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How do you use the coefficient of friction and the limiting friction model on rough surfaces?

Model friction on rough surfaces: use the relationship F is at most mu R, find limiting friction, and analyse motion and equilibrium on rough horizontal surfaces.

A CCEA GCSE Further Mathematics answer on friction, covering the coefficient of friction, limiting friction F equals mu R, the normal reaction, and analysing whether an object moves or stays in equilibrium on a rough surface in the Mechanics unit.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. How friction behaves
  3. The limiting friction model
  4. Will it move?
  5. Motion under friction
  6. Why this matters

What this dot point is asking

Friction is the force that resists sliding between surfaces, and CCEA GCSE Further Mathematics models it on rough horizontal surfaces. You must use the relationship between friction and the normal reaction, find the limiting (maximum) friction with the coefficient of friction, and decide whether an object stays in equilibrium or moves, then find its acceleration if it does. This builds directly on Newton's laws and the normal reaction.

How friction behaves

Friction is a reactive force: it acts only to oppose movement or the tendency to move, and it adjusts itself up to a maximum. While an object is stationary, friction is exactly as large as it needs to be to prevent sliding, up to its limit. Once that limit is exceeded, the object slips and friction stays at the maximum value.

The limiting friction model

The key relationship ties friction to the normal reaction. The larger the force pressing the surfaces together, the larger the available friction.

The normal reaction must be found first, because the friction depends on it. On a horizontal floor with only a horizontal push, R=mgR = mg, but if a force has a vertical component, that changes RR and therefore the friction.

Will it move?

The central decision is whether an applied force is enough to overcome friction. Compare the applied force with the limiting friction.

Motion under friction

Once an object is moving, friction acts at its limiting value μR\mu R directly against the motion, and you use Newton's second law with friction as a resisting force. The resultant force is the driving force minus the friction, and dividing by the mass gives the acceleration. If the driving force is removed while the object moves, friction alone decelerates it, and the suvat equations then find how far it travels before stopping.

Why this matters

Friction makes the dynamics realistic, moving beyond the smooth surfaces of the basic Newton's-laws problems. The skill of finding the normal reaction, computing limiting friction, and deciding between equilibrium and motion runs through the connected-particle problems too, where a block on a rough table is joined to a hanging mass. It combines the normal reaction from the forces topic with the equation of motion in one model.

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 Unit 2 (style)4 marksA block of mass 44 kg rests on a rough horizontal floor with coefficient of friction 0.30.3. Taking g=9.8g = 9.8 m/s2^2, find the maximum (limiting) friction force.
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The normal reaction on a horizontal floor balances the weight: R=mg=4×9.8=39.2R = mg = 4 \times 9.8 = 39.2 N.

Limiting friction is the largest friction available: Fmax=μR=0.3×39.2=11.76F_{\max} = \mu R = 0.3 \times 39.2 = 11.76 N.

So the maximum friction is 11.7611.76 N.

Marks are for the normal reaction, the use of F=μRF = \mu R, and the value. Using mass instead of weight for RR is the common error.

CCEA Unit 2 (style)5 marksA 66 kg block on a rough floor (μ=0.25\mu = 0.25) is pushed by a horizontal force of 2020 N. Taking g=9.8g = 9.8 m/s2^2, determine whether it moves, and if so find its acceleration.
Show worked answer →

Normal reaction: R=mg=6×9.8=58.8R = mg = 6 \times 9.8 = 58.8 N. Limiting friction: Fmax=μR=0.25×58.8=14.7F_{\max} = \mu R = 0.25 \times 58.8 = 14.7 N.

The applied force 2020 N exceeds the maximum friction 14.714.7 N, so the block moves.

Once moving, friction acts at its limiting value against the motion. Resultant force =2014.7=5.3= 20 - 14.7 = 5.3 N, so a=5.36=0.88a = \dfrac{5.3}{6} = 0.88 m/s2^2.

Marks are for the friction, the comparison, the resultant, and the acceleration. Forgetting to subtract friction when finding the acceleration is the usual slip.

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