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SPH 4U Unit #1 Dynamics Topic #9: Friction (Teacher) 1.9.1 Introduction The study of friction is called tribology. The force of friction is defined as: the force that acts between objects in contact to oppose their relative motion. The force of friction acting between two surfaces has three properties: i) It opposes their relative motion ii) It depends on the strength of the force pushing the surfaces together (the normal). iii) (don’t mention this one until after the lab is handed in) does not depend on the surface area in contact. Consider the case of a 100 N. mass on a horizontal surface as shown below: If we apply a horizontal force on the mass as shown, a static frictional force from the interlocking of the irregularities of two surfaces will resist the motion. If the applied force is smaller than the maximum friction force that could be developed, the static friction force will prevent any relative motion. When the applied force becomes sufficiently large, there will be a net force and the mass will begin to accelerate. A graph of this is shown below: Page 1 of 10 SPH 4U Unit #1 Dynamics Topic #9: Friction (Teacher) It is that threshold of motion (i.e. the starting friction force) which can be used to calculate the coefficient of static friction. Once the block begins to move, the friction force is reduced as a result of a smaller coefficient of friction. As a result, the applied force that is required to maintain constant velocity becomes smaller. In general, the force due to friction is calculated as: for static friction and for kinetic friction The coefficients of friction depend on the nature of both surfaces in contact. The static friction coefficient is larger than the corresponding kinetic friction coefficient. A chart of friction coefficients is shown below: Surfaces in Contact Coefficient of Static Friction Coefficient of Kinetic Friction Oak on oak, dry 0.40 0.30 Waxed hickory on snow, dry 0.06 0.04 Waxed hickory on snow, wet 0.20 0.14 Steel on steel, dry 0.60 0.41 Steel on steel, greasy 0.15 0.06 Steel on ice 0.013 0.011 Rubber on concrete, dry 1.0 0.8 Rubber on concrete, wet 0.7 0.5 Rubber on asphalt, dry 1.2 0.8 Rubber on asphalt, wet 0.6 0.5 0.006 0.005 Ice on ice 0.1 0.03 Teflon on teflon 0.04 0.04 Synovial joints in humans 0.01 0.003 Glass on glass 0.94 0.40 Rubber on ice Page 2 of 10 SPH 4U Unit #1 Dynamics Topic #9: Friction (Teacher) The coefficient of friction allows us to calculate the magnitude of the friction force if we know the magnitude of the normal force. We can do this using the formula: i.e. (1.9) Note that this is not a vector equation. (Why not?) Eg.#1 A 10. kg. block of steel is moving on steel with motion to the right as shown. What is the force due to friction? First, we’ll find the magnitude of the Normal force: Next, we determine the coefficient of kinetic friction from the table. For dry steel on steel, Now, we can determine the magnitude of the friction force using the formula above Page 3 of 10 SPH 4U Unit #1 Dynamics Topic #9: Friction (Teacher) Eg.#2 A 1.2 tonne car has its brakes locked and comes to a stop on an asphalt road. a) Calculate the friction force if the road surface is dry b) Calculate the friction force if the road surface is wet c) Comment on the results obtained. A car trying to stop on a wet road will require a greater distance in order to stop. 1.9.2 Kinetic Friction There are three different types of kinetic friction: 1.9.2a Rolling friction The resistance to motion which occurs between a rolling wheel and the surface on which it rotates Caused by small distortions of the two surfaces 1.9.2b Sliding Friction The resistance to motion which occurs when one surface slides over another with which it is in contact 1.9.2c Fluid Friction Friction between an object and the liquid through which it moves, whether that fluid be a liquid, gas or vapor. To recap, static friction is present whenever the frictional force opposes placing a body at rest into motion. Kinetic friction is present whenever the frictional force tends to slow a body that is in motion. In general, the maximum static friction force is greater than the kinetic friction force. Page 4 of 10 SPH 4U Unit #1 Dynamics Topic #9: Friction (Teacher) 1.9.3 Inclined Planes 1.9.3a Static Equilibrium on an Inclined Plane Consider a mass that remains stationary on an inclined plane as shown. An interesting