FORCE DUE TO FRICTION

What do you think? How would you describe friction? How is friction related to: heat, energy, force, velocity When asking students to express their ideas, you might try one of the following methods. (1) You could ask them to write their answers in their notebook and then discuss them. (2) You could ask them to first write their ideas and then share them with a small group of 3 or 4 students. At that time you can have each group present their consensus idea. This can be facilitated with the use of whiteboards for the groups. The most important aspect of eliciting student’s ideas is the acceptance of all ideas as valid. Do not correct or judge them. You might want to ask questions to help clarify their answers. You do not want to discourage students from thinking about these questions and just waiting for the correct answer from the teacher. Thank them for sharing their ideas. Misconceptions are common and can be dealt with if they are first expressed in writing and orally. Weight and mass are often confused. Students learned earlier that mass was the amount of matter in an object and weight was the force of gravity, but they often still confuse the issue. When eliciting their responses, ask them to discuss appropriate units for each. You might discuss “weightlessness” and ask if objects can be massless as well. Friction is often confused with heat or thermal energy. Students likely will think of friction as being related to many of the quantities listed above.

Review: Weight and Mass Mass is the amount of matter in an object Kilograms, slugs Weight is a measure of the gravitational force on an object Newtons, pounds Depends on the acceleration of gravity Weight = mass  acceleration of gravity W = mag where ag = 9.81 m/s2 on Earth Depends on location ag varies slightly with location on Earth ag is different on other planets Mention that weight is less on the moon because ag on the moon is 1.6 m/s2 . Reinforce that converting between mass and weight is simple, just multiply or divide by 9.81 m/s2 . Point out that each kg has a weight of 9.81 N on Earth.

Review: Normal Force Force on an object perpendicular to the surface (Fn) It may equal the weight (Fg), as it does here It does not always equal the weight (Fg), as in the second example Fn = mg cos  Point out that the equation for normal force applies to the first example also. Because cos(0)=1, the equation reduces to Fn = mg when the forces are directly opposite one another.

Static Friction How does the applied force (F) compare to the frictional force (Fs)? Would Fs change if F was reduced? If so, how? If F is increased significantly, will Fs change? If so, how? Are there any limits on the value for Fs? Force that prevents motion Abbreviated Fs These questions should help students understand that static friction balances the external force (F), so it increases and decreases as F increases and decreases. Eventually, F will be so large that the static frictional force (Fs) will no longer be able to balance it, and the net force will cause the object to slide. At this point, frictional forces become kinetic (see next slide).

Kinetic Friction Force between surfaces that opposes movement Abbreviated Fk Does not depend on the speed Using the picture, describe the motion you would observe The jug will accelerate How could the person push the jug at a constant speed? Reduce F so it equals Fk Ask students if it requires more force to get an object moving when it is at rest or to keep it moving once it is already in motion. When pushing an object, we exert enough force to overcome static friction. At that point the object moves. The opposing force is now kinetic friction, which is less than static friction. Therefore, in order to maintain a constant speed and not accelerate, the force pushing the object is reduced.

Calculating the Force of Friction (Ff) Ff is directly proportional to Fn (normal force) Coefficient of friction (): Determined by the nature of the two surfaces s is for static friction k is for kinetic friction s > k Point out to students that Ff is the general term for both static friction (Fs) and kinetic friction (Fk).

Typical Coefficients of Friction Values for  have no units and are approximate Point out that static is greater than kinetic for each example. Also explain that the coefficient is generally less than 1 but there could be sticky surfaces where the frictional force was greater than the normal force. This would lead to coefficients greater than 1.

Classroom Practice Problem A 24 kg crate initially at rest on a horizontal floor requires a 75 N horizontal force to set it in motion. Find the coefficient of static friction between the crate and the floor. Draw a free-body diagram and use it to find: the weight the normal force (Fn) the force of friction (Ff) Find the coefficient of friction. Answer: s = 0.32 This is a relatively simple example from the book (Sample Problem D). Ask students to follow the steps. It is easy to get the answer by skipping the free-body diagram, but they need this diagram to understand that normal force = weight, and the 75 N horizontal push is equal to the force of friction. More complicated problems (next slide) can’t be solved without a free- body diagram.

Classroom Practice Problem A student attaches a rope to a 20.0 kg box of books. He pulls with a force of 90.0 N at an angle of 30.0˚ with the horizontal. The coefficient of kinetic friction between the box and the sidewalk is 0.500. Find the magnitude of the acceleration of the box. Start with a free-body diagram. Determine the net force. Find the acceleration. Answer: a = 0.12 m/s2 This is Sample Problem E from the book. The free-body diagram is essential to solving this problem. Students often make the mistake of assuming the normal force equals the weight. These two forces are not equal because the student is pulling upward on the box, thus reducing the normal force. So, Fn = weight - (90.0 N)(sin 30)°. Students can then determine the value for Fk and subtract it from (90.0 N)(cos 30°) to get the net force. At this point, they can use Newton’s second law to find the acceleration.