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5/9/2015Dr. Sasho MacKenzie - HK 3761 Friction Pages 23-26 in text.

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Presentation on theme: "5/9/2015Dr. Sasho MacKenzie - HK 3761 Friction Pages 23-26 in text."— Presentation transcript:

1 5/9/2015Dr. Sasho MacKenzie - HK 3761 Friction Pages in text

2 5/9/2015Dr. Sasho MacKenzie - HK 3762 What is Friction Friction is a forceFriction is a force A frictional force can exist when two substances contact each other.A frictional force can exist when two substances contact each other. The molecules of each surface interact according to Newton’s Laws of Motion.The molecules of each surface interact according to Newton’s Laws of Motion. Friction always opposes motion, i.e., it is opposite to the direction of velocity.Friction always opposes motion, i.e., it is opposite to the direction of velocity. If there is no motion, then friction opposes the sum of all the other forces which are parallel to the surfaces in contact.If there is no motion, then friction opposes the sum of all the other forces which are parallel to the surfaces in contact.

3 5/9/2015Dr. Sasho MacKenzie - HK 3763 Types of Friction Dry Friction Occurs between the non-lubricated surfaces of solid objects Fluid Friction Occurs with fluids,or lubricated surfaces Dynamic Friction When dry friction acts between two surfaces that are moving relative to each other > Static Friction When dry friction acts between two surfaces that are not moving relative to each other

4 5/9/2015Dr. Sasho MacKenzie - HK 3764 Contact Force Force that occurs between objects that are in contact with each other.Force that occurs between objects that are in contact with each other. Contact forces can be resolved into components that are perpendicular and parallel to the surfaces in contact.Contact forces can be resolved into components that are perpendicular and parallel to the surfaces in contact. The perpendicular component is called the normal force.The perpendicular component is called the normal force. The parallel component is called friction.The parallel component is called friction.

5 5/9/2015Dr. Sasho MacKenzie - HK 3765 Contact Force in Running Normal Force Friction Force During the push off phase in running, the normal force acts upward on the runner, while the friction force acts forward on the runner. The friction force is the only force capable of moving the runner horizontally down the track. The normal force can only accelerate the runner upwards. runner’s pushResultant force on runner

6 5/9/2015Dr. Sasho MacKenzie - HK 3766 Friction and the Normal Force The maximum frictional force is proportional to the normal contact force.The maximum frictional force is proportional to the normal contact force. An increase in the normal force results in an increase in the maximum friction.An increase in the normal force results in an increase in the maximum friction. This is because the molecules on the two surfaces are pushed together more, thus increasing their interactions.This is because the molecules on the two surfaces are pushed together more, thus increasing their interactions.

7 5/9/2015Dr. Sasho MacKenzie - HK 3767  Weight means  Normal Force, and therefore,  Maximum Friction 5 kg 10 kg Surfaces are more compressed together and there are more interactions between molecules

8 5/9/2015Dr. Sasho MacKenzie - HK 3768 Friction and Surface Area Friction is not affected by the size of the surface area in contact.Friction is not affected by the size of the surface area in contact. If the normal force remains constant, but the contacting surface area is increased, then the normal force is spread out over more molecules, thus the force on each molecule is reduced.If the normal force remains constant, but the contacting surface area is increased, then the normal force is spread out over more molecules, thus the force on each molecule is reduced. –Amontons (1699) What about race car tires?What about race car tires?

9 5/9/2015Dr. Sasho MacKenzie - HK 3769 Calculating Friction F f_max =  F NF f_max =  F N F f_max is the maximum force of friction  (Mu) is the coefficient of friction F N is the normal force Friction can range in value from -F f_max to +F f_max  depends on the types of surfaces that are interacting. It would be low for rubber on ice, but high for rubber on asphalt. It also depends on whether the surfaces are moving relative to each other (  static or  dynamic )

10 N 50 N 25 N 50 N 25 N10 N Friction is not always = F F_max 5/9/2015Dr. Sasho MacKenzie - HK kg F applied Dynamic Friction F Friction Static Friction F applied 0 N F friction (magnitude) 0 N 50 N 40 N What is the acceleration? What is the normal force? Calculate  S ? Calculate  D ? Assume: F F_max = 50 N 0 10