result is obtained by applying Newton’s Second Law to this situation. In order that the mass doesn’t slide down the plane, the surface must exert two forces: • • a normal force perpendicular to the plane a friction force parallel to the plane. These two forces when combined will be equal in magnitude and opposite in direction to the force of gravity. Before we apply Newton’s Second Law to this situation, we must resolve the weight into two components, one parallel to the surface and one perpendicular to the surface. The component of the weight acting at 90/ to the surface is The component of the weight acting parallel to the surface is A diagram that shows all of the forces we have discussed is shown below: Page 5 of 10 SPH 4U Unit #1 Dynamics Topic #9: Friction (Teacher) Newton’s Second Law, when applied to the forces perpendicular to the plane yields: When applied to the forces acting along the plane, Newton’s Second Law yields: As long as the block remains stationary, This shows us that, if we know for the surfaces in contact, we can elevate an inclined plane until the tangent of the angle is equal to or less than the friction coefficient and the object will not slide down the plane. If we complete the same analysis with the block moving at a constant velocity ( the analysis will show that Eg. #4. How does the angle of the inclined plane compare when the object moves at constant velocity to the angle it has when the object is stationary? The angle will be less because the coefficient is less. Page 6 of 10 ), SPH 4U Unit #1 Dynamics Topic #9: Friction (Teacher) Eg.#5 A 75.0 kg. crate sitting on a variable angle ramp remains stationary if the angle of the ramp is 37° or less. a) What type of friction is at work here? b) What is the coefficient of friction? Static friction c) Once the crate begins moving, it will continue moving only if the angle of the ramp is 33° or more. What is the type of friction at work here? Kinetic friction d) What is the new coefficient of friction? Page 7 of 10 SPH 4U Unit #1 Dynamics Topic #9: Friction (Teacher) Worksheet 1.9 1. A car accelerates southward because there is a force exerted on the tires by the pavement. a) What is the direction of this force? b) Why does this force exist? [S] It is a reaction force to the force that is applied by the tires onto the road c) Is this friction force static or kinetic? Explain your answer. It is a static friction force because the tires do not move relative to the road surface. 2. The coefficients of kinetic and static friction between a 23 kg. exercise mat and the floor are: and . a) Determine the magnitude of the b) Once the mat is moving what minimum horizontal force that would be minimum force is required in order to needed to get this mat moving. keep it moving? 3. A musician applies a horizontal force of 17 N.[W] to an instrument case of mass 5.1kg. The case slides across a table with an acceleration of 0.39m./s.2[W]. What is the coefficient of kinetic friction between the case and the table? Page 8 of 10 SPH 4U Unit #1 Dynamics Topic #9: Friction (Teacher) 4. A small box is resting on a larger box sitting on a horizontal surface. When a horizontal force is applied to the larger box, both boxes accelerate together. The small box does not slip on the larger box. a) Draw a free body diagram of the small box b) What force causes the small box to accelerate horizontally? The force of static friction between the boxes causes the small box to accelerate horizontally. c) If the two boxes accelerate at 2.5m./s.2, determine the smallest coefficient of friction between the boxes that will prevent slippage d) Draw a free body diagram of the larger box when it is accelerating Page 9 of 10 SPH 4U Unit #1 Dynamics Topic #9: Friction (Teacher) 5. An adult is pulling two smaller children in a sleigh up a snow covered hill. The sleigh and children have a total mass of 47 kg. The hill makes an angle of 23 degrees with the horizontal. The coefficient of kinetic friction between the sleigh and the snow is 0.11. Calculate the magnitude of the tension in the rope needed to keep the sleigh moving at a constant velocity in the snow. 6. A race car accelerates on a level track. The coefficient of friction between the tires and the track is 0.88 and the tires are not slipping. a) Draw a free body diagram of the car b) Determine the maximum possible acceleration if the car is to travel without slipping. Page 10 of 10