Friction Lab

FRICTION The goal of this lab is to understand the relationship between frictional force and normal force. What is the formula for FG (weight)? FG=m*g where g=9.8m/s2 and m = mass in kg Define the following three terms: FN-Normal Force: A support force exerted on an object that is in contact with another stationary object. This force acts perpendicular to the surface the object is in contact with. Ff-Kinetic Friction: Frictional force which occurs when an object is moving on a surface and is opposite the direction of movement Ffs-Static Friction: Frictional force that occurs between two objects that are stationary. This force must be overcome before the object will begin moving. LAB Part 1: KINETIC FRICTION  CLICK FORCES, SUM OF FORCES, VALUES, and MASSES For each mass in the chart, push the object until it is moving fast then let go. The friction force by the red arrow will be your kinetic friction value (your first object should be 94 N) Solve for the Force of Gravity (weight) for each mass See image below for an example. Object(s) Mass (kg) 1 box 50 Man 80 Girl + 1 Box Trash can 90 100 e (N) Kinetic Friction Force (N) 94 150 169 188 Man + 1 box 130 1274 1274 244 Box + trashcan 150 1470 1470 281 Man + trashcan 180 1760 1760 338 Refrigerator 200 1960 1960 XXXXX *IF THE OBJECT DOES NOT MOVE EVEN WITH 500 N of force, place an X through the data as it means we cannot measure the kinetic friction (not moving) or find the static friction’s maximum value. Use Excel, or similar, to make a graph of Friction force (y) vs Normal Force (x). Do a linear best fit and include the equation on the graph. Copy and paste your graph and equation below. Sketch the graph after adding your line of best fit Record the m (slope) 400 and b (y-int) value 350 300 Write your y=mx+b model: y = 0.1918x – 0.1937 250 The slope here represents our 200 coefficient of friction ( µ ) which you 150 100 50 0 400 600 800 1000 1200 1400 1600 1800 can think of as a measure of t y= 0.1918x – stickiness between two surfac 0.1937 measure of the ratio of friction and normal force. The subscript ‘k’ next to the µ indicates we are talking about kinetic friction’s coefficient, and an 2000 object that is in motion. The formula from the graph is: Ff = µk(FN) Kinetic Friction = coefficient of kinetic friction x Normal Force Your slope (m value) in the graph is the coefficient. Write it down µk= 0.1918 1. Using your coefficient from the graph (slope), multiply it by the normal force for the box (490 N). What do you get? Fk = (0.1918)*(490) = 93.982 2. Does your answer above come out to be about 94N? Yes 3. If your coefficient (µ) were 0.25 and the Normal Force were 400 N, what would be the Friction force? Fk = µ*FN = (0.25)*(400) = 100N Part 2: STATIC FRICTION CLICK FORCES, SUM OF FORCES, VALUES, and MASSES For each mass in the chart, slide the slider SLOWLY until the mass starts to move and record it o your first object should be between 130 N and 140 N This pushing force at the moment the mass starts to move at balances out the maximum static force, as such we can use that force as our static force. Solve for the Force of Gravity (weight) for each mass. NOTE: Since we are attempting to apply a force slowly until we notice motion, we will have error in this experiment. Your graph will have data points slightly off the linear fit line. Object(s) Mass (kg) Force of Gravity (N) Fg=m(9.8 m/s2) 1 box 50 Man 80 Girl + 1 Box 90 Static Friction Force (N) 490 Normal Force (N) Cancels gravity in this situation so same value as Fg 490 784 784 201 882 882 126 226 Trash can 100 980 980 251 Man + 1 box 130 1274 1274 326 Box + trashcan 150 1470 1470 376 Man + trashcan 180 1760 1760 451 Refrigerator 200 1960 1960 XXXXX *IF THE OBJECT DOES NOT MOVE EVEN WITH 500 N of force, place an X through the data as it means we cannot measure the kinetic friction (not moving) or find the static friction’s maximum value. Use Excel, or similar, to make a graph of Friction force (y) vs Normal Force (x). Do a linear best fit and include the equation on the graph. Copy and paste your graph and equation below. Sketch the graph after adding your line of best fit Record the m (slope) and b (y-int) value 500 450 400 Write your y=mx+b model: y= 0.2557x + 0.4904 350 The slope here represents our 300 coefficient of friction ( µ ) which you can think of as a measure of the stickiness between two surfac y= 0.2557x + measure of the ratio of frictio 0.4904 250 200 150 100 50 0 400 600 800 1000 1200 1400 1600 1800 2000 normal force. The subscript ‘s’ next to the µ indicates we are talking about static friction and an object not in motion. We can specify it as the formula below. Ffs = µs(FN) Static Friction = coefficient of static friction x Normal Force Your slope (m value) in the graph is the coefficient. Write it down µs= 0.2557 1. Using your coefficient from the graph (slope), multiply it by the normal force for the box (490 N). What do you get? Ffs = µs(FN) = (0.2557)(490) = 125.3N 2. Does your answer above come out to be in the 130 N to 140 N range? No, using the simulator, the force to overcome the static friction for the box (490N) was only 126N. 3. If your coefficient (µ) were 0.45 and the Normal Force were 400 N, what would be the Friction force? Ffs = (0.45)(400) = 180N Questions: 1. Write your coefficient (µ) from each friction. Coefficient for KINETIC (first graph): 0.1918 Coefficient for STATIC (second graph): 0.2557 2. Both coefficients were decimals, however which force had the coefficient with the larger value? Static Friction 3. The same objects were used in both situations. Compare the static friction forces to kinetic friction in the data tables. What frictional force was always larger? Static Friction is larger 4. What will always be more difficult, to start moving an object that is still with static friction, or to keep moving an object that is already moving that has kinetic friction? It will always be more difficult to start moving an object that is still and overcome static friction. We can rewrite our two variations of the friction equation to be: Ff = µFN Friction Force = Coefficient of Friction x Normal Force Remember, if it is NOT MOVING, it is a static force and we must use the static coefficient (µs) Remember, if it is MOVING, it is a kinetic force and we must use the kinetic coefficient (µk) If we have a situation that just states friction and does not define the type of friction or state of motion, we can use the generic version shown above. We can use the generalized formula Ff = µFN 1. An object has a coefficient of kinetic friction of 0.2 and a normal force of 30N. Find the force of kinetic friction. FK = 0.2*30 = 6N 2. An object has a coefficient of static friction of 0.3 and a normal force of 30N. Find the force of static friction. FS = 0.3*30 = 9N 3. An object has 45 N of static friction and a normal force of 450. What is the coefficient? µS = FS/FN = 45/450 = 0.1 4. There are 80 N of kinetic Friction and a coefficient of 0.25. What is the Normal Force? FN = FK/µK = 80/0.25 = 320N 5. An object has a mass of 20 kg and a coefficient of friction of 0.4. a. Find the force of gravity (weight) for the mass. FG = mg = 20*9.8 = 196N b. If gravity and normal force cancel, what is the normal force? FN = FG = 196N c. Find the force of friction in this situation. Ff = µFN = (0.4)(196) = 78.4N Static: Ffs = µs(FN) Kinetic: Ff = µk(FN) 6. An object is known to have a coefficient of kinetic friction (µk) of 0.167 and a coefficient of static friction (µS) of 0.42. If the normal force is 200 N, how much frictional force will it encounter while it is moving? Fk = µ(FN) = (0.167)(200) = 33.4N 7. An 80 kg object has a µk = 0.35 and a µs = 0.60. Assuming it is on a flat surface a. What is the normal force on the object (draw a diagram if needed) FN = mg = (80)(9.8) = 784N b. How much force is required to get the object to start to move from rest? Fs = µS(FN) = (0.60)(784) = 470.4N c. If the above object is moving already, and a tension force of 15 N to the right is pulling it, what will be the NET Force on the object? Force is a vector so direction should be included. FNET = FApp + Ff = 15N - µK(FN) = 15 – (0.35)(784) = - 259.4N or 259.4N to the left d. What is the acceleration (with direction) of the object based on your answer for part c? ax = F/m = -259.4 / 80 = -3.2425 m/s2 or 3.2425 m/s2 to the left