Friction equation

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Friction equation
Friction equation | Image: Learnool

The friction equation establishes a relationship between the friction force (Ff), the coefficient of friction (µ), and the normal force (FN). Mathematically, it is represented as Ff = µ × FN. This equation enables the determination of the frictional force acting on an object by taking into account the coefficient of friction and the corresponding normal force.

Practice problems

Problem #1

Calculate the friction force acting on a box of mass 9 kg as it is pushed along the floor. The coefficient of friction between the box and the floor is 0.09.

Solution

Given data:

  • Friction force acting on a box, Ff = ?
  • Mass of a box, m = 9 kg
  • Coefficient of friction between the box and floor, µ = 0.09
  • Assuming gravitational acceleration, g = 9.81 m/s2

Applying the formula:

Therefore, the friction force acting on a box is 7.94 N.

Problem #2

Determine the value of the friction force acting on an ice slab of mass 12 kg as it is pulled along a cement floor. The coefficient of friction between the ice slab and the cement floor is 0.35.

Solution

Given data:

  • Friction force acting on an ice slab, Ff = ?
  • Mass of an ice slab, m = 12 kg
  • Coefficient of friction between the ice slab and cement floor, µ = 0.35
  • Assuming gravitational acceleration, g = 9.81 m/s2

Applying the formula:

Therefore, the friction force acting on an ice slab is 41.20 N.

Problem #3

A 2 kg book resting on a desk is pushed with a hand. Calculate the friction force acting on the book, given that the coefficient of friction between the book and the desk is 0.24.

Solution

Given data:

  • Mass of a book, m = 2 kg
  • Friction force acting on a book, Ff = ?
  • Coefficient of friction between the book and desk, µ = 0.24
  • Assuming gravitational acceleration, g = 9.81 m/s2

Applying the formula:

Therefore, the friction force acting on a book is 4.70 N.

Problem #4

Calculate the friction force acting on a 500 gm knife as it is sharpened on a knife sharpening stone. The coefficient of friction between the knife and the knife sharpening stone is 0.75.

Solution

Given data:

  • Friction force acting on a knife, Ff = ?
  • Mass of a knife, m = 500 gm = 0.5 kg
  • Coefficient of friction between a knife and knife sharpening stone, µ = 0.75
  • Assuming gravitational acceleration, g = 9.81 m/s2

Applying the formula:

Therefore, the friction force acting on a knife is 3.67 N.

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Deep

Learnool.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.

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