This document discusses the laws of friction and provides examples of how to calculate coefficients of friction and frictional forces. It defines key terms like coefficient of friction, normal force, static friction, and kinetic friction. Formulas are provided to calculate the coefficient of friction and frictional force. Two sample problems demonstrate how to apply the formulas to find the coefficient of friction or frictional force given values for normal force, applied force, mass, and angle of inclination. Finally, two practice exercises are presented.

1. APPLICATION OF
FRICTION
Laws of Friction
Problems involving Static and Kinetic Friction

2. LAWS OF FRICTION:
1. Friction is parallel to the surfaces that are in contact and is in
the opposite direction of the motion of the object.
2. Friction depends on the nature of the materials in contact.
3. The sliding friction is usually less than the starting friction.
4. Sliding friction is independent of speed.
5. Friction is practically independent of the area of contact.
6. Friction is directly proportional to the force pressing
surfaces together.

3. COEFFICIENT OF
FRICTION

4. DEFINITION OF TERMS
Coefficient of friction- ratio of frictional force and
the normal force
Normal force- perpendicular force pressing the
surfaces together

5. FORMULA
For static friction:
𝜇 =
𝐹𝑓
𝐹𝑛
For kinetic friction:
𝐹𝑛 = 𝑚𝑔 𝑐𝑜𝑠𝜃
𝐹𝑤 = 𝑚𝑔

6. SAMPLE PROBLEM #1
As Nestor is taking a bath, the soap falls out of the soap dish and
he steps on it with a force of 400 N. If Nestor slides forward and the
frictional force between the soap and the floor in 40N, what is the
coefficient of the friction between these two surfaces.
Given: Fn= 400 NFf= 40 N
Solution: 𝜇 =
𝐹𝑓
𝐹𝑛
𝜇 =
40𝑁
400𝑁
𝜇 = 0.1

7. SAMPLE PROBLEM #2
At an animal circus show, a 900kg sea lion slides down a wet slide
inclined at an angle of 28° to the horizontal. The coefficient of
friction between the sea lion and the slide is 0.40. What frictional
force slows the sea lion’s motion down the slide?
Given: m= 900kg 𝜃= 28° g= 9.8m/s2 𝜇= 0.40
Solution: 𝐹𝑛 = 𝑚𝑔 𝑐𝑜𝑠𝜃 Fn= (900)(9.8)cos 28° Fn= 7787.60N
𝜇 =
𝐹𝑓
𝐹𝑛
Ff= 𝜇Fn Ff= (0.40)(7787.60) Ff= 3115.04 N

8. PRACTICE EXERCISE
1. A man prevents a 15kg wood from falling by pressing it
against a vertical wall. What force must he use if the
coefficient friction is 0.3?
2. The coefficient sliding friction between a loaded
refrigerator and the floor is 0.40. If a normal force of 60
N acting 35° above the horizontal gives the refrigerator
a constant velocity, what is the weight of the loaded
refrigerator?