starts from rest. Its engine gives a driving force of 2,400 N and resistance to motion is 600 N. Find its speed after 6 s.

1
Resultant force in the direction of motion: $2400 - 600 = 1800$ 2400 − 600 = 1800 N
2
Newton's second law: $1800 = 900 a$ 1800 = 900a, so $a = 2$ a = 2
3
Constant acceleration, so use suvat: $v = u + a t = 0 + 2 \times 6$ v = u + at = 0 + 2 × 6

Answer

12 m/s

⚠️ Watch out

F

in

F = ma

is the resultant force. Using 2,400 N alone gives

a = 2.67

.

Quick check

?A book rests on a table. By Newton's third law, which force is paired with the book's weight?

Mechanics · Newton's laws

Lifts and apparent weight

A passenger in a lift has two forces on them: weight $m g$ mg down and the floor's reaction $R$ R up.
Taking up as positive: $R - m g = ma$ R − mg = ma, so $R = m (g + a)$ R = m(g + a).
Bathroom scales measure $R$ R and display $\frac{R}{g}$ R/g as a mass.
Accelerating upwards (or slowing on the way down): $R > m g$ R > mg. Accelerating downwards (or slowing on the way up): $R < m g$ R < mg. Constant velocity: $R = m g$ R = mg.
For the lift cable, treat lift and passenger as one particle: $T - (M + m) g = (M + m) a$ T − (M + m)g = (M + m)a.

Calculate

?A lift is moving upwards and slowing down at 1.2 m s

^{- 2}

. A passenger of mass 55 kg stands in it. Find the normal reaction of the lift floor on the passenger.

newtons

Mechanics · Newton's laws

Connected particles

Particles joined by a light inextensible string (or a rigid tow bar) have the same acceleration.
A light string over a smooth pulley has the same tension on both sides.
Find the acceleration from the whole system: the tension is internal and cancels.
Then use one particle on its own to find the tension (or the thrust in a tow bar when braking).

Worked example Mechanics · Newton's laws

Car and trailer

Question

A car of mass 1,000 kg tows a trailer of mass 400 kg along a straight horizontal road with a light rigid tow bar. The driving force is 2,000 N. Resistances are 300 N on the car and 160 N on the trailer. Find the acceleration and the tension in the tow bar.

1
Whole system: $2000 - 300 - 160 = 1400 a$ 2000 − 300 − 160 = 1400a, so $a = 1.1$ a = 1.1
2
Trailer alone: $T - 160 = 400 \times 1.1$ T − 160 = 400 × 1.1
3
Solve: $T = 440 + 160 = 600$ T = 440 + 160 = 600
4
Check with the car alone: $2000 - 300 - 600 = 1100 = 1000 \times 1.1$ 2000 − 300 − 600 = 1100 = 1000 × 1.1

Answer

a = 1.1

m s

^{- 2}

; tension 600 N

⚠️ Watch out

the tension is not just

400 a = 440

N. The tow bar must also overcome the trailer's 160 N resistance.

Calculate

?Particles

A

of mass 6 kg and

B

of mass 4 kg are attached to the ends of a light inextensible string that passes over a smooth fixed pulley. They are released from rest with the string taut and both parts vertical. Find the tension in the string. Give your answer to 3 significant figures.

newtons

Quick check

?Forces

(3 i - 2 j)

N and

(5 i + 8 j)

N are the only forces acting on a particle of mass 2 kg. Find the magnitude of its acceleration.

Mission check

?On scales in a lift, what do they read as it speeds up going up? A 60 kg person stands on them while the lift accelerates upwards at 0.5 m s

^{- 2}

.

Mission 2

Tilt a slope until a box slides. Why doesn't its weight matter?

Make a guess: Tilt a slope until a box slides. Why doesn't its weight matter?

Mechanics · resolving and friction

Resolving a force (A-level only)

A force $F$ F at angle $θ$ θ to a direction has component $F cos θ$ Fcos θ along it and $F sin θ$ Fsin θ perpendicular to it.
On a slope at $θ$ θ to the horizontal, resolve parallel and perpendicular to the slope: the weight gives $m g sin θ$ mgsin θ down the slope and $m g cos θ$ mgcos θ into the slope.
A particle is in equilibrium if and only if the resolved parts in every direction sum to zero. Two perpendicular directions are enough.
Choose directions that avoid an unknown force where you can.

Worked example Mechanics · resolving and friction

A particle held on a smooth slope

Question

A particle of mass 2 kg rests on a smooth plane inclined at

3 0^{\circ}

to the horizontal. It is held in equilibrium by a horizontal force

P

N. Find

P

and the normal reaction

R

.

