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CIE A-Level Maths Study Notes

3.1.5 Frictional Forces and Limiting Equilibrium

In exploring the concepts of frictional forces and limiting equilibrium, this discussion aims to provide a thorough understanding of these essential mechanics topics, which are vital for students.

Understanding Frictional Forces

  • Friction: A force that stops objects from moving easily against each other.
  • Two Types:
    • Static Friction: Stops objects from starting to move.
    • Kinetic Friction: Slows down moving objects, usually less than static friction.
  • Depends on: Surface types and the force pushing the surfaces together.
Frictional Forces

Image courtesy of Jackwestin

Coefficient of Friction (μ)

  • What it is: A number showing how much friction surfaces produce.
  • Formula: μ = Friction Force / Force pushing surfaces together.
  • Key Point: Bigger μ means more friction.

Limiting Equilibrium

  • The point where an object is about to move.
  • Max Friction = μ times the force pushing surfaces together.

Example Problems

Problem 1: Calculating Limiting Friction

Question: Find the limiting friction for a 10 kg box on a surface, with a static friction coefficient of 0.5.

Solution:

  • Normal Reaction (R): R=mass×gravity=10 kg×9.8 m/s2=98 N R = \text{mass} \times \text{gravity} = 10 \text{ kg} \times 9.8 \text{ m/s}^2 = 98 \text{ N}
  • Limiting Friction (F): F=μ×R=0.5×98 N=49 N F = \mu \times R = 0.5 \times 98 \text{ N} = 49 \text{ N}
  • Result: Limiting friction is 49N49 N.
  • Graph:
Limiting Friction

Problem 2: Limiting Equilibrium on an Incline

Question: Check if a 5 kg box on a 30° incline (static friction coefficient 0.3) is in limiting equilibrium.

Solution:

  • Normal Reaction (R): R=mass×gravity×cos(θ)=5 kg×9.8 m/s2×cos(30°)42.43 NR = \text{mass} \times \text{gravity} \times \cos(\theta) = 5 \text{ kg} \times 9.8 \text{ m/s}^2 \times \cos(30°) \approx 42.43 \text{ N}
  • Parallel Force: =mass×gravity×sin(θ)=5 kg×9.8 m/s2×sin(30°)24.5 N= \text{mass} \times \text{gravity} \times \sin(\theta) = 5 \text{ kg} \times 9.8 \text{ m/s}^2 \times \sin(30°) \approx 24.5 \text{ N}
  • Maximum Frictional Force (F): F=μ×R=0.3×42.43 N12.73 NF = \mu \times R = 0.3 \times 42.43 \text{ N} \approx 12.73 \text{ N}
  • Check: Parallel Force 24.5N24.5 N > Max Friction 12.73N12.73 N, so not in limiting equilibrium.
  • Result: The box is not in limiting equilibrium.
  • Graph:
 Limiting Equilibrium on an Incline
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
LinkedIn
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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