# Friction equation

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.

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## 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.