friction f friction = μf normal. f f = μf n f f = friction force (direction is always the opposite...
TRANSCRIPT
![Page 1: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/1.jpg)
Friction
Ffriction = μFNormal
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Ff = μFN
• Ff = friction force (direction is always the opposite of velocity causing it to slow objects down)
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Ff = μFN
• Ff = friction force (direction is always the opposite of velocity causing it to slow objects down)
• μ = coefficient of friction (experimentally determined value based on the two surfaces in contact with one another)
![Page 4: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/4.jpg)
Ff = μFN
• Ff = friction force (direction is always the opposite of velocity causing it to slow objects down)
• μ = coefficient of friction (experimentally determined value based on the two surfaces in contact with one another)
• FN = normal force (force perpendicular to the surface boundary)
![Page 5: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/5.jpg)
Types of Friction
• static friction – friction that prevents an object from moving.
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Types of Friction
• static friction – friction that prevents an object from moving.
• kinetic (sliding) friction – friction that slows down an object in motion.
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Types of Friction
• static friction – friction that prevents an object from moving.
• kinetic (sliding) friction – friction that slows down an object in motion.
• static friction is always greater than kinetic friction for the same object.
![Page 8: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/8.jpg)
Types of Friction
• static friction – friction that prevents an object from moving.
• kinetic (sliding) friction – friction that slows down an object in motion.
• static friction is always greater than kinetic friction for the same object.
• This means that it takes more force to get an object moving than it takes to keep it moving.
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Friction on Horizontal Surfaces
• On a horizontal surface, the normal force will always be equal to the weight of the object.
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Friction on Horizontal Surfaces
• On a horizontal surface, the normal force will always be equal to the weight of the object.
• Example: You push a 25.0 kg wooden box across a wooden floor at a constant speed. How much force do you exert on the box?
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Example: You push a 25.0 kg wooden box across a wooden floor at a constant speed. How much force do you exert on the box?• Start by drawing a free-body diagram.
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Example: You push a 25.0 kg wooden box across a wooden floor at a constant speed. How much force do you exert on the box?• Start by drawing a free-body diagram.
![Page 13: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/13.jpg)
Example: You push a 25.0 kg wooden box across a wooden floor at a constant speed. How much force do you exert on the box?• Start by drawing a free-body diagram.
• All of the forces must cancel out because the box is moving at a constant speed.
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• This means that the pushing force must be the same magnitude as the friction force.
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• This means that the pushing force must be the same magnitude as the friction force.
• The coefficient of friction can be obtained from the table on p.129. We want the value for kinetic friction since the box is moving.
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• This means that the pushing force must be the same magnitude as the friction force.
• The coefficient of friction can be obtained from the table on p.129. We want the value for kinetic friction since the box is moving.
• The normal force will be the weight of the box since the surface boundary is horizontal.
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• This means that the pushing force must be the same magnitude as the friction force.
• The coefficient of friction can be obtained from the table on p.129. We want the value for kinetic friction since the box is moving.
• The normal force will be the weight of the box since the surface boundary is horizontal.
• Ff = (0.20)(25.0 kg x 9.80 m/s2) = 49.0 N
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Calculate the minimum force required to get the box moving from rest in the previous example.
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Calculate the minimum force required to get the box moving from rest in the previous example.
• This time we must use the coefficient of static friction from the table. The normal force remains the same.
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Calculate the minimum force required to get the box moving from rest in the previous example.
• This time we must use the coefficient of static friction from the table. The normal force remains the same.
• Ff = (0.50)(25.0 kg x 9.80 m/s2) = 123 N
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Calculate the minimum force required to get the box moving from rest in the previous example.
• This time we must use the coefficient of static friction from the table. The normal force remains the same.
• Ff = (0.50)(25.0 kg x 9.80 m/s2) = 123 N
• It takes 123 N to get the box moving, but just 49 N to keep it moving.
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Why do you suppose it is bad to lock the brakes on a moving vehicle?
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Why do you suppose it is bad to lock the brakes on a moving vehicle?
• When the wheel is not spinning, the friction produced is kinetic friction.
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Why do you suppose it is bad to lock the brakes on a moving vehicle?
• When the wheel is not spinning, the friction produced is kinetic friction.
• When the wheel is spinning, static friction is taking place.
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Why do you suppose it is bad to lock the brakes on a moving vehicle?
• When the wheel is not spinning, the friction produced is kinetic friction.
• When the wheel is spinning, static friction is taking place.
• Because static friction is always greater, there is more friction between your tires and the road when the wheels are spinning. This gives you greater control over the vehicle.
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What pushing force is needed to accelerate the 25.0 kg box at a rate of 1.0 m/s2?
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What pushing force is needed to accelerate the 25.0 kg box at a rate of 1.0 m/s2?
• Draw another free body diagram.
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What pushing force is needed to accelerate the 25.0 kg box at a rate of 1.0 m/s2?
• Draw another free body diagram.
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What pushing force is needed to accelerate the 25.0 kg box at a rate of 1.0 m/s2?
• Draw another free body diagram.
• To get the box to accelerate, the pushing force has to be greater than the friction force.
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• The net force can be calculated using Newton’s 2nd Law (Fnet = ma)
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• The net force can be calculated using Newton’s 2nd Law (Fnet = ma)
• Fnet = (25.0 kg)(1.0 m/s2) = 25 N
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• The net force can be calculated using Newton’s 2nd Law (Fnet = ma)
• Fnet = (25.0 kg)(1.0 m/s2) = 25 N
• This means that the pushing force is 25 N greater than the friction force.