11 5/9/2015Dr. Sasho MacKenzie - HK 37611

12 5/9/2015Dr. Sasho MacKenzie - HK Friction Example A 5 kg block of wood rests on a ceramic counter. If the coefficient of static friction between the block and the counter is 0.4, what horizontal force is necessary to move the block. 5 kg FhFh Normal force = F N = mg = 5 x 9.81 = 49 N FNFN mg FfFf FhFh Free body diagram F h = Friction force =  F N = 0.4 x 49 = 19.6 N  F x = ma x F h – F f = ma x = 0 F h = F f  F y = ma y F N – mg = ma y = 0 F N = mg

13 5/9/2015Dr. Sasho MacKenzie - HK Horse Pulling Cart According to Newton’s 3 rd Law, these forces are equal and opposite. So, if the horse pulls forward on the cart with the same force as the cart pulls back on the horse, how will the horse ever move the cart?

14 5/9/2015Dr. Sasho MacKenzie - HK Solution Friction acts on the horse’s feet but very little acts on the wheels of the cart. Drawing a free body diagram reveals the answer. The horse and cart are one system so the forces in between them are internal and cannot produce a change in motion of the system. mg Friction force resulting from the horse pulling back on the ground Force of friction on the wheel which opposes the motion of the horse-cart system N N

15 5/9/2015Dr. Sasho MacKenzie - HK Tug of War Fat Bastard vs. Phil Pfister Fat Bastard Pull Force = 3000 N Mass = 210 kg Height = 1.8 m Pfister Pull Force = 3000 N Mass = 120 kg Height = 1.8 m Both competitors are wearing the same footwear which has a coefficient of friction of 1.5 with the rubber floor they are competing on. If both men employ the same technique, who wins?

16 5/9/2015Dr. Sasho MacKenzie - HK Two Free Body Diagrams Fat BastardPfister 3000 N 2060 N F f =  F N = 1.5 x 2060 = 3090 N 2060 N 1180 N F f =  F N = 1.5 x 1180 = 1770 N

17 5/9/2015Dr. Sasho MacKenzie - HK Fat Bastard Wins Both competitors have a force of 3000 N pulling on them from the rope.Both competitors have a force of 3000 N pulling on them from the rope. Fat Bastard’s extra mass gives him a potential friction force (3090 N) which is greater than the force of the rope, so he doesn’t move.Fat Bastard’s extra mass gives him a potential friction force (3090 N) which is greater than the force of the rope, so he doesn’t move. Pfister’s maximum friction force (1170 N) is less than the force of rope, so he is pulled toward Fat Bastard.Pfister’s maximum friction force (1170 N) is less than the force of rope, so he is pulled toward Fat Bastard.

18 5/9/2015Dr. Sasho MacKenzie - HK Would it be better to pull up or down on the rope? Suppose competitor A was taller than competitor B.Suppose competitor A was taller than competitor B. A would be pulling on an upward angle, while B would be pulling on a downward angle.A would be pulling on an upward angle, while B would be pulling on a downward angle. Who has the advantage?Who has the advantage?

19 5/9/2015Dr. Sasho MacKenzie - HK Pulling Up On The Rope mg FNFN F x1 rope F x2 F y1 This component increases N  F y = ma y F N – F y1 – mg = ma y = 0 F N = mg + F y1 Friction force = F x1 =  F N Bigger N, means larger friction force

20 5/9/2015Dr. Sasho MacKenzie - HK Pulling Down On The Rope mg FNFN F x1 rope F x2 F y1 This component decreases N  F y = ma y F N + F y1 – mg = ma y = 0 F N = mg – F y1 Friction force = F x1 =  F N Smaller N, means less friction force

21 5/9/2015Dr. Sasho MacKenzie - HK Midterm Example Question 5 kg 40  y x F x1 A 5 kg box is being pushed up a 40  incline with an acceleration of -2 m/s/s. If the coefficient of dynamic friction between the incline and box is 0.2, then what is the value of F x1 ? Remember that friction always opposes the direction of motion.


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