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APPLICATION OF
FRICTION
Laws of Friction
Problems involving Static and Kinetic Friction
LAWS OF FRICTION:
1. Friction is parallel to the surfaces that are in contact and is in
the opposite direction of the moti...
COEFFICIENT OF
FRICTION
DEFINITION OF TERMS
Coefficient of friction- ratio of frictional force and
the normal force
Normal force- perpendicular fo...
FORMULA
For static friction:
𝜇 =
𝐹𝑓
𝐹𝑛
For kinetic friction:
𝐹𝑛 = 𝑚𝑔 𝑐𝑜𝑠𝜃
𝐹𝑤 = 𝑚𝑔
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...
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 ho...
PRACTICE EXERCISE
1. A man prevents a 15kg wood from falling by pressing it
against a vertical wall. What force must he us...
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Application of friction

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Application of friction

  1. 1. APPLICATION OF FRICTION Laws of Friction Problems involving Static and Kinetic Friction
  2. 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. 3. COEFFICIENT OF FRICTION
  4. 4. DEFINITION OF TERMS Coefficient of friction- ratio of frictional force and the normal force Normal force- perpendicular force pressing the surfaces together
  5. 5. FORMULA For static friction: 𝜇 = 𝐹𝑓 𝐹𝑛 For kinetic friction: 𝐹𝑛 = 𝑚𝑔 𝑐𝑜𝑠𝜃 𝐹𝑤 = 𝑚𝑔
  6. 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. 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. 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?

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