Friction Pages 23-26 in text 4/9/2019 Dr. Sasho MacKenzie - HK 376

What is Friction Friction is a force A frictional force can exist when two substances contact each other. 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. If there is no motion, then friction opposes the sum of all the other forces which are parallel to the surfaces in contact. 4/9/2019 Dr. Sasho MacKenzie - HK 376

> Types of Friction Dry Friction Fluid Friction Static Friction Occurs with fluids,or lubricated surfaces Dry Friction Occurs between the non-lubricated surfaces of solid objects Static Friction When dry friction acts between two surfaces that are not moving relative to each other Dynamic Friction When dry friction acts between two surfaces that are moving relative to each other > 4/9/2019 Dr. Sasho MacKenzie - HK 376

Contact Force 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. The perpendicular component is called the normal force. The parallel component is called friction. 4/9/2019 Dr. Sasho MacKenzie - HK 376

Contact Force in Running Resultant force on runner runner’s push 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. 4/9/2019 Dr. Sasho MacKenzie - HK 376

Friction and the Normal 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. This is because the molecules on the two surfaces are pushed together more, thus increasing their interactions. 4/9/2019 Dr. Sasho MacKenzie - HK 376

 Weight means  Normal Force, and therefore,  Maximum Friction 10 kg 5 kg Surfaces are more compressed together and there are more interactions between molecules 4/9/2019 Dr. Sasho MacKenzie - HK 376

Friction and Surface Area 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. Amontons (1699) What about race car tires? 4/9/2019 Dr. Sasho MacKenzie - HK 376

Calculating Friction Ff_max = FN Ff_max is the maximum force of friction  (Mu) is the coefficient of friction FN is the normal force Friction can range in value from -Ff_max to +Ff_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 ) 4/9/2019 Dr. Sasho MacKenzie - HK 376

Friction is not always = FF_max Fapplied Assume: FF_max = 50 N 10.2 kg FFriction 50 N 50 N 10 N 25 N 0 N 50 N 40 N 25 N 10 N 0 N Static Friction Dynamic Friction 50 Ffriction (magnitude) What is the acceleration? 25 What is the normal force? Calculate S? 10 Calculate D? Fapplied 4/9/2019 Dr. Sasho MacKenzie - HK 376

4/9/2019 Dr. Sasho MacKenzie - HK 376

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. FN mg Ff Fh Free body diagram 5 kg Fh Fy = may FN – mg = may = 0 FN = mg Normal force = FN = mg = 5 x 9.81 = 49 N Fx = max Fh – Ff = max = 0 Fh = Ff Fh= Friction force = FN = 0.4 x 49 = 19.6 N 4/9/2019 Dr. Sasho MacKenzie - HK 376

Horse Pulling Cart According to Newton’s 3rd 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? 4/9/2019 Dr. Sasho MacKenzie - HK 376

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 N Force of friction on the wheel which opposes the motion of the horse-cart system 4/9/2019 Dr. Sasho MacKenzie - HK 376

Tug of War Fat Bastard vs. Phil Pfister 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? 4/9/2019 Dr. Sasho MacKenzie - HK 376

Two Free Body Diagrams Fat Bastard Pfister 3000 N 3000 N 2060 N 1180 N Ff = FN = 1.5 x 1180 = 1770 N 1180 N 2060 N Ff = FN = 1.5 x 2060 = 3090 N 4/9/2019 Dr. Sasho MacKenzie - HK 376

Fat Bastard Wins 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. Pfister’s maximum friction force (1170 N) is less than the force of rope, so he is pulled toward Fat Bastard. 4/9/2019 Dr. Sasho MacKenzie - HK 376

Would it be better to pull up or down on the rope? 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. Who has the advantage? 4/9/2019 Dr. Sasho MacKenzie - HK 376

Pulling Up On The Rope Fy = may rope FN – Fy1 – mg = may = 0 FN = mg + Fy1 Friction force = Fx1 = FN Fx2 Bigger N, means larger friction force This component increases N mg Fy1 FN Fx1 4/9/2019 Dr. Sasho MacKenzie - HK 376

Pulling Down On The Rope Fy = may FN + Fy1 – mg = may = 0 FN = mg – Fy1 This component decreases N Fy1 Friction force = Fx1 = FN Fx2 Smaller N, means less friction force mg FN Fx1 4/9/2019 Dr. Sasho MacKenzie - HK 376

Midterm Example Question 5 kg y Fx1 x 40 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 Fx1? Remember that friction always opposes the direction of motion. 4/9/2019 Dr. Sasho MacKenzie - HK 376