A block rests on a rough horizontal surface with μ = 0.4 and weight 50 N. Find the maximum friction before it slides. [2 marks]

ScotlandMathematics of MechanicsSyllabus dot point

How do Newton's laws relate force and motion, and how do you analyse equilibrium, friction, inclined planes and connected particles?

Apply Newton's three laws of motion; draw free-body diagrams; resolve forces; analyse equilibrium, friction, motion on inclined planes, and systems of connected particles.

A focused answer to the SQA Advanced Higher Mathematics of Mechanics content on dynamics, covering Newton's three laws, free-body diagrams, resolving forces, equilibrium, friction with the coefficient of friction, motion on inclined planes, and connected-particle systems using the equation of motion F equals ma.

Generated by Claude Opus 4.814 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
Newton's three laws
Free-body diagrams and resolving
Friction and the inclined plane
Connected particles
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What this dot point is asking

Kinematics describes motion; dynamics explains it. The SQA wants you to use Newton's three laws to connect the forces on a body to its acceleration, to draw a clear free-body diagram, to resolve forces into components, and to handle the standard situations: a body in equilibrium, friction at a surface, a body on an inclined plane, and particles connected by strings over pulleys.

Newton's three laws

The three laws are the foundation of the whole mechanics course.

The second law is the workhorse. The first law is the special case $\mathbf{a} = \mathbf{0}$ (equilibrium). The third law is what lets you treat the tension in a string as the same magnitude at both ends and the normal reaction as a genuine force from the surface.

Free-body diagrams and resolving

The reliable method for any dynamics problem is the same: isolate one body, mark every force acting on it, choose two perpendicular directions, and apply $\mathbf{F} = m\mathbf{a}$ in each.

Box pulled at an angle on smooth ground

A $4$ kg box is pulled along smooth horizontal ground by a force of $20$ N at $60^\circ$ above the horizontal. Find its acceleration ( $g = 9.8$ ).

Step 1: Draw the free-body diagram and identify forces

Isolate the box and mark every force acting on it: weight $4g$ downward, normal reaction $R$ upward, and the applied pull of $20$ N at $60^\circ$ above the horizontal. The ground is smooth so there is no friction.

Step 2: Resolve horizontally and apply the second law

The horizontal component of the pull drives the acceleration. The vertical component does not accelerate the box horizontally, so it plays no role here:

20\cos 60^\circ = 10\ \text{N} = 4a, \quad a = 2.5\ \text{m s}^{-2}.

Step 3: Check the vertical equation

Vertically the box is in equilibrium, which confirms it does not leave the ground:

R + 20\sin 60^\circ = 4g, \quad R = 4(9.8) - 17.3 = 21.9\ \text{N.}

Final answer: the acceleration of the box is $2.5\ \text{m s}^{-2}$ horizontally.

Friction and the inclined plane

A rough surface resists relative sliding. Friction acts along the surface, opposing the direction the body tends to move, and cannot exceed a limiting value set by the normal reaction.

On an inclined plane the natural directions are along the slope and perpendicular to it. The weight $mg$ resolves into $mg\sin\theta$ down the slope and $mg\cos\theta$ into the slope, so $R = mg\cos\theta$ for a body on the plane with no other perpendicular force. Comparing $mg\sin\theta$ (the driving component) with $\mu R$ (the maximum friction) decides whether the body slips.

Connected particles

When two particles are joined by a light inextensible string over a smooth pulley, they share the same magnitude of acceleration and the string tension is the same throughout. Treat each particle separately.

Connected masses over a smooth pulley

Masses $m_1 > m_2$ hang over a smooth pulley. Find the acceleration in terms of $g$ .

Step 1: Write the equation of motion for the heavier mass

Treat each particle separately. For $m_1$ , which descends, taking down as positive gives resultant force $m_1 g - T$ where $T$ is the tension, common to both ends of the string:

m_1 g - T = m_1 a.

Step 2: Write the equation of motion for the lighter mass

For $m_2$ , which rises, the net upward force is $T - m_2 g$ :

T - m_2 g = m_2 a.

Step 3: Add the equations to eliminate the tension

Adding the two equations removes $T$ , leaving an equation in $a$ alone:

(m_1 - m_2)g = (m_1 + m_2)a, \quad a = \frac{(m_1 - m_2)g}{m_1 + m_2}.

Final answer: the acceleration of the system is $\dfrac{(m_1 - m_2)g}{m_1 + m_2}$ , which is less than $g$ because the lighter mass resists the fall.

Try this

Q1. A $2$ kg mass on smooth horizontal ground is pushed by a $6$ N horizontal force. Find the acceleration. [2 marks]

Cue. $F = ma$ : $6 = 2a$ , so $a = 3$ m s $^{-2}$ .

Q2. A block rests on a rough horizontal surface with $\mu = 0.4$ and weight $50$ N. Find the maximum friction before it slides. [2 marks]

Cue. $R = 50$ N (horizontal surface), so $F_{\max} = \mu R = 0.4 \times 50 = 20$ N.

Exam-style practice questions

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

AH style: inclined plane6 marksA block of mass

5

kg rests on a rough plane inclined at

30^\circ

to the horizontal. The coefficient of friction is

\mu = 0.3

. Find the frictional force when the block is on the point of slipping down, and determine whether it does slip. Take

g = 9.8

m s

^{-2}

Show worked answer →

Resolve perpendicular to the plane: the normal reaction is $R = mg\cos 30^\circ = 5(9.8)(0.866) = 42.4$ N (2 marks).

Maximum (limiting) friction is $F_{\max} = \mu R = 0.3 \times 42.4 = 12.7$ N (1 mark).

The component of weight down the plane is $mg\sin 30^\circ = 5(9.8)(0.5) = 24.5$ N (2 marks).

Since $24.5 > 12.7$ , the driving force exceeds the maximum friction, so the block slips down the plane (1 mark). Markers reward resolving both perpendicular and parallel to the plane and comparing the weight component with limiting friction.

AH style: connected particles5 marksMasses of

3

kg and

2

kg hang from a light inextensible string over a smooth pulley. Find the acceleration of the system and the tension in the string. Take

g = 9.8

m s

^{-2}

Show worked answer →

For the $3$ kg mass (descending): $3g - T = 3a$ . For the $2$ kg mass (rising): $T - 2g = 2a$ (2 marks).

Add the equations to eliminate $T$ : $3g - 2g = 5a$ , so $g = 5a$ , giving $a = \dfrac{9.8}{5} = 1.96$ m s $^{-2}$ (2 marks).

Substitute back: $T = 2g + 2a = 2(9.8) + 2(1.96) = 23.5$ N (1 mark). Markers reward an equation of motion for each particle in its direction of motion, eliminating $T$ , then solving for the tension.

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Sources & how we know this

Advanced Higher Mathematics of Mechanics Course Specification (C802 77) — SQA (2019)