A-Level Mathematics May/June 2025 Q5(b): Find the coefficient of friction between A and the inclined plane.

1 answer

• ▲0
  accepted ✓

  Let $\mu$ be the coefficient of friction between particle A and the plane. We are given the mass of A, $m_A = 3$ kg, the angle of inclination, $\theta = 30°$, and the acceleration up the plane, $a = 2$ m s$^{-2}$. From part (a), the tension in the string is $T_{AB} = 32$ N.

  First, resolve forces perpendicular to the inclined plane to find the normal contact force, $R$.

  $R - 3g \cos 30° = 0$ $R = 3g \cos 30°$ Using $g=10$ m s$^{-2}$: $R = 3(10) \left(\frac{\sqrt{3}}{2}\right)$ $R = 15\sqrt{3}$ N

  Next, apply Newton's Second Law ($F=ma$) for particle A, parallel to the plane. The particle is accelerating up the plane, so the net force is in the upward direction. The frictional force, $F$, opposes the motion and therefore acts down the plane.

  $T_{AB} - F - 3g \sin 30° = 3a$

  Substitute the known values: $32 - F - 3(10) \sin 30° = 3(2)$ $32 - F - 3(10)(0.5) = 6$ $32 - F - 15 = 6$ $17 - F = 6$ $F = 11$ N

  Since the particle is moving, friction is limiting, so we use the formula $F = \mu R$.

  $11 = \mu (15\sqrt{3})$

  Now, solve for $\mu$: $\mu = \frac{11}{15\sqrt{3}}$

  To rationalise the denominator: $\mu = \frac{11 \times \sqrt{3}}{15\sqrt{3} \times \sqrt{3}} = \frac{11\sqrt{3}}{15 \times 3} = \frac{11\sqrt{3}}{45}$

  As a decimal to 3 significant figures: $\mu = 0.4233... \approx 0.423$

  So, the coefficient of friction is $\frac{11\sqrt{3}}{45}$ or $0.423$.

  ---

  How the marks are awarded

  • B1 — For correctly resolving forces perpendicular to the plane to find the correct expression for the normal contact force, $R = 3g \cos 30°$.
  • M1 — For applying Newton's Second Law parallel to the plane for particle A. The equation must have the correct four terms (Tension, Friction, Weight Component, $ma$), even with sign errors. The value of tension is taken from the previous part of the question.
  • DM1 — For using the relationship for limiting friction, $F = \mu R$, and substituting the expressions for $F$ (from the parallel motion equation) and $R$ (from the perpendicular resolution) to form and solve an equation for $\mu$.
  • A1 — For the correct final answer, either in exact form as $11\sqrt{3}/45$ or as a decimal rounded to 3 s.f., $0.423$.

  Common mistakes

  • Incorrectly setting the direction of the frictional force. Since the particle accelerates up the plane, friction must act down the plane to oppose the motion.
  • Mixing up sine and cosine when resolving the weight component. The component parallel to the slope is $mg \sin \theta$ and the component perpendicular is $mg \cos \theta$.
  • Sign errors when setting up the equation of motion, for example, adding the frictional force or weight component to the tension instead of subtracting them.
  • Incorrectly assuming the normal reaction force $R$ is equal to $mg$, instead of resolving forces perpendicular to the inclined plane to find $R = mg \cos \theta$.

  Examiner tip: Always draw a clear force diagram for each particle and resolve forces parallel and perpendicular to the direction of acceleration.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers

Your answer

Sign in to answer this question.