Application of friction
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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.
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Friction is an important term, By this presentation I have learned lot about friction, types of friction, major types, application of friction, static and kinetic friction, law of friction, Co efficient of frictions.
I think this all information help you all to understand friction. This information is given by lot of legal sites.
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Application of friction
- 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.
- 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?