Skip to main content
WalesMathsSyllabus dot point

How do Newton's laws relate force to acceleration, and how do we handle weight, tension, friction and connected particles?

Newton's three laws, force diagrams, weight, normal reaction, tension, friction, and connected particles over a pulley.

A focused answer to WJEC AS Unit 2 forces, covering Newton's three laws, force diagrams, weight, normal reaction, tension, friction, and the motion of connected particles over a pulley.

Generated by Claude Opus 4.813 min answer

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

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

WJEC wants you to state and apply Newton's three laws, draw clear force diagrams, identify weight, normal reaction, tension and friction, and solve problems involving connected particles linked by a string over a smooth pulley. This is the heart of the mechanics section and combines with the kinematics equations to find accelerations, speeds and distances.

The answer

Newton's three laws

The second law is the workhorse: resolve forces into a chosen direction, find the resultant, and set it equal to mama.

Force diagrams and the standard forces

A force diagram shows every force acting on the body as an arrow. The forces you meet at AS are:

On a horizontal surface with no vertical acceleration, the normal reaction balances the weight, so R=mgR = mg.

Friction

A body on the point of moving, or already moving, has friction at its maximum value μR\mu R.

Connected particles

For two particles joined by a light inextensible string over a smooth pulley, the string has the same tension throughout and the particles share the same acceleration magnitude.

Examples in context

Example 1. A lift accelerating. A person of mass 70kg70\,\text{kg} stands in a lift accelerating upward at 2m s22\,\text{m s}^{-2}. The normal reaction (what a scale reads) satisfies Rmg=maR - mg = ma, so R=m(g+a)=70(9.8+2)=826NR = m(g + a) = 70(9.8 + 2) = 826\,\text{N}, more than their weight of 686N686\,\text{N}. Newton's second law explains the heavy feeling on acceleration.

Example 2. Friction limiting motion. A 10kg10\,\text{kg} crate sits on a floor with μ=0.3\mu = 0.3. The maximum friction is μR=0.3×10×9.8=29.4N\mu R = 0.3 \times 10 \times 9.8 = 29.4\,\text{N}. A horizontal push of 25N25\,\text{N} is below this, so the crate does not move; a push of 35N35\,\text{N} exceeds it, so the crate accelerates. The friction inequality decides whether motion starts.

Try this

Q1. A 2kg2\,\text{kg} object has a resultant force of 10N10\,\text{N} acting on it. Find its acceleration. [2 marks]

  • Cue. F=maF = ma: a=102=5m s2a = \dfrac{10}{2} = 5\,\text{m s}^{-2}.

Q2. Find the weight of a 6kg6\,\text{kg} mass (take g=9.8m s2g = 9.8\,\text{m s}^{-2}). [1 mark]

  • Cue. W=mg=6×9.8=58.8NW = mg = 6 \times 9.8 = 58.8\,\text{N}.

Q3. A block on a rough horizontal surface has normal reaction 40N40\,\text{N} and μ=0.25\mu = 0.25. Find the maximum friction. [2 marks]

  • Cue. F=μR=0.25×40=10NF = \mu R = 0.25 \times 40 = 10\,\text{N}.

Exam-style practice questions

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

WJEC AS style5 marksA box of mass 5kg5\,\text{kg} is pulled along a rough horizontal floor by a horizontal force of 30N30\,\text{N}. The frictional force is 12N12\,\text{N}. Taking g=9.8m s2g = 9.8\,\text{m s}^{-2}, find the acceleration of the box.
Show worked answer →

Apply Newton's second law along the direction of motion, using the resultant horizontal force.

Resultant horizontal force =3012=18N= 30 - 12 = 18\,\text{N} (pull minus friction).

Newton's second law: F=maF = ma, so 18=5a18 = 5a.

a=185=3.6m s2a = \dfrac{18}{5} = 3.6\,\text{m s}^{-2}.

Markers reward finding the resultant force by subtracting friction from the pull, applying F=maF = ma with the correct mass, and a final answer of 3.6m s23.6\,\text{m s}^{-2}. The weight and normal reaction are vertical and balance, so they do not enter the horizontal equation.

WJEC AS style6 marksTwo particles of mass 3kg3\,\text{kg} and 2kg2\,\text{kg} are connected by a light inextensible string over a smooth pulley. Find the acceleration of the system and the tension in the string. Take g=9.8m s2g = 9.8\,\text{m s}^{-2}.
Show worked answer →

Write Newton's second law for each particle, taking the heavier one as descending, then solve simultaneously.

For the 3kg3\,\text{kg} mass (descending): 3gT=3a3g - T = 3a.
For the 2kg2\,\text{kg} mass (rising): T2g=2aT - 2g = 2a.

Add the equations: 3g2g=5a3g - 2g = 5a, so g=5ag = 5a and a=9.85=1.96m s2a = \dfrac{9.8}{5} = 1.96\,\text{m s}^{-2}.

Substitute into the second equation: T=2g+2a=2(9.8)+2(1.96)=19.6+3.92=23.5NT = 2g + 2a = 2(9.8) + 2(1.96) = 19.6 + 3.92 = 23.5\,\text{N}.

Markers reward two correct equations of motion (one per particle, with consistent acceleration direction), adding to eliminate TT, and finding both a=1.96m s2a = 1.96\,\text{m s}^{-2} and T=23.5NT = 23.5\,\text{N}. The tension is the same throughout a light string over a smooth pulley.

Related dot points

Sources & how we know this