![Page 33: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/33.jpg)
• The net force can be calculated using Newton’s 2nd Law (Fnet = ma)
• Fnet = (25.0 kg)(1.0 m/s2) = 25 N
• This means that the pushing force is 25 N greater than the friction force.
• The friction force is always the same for this box on this surface Ff = (0.20)(25.0 kg x 9.80 m/s2) = 49.0 N
![Page 34: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/34.jpg)
• The net force can be calculated using Newton’s 2nd Law (Fnet = ma)
• Fnet = (25.0 kg)(1.0 m/s2) = 25 N
• This means that the pushing force is 25 N greater than the friction force.
• The friction force is always the same for this box on this surface Ff = (0.20)(25.0 kg x 9.80 m/s2) = 49.0 N
• The pushing force must be 74.0 N (49 + 25).
![Page 35: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/35.jpg)
Homework
• Read section 5.2 (p.126-130)• Problems 17-26 beginning on p.128
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Friction on an incline
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Friction on an incline
• The normal force is no longer the weight of the object.
![Page 38: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/38.jpg)
Friction on an incline
• The normal force is no longer the weight of the object.
• The weight vector can be broken down into 2 components, parallel to the incline and perpendicular to the incline.
![Page 39: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/39.jpg)
Friction on an incline
• The normal force is no longer the weight of the object.
• The weight vector can be broken down into 2 components, parallel to the incline and perpendicular to the incline.
![Page 40: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/40.jpg)
Friction on an incline
• The normal force is no longer the weight of the object.
• The weight vector can be broken down into 2 components, parallel to the incline and perpendicular to the incline.
• The perpendicular component is the normal force.
![Page 41: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/41.jpg)
Friction on an incline
• The normal force is no longer the weight of the object.
• The weight vector can be broken down into 2 components, parallel to the incline and perpendicular to the incline.
• The perpendicular component is the normal force.• The parallel component represents a pulling force.
![Page 42: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/42.jpg)
Calculating normal force
• SOH CAH TOA can be used to determine both the normal force and the pulling force.
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Calculating normal force
• SOH CAH TOA can be used to determine both the normal force and the pulling force.
• cos21o = FN/weight (w = mg)
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Calculating normal force
• SOH CAH TOA can be used to determine both the normal force and the pulling force.
• cos21o = FN/weight (w = mg)• sin21o = Fpull/weight (w = mg)
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Determine the coefficient of static friction for a metal box on a wooden ramp if the box has a mass of 13.5 kg and begins to slide down the
ramp when the incline reaches 38o.
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Determine the coefficient of static friction for a metal box on a wooden ramp if the box has a mass of 13.5 kg and begins to slide down the
ramp when the incline reaches 38o.• Start by drawing a free-body diagram. It is
helpful to draw the weight force as 2 component forces (parallel and perpendicular)
![Page 47: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/47.jpg)
Determine the coefficient of static friction for a metal box on a wooden ramp if the box has a mass of 13.5 kg and begins to slide down the
ramp when the incline reaches 38o.• Start by drawing a free-body diagram. It is
helpful to draw the weight force as 2 component forces (parallel and perpendicular)
• The box will begin to slide when the friction force and the parallel force are equal.
![Page 48: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/48.jpg)
Determine the coefficient of static friction for a metal box on a wooden ramp if the box has a mass of 13.5 kg and begins to slide down the
ramp when the incline reaches 38o.• Start by drawing a free-body diagram. It is
helpful to draw the weight force as 2 component forces (parallel and perpendicular)
• The box will begin to slide when the friction force and the parallel force are equal.
• sin38o = (F||)/(132 N)
![Page 49: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/49.jpg)
Determine the coefficient of static friction for a metal box on a wooden ramp if the box has a mass of 13.5 kg and begins to slide down the
ramp when the incline reaches 38o.• Start by drawing a free-body diagram. It is
helpful to draw the weight force as 2 component forces (parallel and perpendicular)
• The box will begin to slide when the friction force and the parallel force are equal.
• sin38o = (F||)/(132 N)
• F|| = 81.3 N
![Page 50: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/50.jpg)
• F|| = 81.3 N
• Next, we need to determine the normal force (perpendicular component) to use in the friction equation.
![Page 51: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/51.jpg)
• F|| = 81.3 N
• Next, we need to determine the normal force (perpendicular component) to use in the friction equation.
• cos38o = (FN)/(132 N)
![Page 52: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/52.jpg)
• F|| = 81.3 N
• Next, we need to determine the normal force (perpendicular component) to use in the friction equation.
• cos38o = (FN)/(132 N)
• FN = 104 N
![Page 53: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/53.jpg)
• F|| = 81.3 N
• Next, we need to determine the normal force (perpendicular component) to use in the friction equation.
• cos38o = (FN)/(132 N)
• FN = 104 N
• Now, we can solve for μ.81.3 N = μ(104 N)
![Page 54: Friction F friction = μF Normal. F f = μF N F f = friction force (direction is always the opposite of velocity causing it to slow objects down)](https://reader036.vdocuments.us/reader036/viewer/2022081503/5697bf721a28abf838c7ec3f/html5/thumbnails/54.jpg)
• F|| = 81.3 N
• Next, we need to determine the normal force (perpendicular component) to use in the friction equation.
• cos38o = (FN)/(132 N)
• FN = 104 N
• Now, we can solve for μ.81.3 N = μ(104 N)μ = 0.782