1
Resolve parallel to the slope (so $R$ R drops out): $P cos 3 0^{\circ} = 2 g sin 3 0^{\circ}$ Pcos 30°=2gsin 30°
2
Solve: $P = 2 g tan 3 0^{\circ} = 11.3$ P=2gtan 30°=11.3 (3 s.f.)
3
Resolve perpendicular to the slope; $P$ P also pushes into it: $R = 2 g cos 3 0^{\circ} + P sin 3 0^{\circ}$ R=2gcos 30°+Psin 30°
4
Evaluate: $R = 16.97 + 5.66 = 22.6$ R=16.97+5.66=22.6 (3 s.f.)

Answer

P = 11.3

N and

R = 22.6

N

⚠️ Watch out

R = m g cos θ

only when no other force has a component perpendicular to the slope; here it would give 17.0 N, which is wrong.

Quick check

?A box of mass 5 kg lies on a smooth slope inclined at

2 0^{\circ}

to the horizontal. What is the component of its weight down the slope?

Mechanics · resolving and friction

The friction model

Friction acts along the surface, opposing motion or the tendency to move.
On a rough surface, $F \leq μ R$ F≤μ R, where $μ$ μ is the coefficient of friction and $R$ R is the normal reaction.
Static friction takes whatever value is needed, up to $μ R$ μ R. When it reaches $μ R$ μ R the body is in limiting equilibrium: about to slide.
When the body is moving, $F = μ R$ F=μ R.

Worked example Mechanics · resolving and friction

On the point of sliding

Question

A crate of mass 4 kg rests on a rough plane inclined at

2 5^{\circ}

to the horizontal and is about to slide down. Find

μ

.

1
Resolve perpendicular to the slope: $R = 4 g cos 2 5^{\circ} = 35.5$ R=4gcos 25°=35.5 N
2
Resolve parallel; friction acts up the slope: $F = 4 g sin 2 5^{\circ} = 16.6$ F=4gsin 25°=16.6 N
3
Limiting equilibrium, so $F = μ R$ F=μ R: $μ = \frac{4 g sin 2 5 ^{\circ}}{4 g cos 2 5 ^{\circ}}$ μ=(4gsin 25°)/(4gcos 25°)
4
Simplify: $μ = tan 2 5^{\circ} = 0.466$ μ=tan 25°=0.466 (3 s.f.)

Answer

μ = 0.466

⚠️ Watch out

friction acts up the slope here because the crate is about to slide down; draw its direction before resolving.

Calculate

?A block of mass 3 kg is pulled up a line of greatest slope of a rough plane inclined at

3 0^{\circ}

to the horizontal by a force of 30 N parallel to the slope. The coefficient of friction is 0.25. Find the acceleration of the block, to 3 significant figures.

m s⁻²

Calculate

?A sledge of mass 12 kg is pulled across rough horizontal ground at constant speed by a rope at

2 0^{\circ}

above the horizontal. The tension in the rope is 40 N. Find the coefficient of friction, to 3 significant figures.

Quick check

?Particle

A

(2 kg) is on a rough plane inclined at

3 0^{\circ}

, with

μ = 0.3

. A light inextensible string from

A

runs up the slope, over a smooth pulley at the top, to particle

B

(3 kg) hanging freely. The system is released from rest. Find the acceleration.

Mission check

?A box is about to slide on a rough slope. Why doesn't the box's weight affect the angle?

Recap

Big ideas

$F = ma$ F = ma uses the resultant force; forces add as vectors.
Lifts: $R - m g = ma$ R − mg = ma with up positive. Scales read $R$ R, not your mass.
Connected particles share one acceleration: use the whole system for $a$ a, then one particle for the tension.
Resolve parallel and perpendicular to a slope: weight gives $m g sin θ$ mgsin θ along and $m g cos θ$ mgcos θ into it; other forces can change $R$ R.
Friction obeys $F \leq μ R$ F≤μ R; $F = μ R$ F=μ R when moving or in limiting equilibrium.
On the point of sliding with no other forces, $μ = tan θ$ μ=tan θ, so mass does not matter.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Forces and Newton's laws for AQA A-level Mathematics (7357). 🎉

Mission 1On scales in a lift, what do they read as it speeds up going up?
Mission 2Tilt a slope until a box slides. Why doesn't its weight matter?

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Specification points and sources for this lesson are listed at the bottom of the page.