A 2\,kg block on a rough horizontal surface has μ = 0.3. Find the limiting friction. Take g = 9.8\,m s^-2.

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How do Newton's laws relate force to motion, and how do you handle friction and connected bodies?

Forces as vectors, resolving and equilibrium, Newton's three laws, the equation of motion F equals ma, friction and the coefficient of friction, and connected particles over pulleys.

A CCEA AS Further Maths Mechanics answer on forces as vectors, resolving and equilibrium, Newton's three laws, the equation of motion, the friction model with the coefficient of friction, and connected particles linked by a string over a pulley.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
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What this dot point is asking

CCEA wants you to treat forces as vectors, resolve them and apply the equilibrium condition, state and use Newton's three laws, write the equation of motion $F = ma$ , model friction with the coefficient $\mu$ , and analyse connected particles joined by a light string over a smooth pulley.

The answer

Forces, resolving and equilibrium

Newton's three laws

Friction and the coefficient of friction

Connected particles

A block on a slope with friction

A box of mass $4\,\text{kg}$ is on a slope inclined at $30^\circ$ , with coefficient of friction $0.25$ . Find its acceleration down the slope. Take $g = 9.8\,\text{m s}^{-2}$ .

step Resolve perpendicular to the slope

The normal reaction balances the perpendicular component of weight:

R = mg\cos 30^\circ = 4(9.8)(0.866) = 33.9\,\text{N}.

step Find the limiting friction

F = \mu R = 0.25 \times 33.9 = 8.49\,\text{N},

acting up the slope, opposing the motion.

step Apply F = ma down the slope

The driving force is $mg\sin 30^\circ = 4(9.8)(0.5) = 19.6\,\text{N}$ . The resultant is $19.6 - 8.49 = 11.1\,\text{N}$ :

11.1 = 4a \Rightarrow a = 2.78\,\text{m s}^{-2}.

Examples in context

Example 1. Anti-lock braking. A car's tyres grip the road through friction limited by $\mu R$ . Anti-lock systems keep the wheels just below the point of skidding, where friction is greatest, which is exactly the limiting-friction model in action.

Example 2. A lift and counterweight. A lift cabin and its counterweight behave like connected particles over a pulley: the motor supplies the extra force, but the same common-tension, shared-acceleration analysis predicts how fast the cabin rises and the load in the cable.

Try this

Q1. State Newton's second law as an equation. [1 mark]

Cue. Resultant force equals mass times acceleration, $F = ma$ .

Q2. A $2\,\text{kg}$ block on a rough horizontal surface has $\mu = 0.3$ . Find the limiting friction. Take $g = 9.8\,\text{m s}^{-2}$ . [2 marks]

Cue. $R = mg = 19.6\,\text{N}$ ; $F = \mu R = 0.3(19.6) = 5.88\,\text{N}$ .

Q3. For a light string over a smooth pulley, what can you say about the tension on the two sides? [1 mark]

Cue. It is the same throughout the string.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA AS 20207 marksA block of mass

5\,\text{kg}

rests on a rough horizontal surface with coefficient of friction

0.4

. A horizontal force

P

is applied. Taking

g = 9.8\,\text{m s}^{-2}

, find the value of

P

that is just sufficient to move the block, and the acceleration if

P = 30\,\text{N}

Show worked answer →

The normal reaction balances the weight: $R = mg = 5 \times 9.8 = 49\,\text{N}$ .

The maximum (limiting) friction is $F_{\max} = \mu R = 0.4 \times 49 = 19.6\,\text{N}$ .

The block is on the point of moving when $P$ equals the limiting friction, so $P = 19.6\,\text{N}$ .

With $P = 30\,\text{N}$ , the block moves and friction acts at its maximum. The resultant force is $30 - 19.6 = 10.4\,\text{N}$ . Using $F = ma$ :

$10.4 = 5a \Rightarrow a = 2.08\,\text{m s}^{-2}.$

Markers reward the normal reaction, the limiting friction, the threshold value of $P$ , and the acceleration from the equation of motion.

CCEA AS 20187 marksTwo particles of masses

3\,\text{kg}

and

2\,\text{kg}

are connected by a light inextensible string over a smooth pulley and released from rest. Taking

g = 9.8\,\text{m s}^{-2}

, find the acceleration of the system and the tension in the string.

Show worked answer →

Let the acceleration be $a$ and the tension $T$ . The heavier $3\,\text{kg}$ mass descends, the $2\,\text{kg}$ mass rises.

For the $3\,\text{kg}$ mass (down positive): $3g - T = 3a$ .
For the $2\,\text{kg}$ mass (up positive): $T - 2g = 2a$ .

Adding the equations eliminates $T$ : $3g - 2g = 5a$ , so $g = 5a$ , giving

$a = \dfrac{9.8}{5} = 1.96\,\text{m s}^{-2}.$

Substitute back: $T = 2g + 2a = 2(9.8) + 2(1.96) = 19.6 + 3.92 = 23.52\,\text{N}.$

Markers reward an equation of motion for each particle, adding to eliminate the tension, and the correct acceleration and tension.

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Sources & how we know this

CCEA GCE Further Mathematics specification — CCEA (2018)