Friction There are many forms of friction. This lesson introduces the force laws for static friction, kinetic friction, and rolling friction. Students learn the meaning and typical range of values for the coefficients of friction. In the investigation, students determine the coefficients of static and kinetic friction between two surfaces.

Objectives Calculate friction forces from equation models for static, kinetic, and rolling friction. Solve one-dimensional force problems that include friction. The lesson objectives describe what a student should know and be able to do.

Assessment A box with a mass of 10 kg is at rest on the floor. The coefficient of static friction between the box and the floor is 0.30. Estimate the force required to start sliding the box. This first assessment is keyed to the first objective: calculate friction forces from equation models for static, kinetic, and rolling friction. It will be repeated at the end of the lesson, followed by the answer.

Assessment A 500 gram puck is sliding at 20 m/s across a level surface. The coefficient of kinetic friction between the puck and surface is 0.20. Draw a free-body diagram for the puck and calculate the magnitude of each force. How long will it take the puck to skid to a stop? The second assessment is keyed to the second objective: solve one-dimensional force problems including friction. It will be repeated at the end of the lesson, followed by the answer.

Physics terms coefficient of friction static friction kinetic friction rolling friction viscous friction air resistance This slide lists new or important terms for this lesson.

Equations Models for friction kinetic friction static friction rolling friction Models for friction The friction force is approximately equal to the normal force multiplied by a coefficient of friction. This slide lists new equations for this lesson.

What is friction? Friction is a “catch-all” term that collectively refers to all forces which act to reduce motion between objects and the matter they contact. Friction often transforms the energy of motion into thermal energy or the wearing away of moving surfaces.

Kinds of friction The term ‘friction” is used to describe a variety of resistive forces. Air friction, for example, is also referred to as air resistance, and as drag. Another type of friction that will be treated in this lesson but is not mentioned in the slide is static friction.

Kinetic friction Kinetic friction is sliding friction. It is a force that resists sliding or skidding motion between two surfaces. If a crate is dragged to the right, friction points left. Friction acts in the opposite direction of the (relative) motion that produced it. This slide describes the direction of the friction force. For kinetic (sliding friction) the direction is easy to determine or describe.

Which takes more force to push over a rough floor? Kinetic friction Which takes more force to push over a rough floor? This slide begins a set of slides that develop the equation model for the magnitude of friction.

Friction and the normal force The board with the bricks, of course! The simplest model of friction states that frictional force is proportional to the normal force between two surfaces. If this weight triples, then the normal force also triples—and the force of friction triples too. Point out that the normal force is not always equal to the weight of an object. If you push a book against the wall and slide it upward, the normal force will be equal to the horizontal component of your push.

A model for kinetic friction The force of kinetic friction Ff between two surfaces equals the coefficient of kinetic friction μk times the normal force FN. direction of motion But what is this coefficient of friction, μk? Ask the students what they think this new constant could be related to.

The coefficient of friction The coefficient of friction is a constant that depends on both materials. Pairs of materials with more friction have a higher μk. direction of motion The μk tells you how many newtons of friction you get per newton of normal force. Do you see why μk has no units?

A model for kinetic friction The coefficient of friction μk is typically between 0 and 1. direction of motion When μk = 0 there is no friction. When μk = 0.5 the friction force equals half the normal force. When μk = 1.0 the friction force equals the normal force. Ask for an example of a high coefficient of friction, and a low coefficient of friction. How do you lower the friction in a machine?

Calculating kinetic friction Consider a 30 N brick sliding across a floor at constant speed. What forces act on the block? Draw the free body diagram.

Calculating kinetic friction Consider a 30 N brick sliding across a floor at constant speed. What is the friction force on the brick if μk = 0.5?

Calculating kinetic friction Consider a 30 N brick sliding across a floor at constant speed. The force F needed to make the board slide at constant speed must also be 15 N. According to the model, the friction force will be 15 N at any (non-zero) speed. But IF the speed is constant, there must be an applied force equal to the friction.

Static friction Static friction is gripping friction. It is a force that prevents relative motion between surfaces in contact with each other. Without static friction between your feet and the floor, you could not walk or run. Your feet would slip. Without static friction between your tires and the road, you could not start or stop a car. Note: Static friction does not actually prevent motion. It prevents relative motion. Demonstration: place a block on top of a pad of a paper. Move the block back and forth by sliding the pad back and forth. it is static friction that keeps the block at rest relative to the pad.

Static friction Static friction prevents this crate from sliding when pushed . . .

Static friction Static friction prevents this crate from sliding when pushed . . . . . . until the pushing force is greater than the maximum static friction force available.

Static friction How much static friction acts in case a? In case b? Point out to students that in a and b the crate remains at rest, so the net force on it must be zero. Therefore the static friction 120 N in a, and 160 N in b. The static friction has a maximum value of 200 N. A good analogy: static friction is like money in the bank; you only take out exactly what you need. If only 120 N is needed to keep an object at rest, this is exactly how much static friction will act on the object. But there is a maximum amount available in the bank.

Static friction How much static friction acts in case a? 120 N In case b? 160 N The crate is at rest so the net force must be zero. The static friction increases exactly as needed to keep the box at rest.

Static friction How much static friction acts in case a? 120 N In case b? 160 N What is the maximum static friction available? Sliding friction is typically LESS than the maximum static friction, so once the crate gets moving the force needed to KEEP it moving drops to 160 N.

Static friction How much static friction acts in case a? 120 N In case b? 160 N What is the maximum static friction available? 200 N Once the maximum static friction is exceeded, the crate begins to move. Sliding friction is typically LESS than the maximum static friction, so once the crate gets moving the force needed to KEEP it moving drops to 160 N.

A model for static friction The maximum static friction force Ff between two surfaces is the coefficient of static friction μs times the normal force FN. direction of applied force When μs = 0 there is no friction. When μs = 0.5 the maximum friction force equals half the normal force. When μs = 1.0 the maximum friction force equals the normal force. It is actually possible but unusual to have a coefficient of friction greater than one.

Calculating static friction A 10 N board is at rest on a table. How much force does it take to start the board sliding if μs = 0.2? Ask yourself: What forces act on the block? Draw the free-body diagram. mg = -10 N FN = +10 N

Calculating static friction A 10 N board is at rest on a table. How much force does it take to start the board sliding if μs = 0.2? The applied force F must be enough to break the grip of static friction. mg = -10 N FN = +10 N

Calculating static friction A 10 N board is at rest on a table. How much force does it take to start the board sliding if μs = 0.2? 2 N is the maximum force of static friction available. 2 N is also the minimum force needed to start the board moving. mg = -10 N FN = +10 N

Typical values of μs and μk Which combination of materials has the highest friction? lowest? Why is it good that rubber on dry concrete has such a high value? How do you reduce the friction between steel parts? What do you notice about the relative values of μs versus μk? Our joints have a very low coefficient of friction for ease of movement. Machinery is oiled to lower the friction. For driving we want a high value of static (gripping) friction to keep us safely on the road.

Typical values of μs and μk These coefficients of friction are only estimates, subject to ± 30% or more uncertainty. Actual experiments are needed in any situation where an accurate value is required. In accident reconstruction, the state police will drag a rubber sled across the pavement to get an accurate value for the friction force involved in a car accident. The length and shape of skid marks and the displacement vectors of projectiles provide addition data for solving the physics problems involved in understanding a traffic accident.

Engaging with the concepts In Investigation 5C you will determine the coefficients of friction between a friction block and table top. The investigation is found on page 157. Students apply their new understanding of friction to an actual situation involving a block on a table.

Investigation Part 1: Coefficient of static friction Set up the stand and pulley near a table. The string passing over the pulley should act along the centerline of the friction block. Tie one end of the string to the friction block and the other end to the cup. The string should be straight and level.

Investigation Part 1: Coefficient of static friction Set up the experiment. Measure all masses to within 1 gram. Add mass to the cup and record the maximum mass at which the block stays at rest. Be sure to brush any dust or grit from the surfaces before each trial. Friction is easily affected by small variations in the surfaces. Keeping surfaces as clean and consistent as possible will improve the lab results.

Investigation Part 1: Coefficient of static friction Add more mass on top of the block and repeat the experiment. Add more mass on top of the block and repeat the experiment a third time. Record all measurements in scientific notation and correct SI units. If possible, use aquarium gravel to increase the mass of the cup. Sand can be used and is very inexpensive, but it tends to be messy. The stopwatch utility is included on the investigation page.

Investigation ? How can you get the coefficient of static friction from the measured masses? The next five slide show the students how to do the calculation for getting the coefficient of static friction.

Investigation: finding μs When the block is on the verge of moving, the static friction must equal the force from the weight of the hanging cup, m1g. Question (b) in the assignment asks the students to draw this free-body diagram.

Investigation: finding μs When the block is on the verge of moving, the static friction must equal the force from the weight of the hanging cup, m1g. The coefficient of static friction equals the ratio of the masses. Question (c) in the student assignment asks them to reproduce this derivation.

Investigation Part 2: Coefficient of kinetic friction Use the friction block arrangement with the largest mass m2 (from part 1). Adjust the mass of the cup until the friction block has a noticeable acceleration across the table. Measure all masses to within 1 gram. The weight of the hanging cup must be sufficient to provide a noticeable acceleration, but the event can’t be TOO fast because the students need to time it.

Investigation Part 2: Coefficient of kinetic friction Measure the height h the cup drops from its maximum possible height directly under the pulley. Mark the table with tape so you can start the block at the same place each time.

Investigation Release the friction block and measure the time it takes for the cup to fall the distance h. Do several trials. Record all measurements in scientific notation and correct SI units.

Investigation: finding μk Step 1: Find the acceleration from the height h and time t. Let down be positive so that h and a will both be positive. In part 2, question (a) asks the students for this derivation. Once they have the equation, they should use it to complete the corresponding column in Table 3.

Investigation: finding μk system boundary Step 2: Find the net force on the system. The total mass of the system: The net force on the system: In part 2, question (b) asks the students for the net force. Once they have the equation, they should use it to complete the corresponding column in Table 3. Note: The force T is not equal to m1g in this situation where the hanging cup is accelerating. Mass m1 accelerates downward because m1g is greater than the tension.

Investigation: finding μk system boundary Step 3: Find the force of friction The weight of the cup speeds the system up, but friction slows it down. Rearrange the equation to solve for the friction. In part 2, question (c) asks the students for the kinetic friction. Once they have the equation, they should use it to complete the corresponding column in Table 3. Point out the the weight and normal force on the block cancel each other out, and so do not contribute to the net force. The forces of tension are INSIDE the system and cancel each other, so they do not contribute to the net force of the system. The net force on the system comes from OUTSIDE the system: the friction force comes from the table, and the weight of the hanging cup comes from the pull of Earth’s gravity.

Investigation: finding μk system boundary Step 4: Solve for μk. In part 2, question (d) asks the students to calculate the coefficient of kinetic friction. Once they have the equation, they should use it to complete the corresponding column in Table 3.

Evaluating models The scientific explanations or models for static and kinetic friction use a constant value for each friction coefficient. Analyze these models for friction by using the percentage variation in your results among trials. Critique these models based on your experimental testing. Evaluate the models by comparing your experimental results to the tabulated values of the coefficients. How precise are the models and coefficients?

Rolling friction Many machines, such as cars and bicycles, experience rolling friction. The equation model for rolling friction is similar to the model for sliding friction.

Coefficient of rolling friction Rolling friction comes mainly from slight deformations of the wheel. It is typically much lower than static or kinetic friction. Larger wheels tend to have lower coefficients of friction. Point out that underinflated tires have more rolling friction and therefore reduce gas mileage.

Viscous friction Fluid friction is the largest source of friction for cars, boats, and aircraft at speeds above 50 mph. There are two main sources of fluid friction: the force required to push the fluid out of the way the resistance of the fluid due to viscosity Viscous friction is complex. It depends on speed, shape, and fluid properties.

Shape factors The drag coefficient describes how easily fluid flows around a particular shape. Blunt objects have high drag coefficients. Aerodynamic, streamlined shapes have low drag coefficients. Relate these shapes to the shapes of cars, raindrops, and airplane wings.

Viscosity Viscosity describes a fluid’s resistance to flow. Air has a very low viscosity. Water also has a low viscosity and pours readily. Honey has a high viscosity and pours slowly.

Assessment A box with a mass of 10 kg is at rest on the floor. The coefficient of static friction between the box and the floor is 0.30. Estimate the force required to start sliding the box. This first assessment is keyed to the first objective: calculate friction forces from equation models for static, kinetic, and rolling friction. The answer appears on the next slide.

Assessment A box with a mass of 10 kg is at rest on the floor. The coefficient of static friction between the box and the floor is 0.30. Estimate the force required to start sliding the box. The required force is about 29 N. This first assessment is keyed to the first objective: calculate friction forces from equation models for static, kinetic, and rolling friction. The answer appears on the next slide.

Assessment A 500-gram puck is sliding at 20 m/s across a level surface. The coefficient of kinetic friction between the puck and surface is 0.20. Draw a free-body diagram for the puck. The second assessment is keyed to the second objective: solve one-dimensional force problems including friction. The answer appears on the next slides.

Assessment A 500-gram puck is sliding at 20 m/s across a level surface. The coefficient of kinetic friction between the puck and surface is 0.20. Draw a free-body diagram for the puck. direction of motion

Assessment A 500-gram puck is sliding at 20 m/s across a level surface. The coefficient of kinetic friction between the puck and surface is 0.20. Draw a free-body diagram for the puck and calculate the magnitude of each force. direction of motion

Assessment A 500-gram puck is sliding at 20 m/s across a level surface. The coefficient of kinetic friction between the puck and surface is 0.20. How long will it take the puck to skid to a stop? Hint: What is the acceleration of the puck? direction of motion

Assessment A 500-gram puck is sliding at 20 m/s across a level surface. The coefficient of kinetic friction between the puck and surface is 0.20. How long will it take the puck to skid to a stop? direction of motion