Friction
Practice
practice problem 1
- A forklift driver decides to push it without lifting it. What force must be applied to just get the pallet moving?
- After a bit of time, the pallet begins to slide. How fast is the pallet moving after sliding under the same force you calculated in part a. for half a second?
- If the forklift stops pushing, how far does the pallet slide before coming to a stop?
solution
Four forces are acting on the pallet: the downward pull of earth's gravity, the normal force of the floor pushing up, the forward push of the forklift, and the backward resistance of friction. Weight and normal are equal throughout this example since the floor is level. Friction changes from static to kinetic — static friction initially since the pallet isn't moving initially, then kinetic friction once the pallet gets going. The push also changes from nothing to the value needed to get the pallet moving, then back to nothing after 0.5 seconds of motion.
To get the pallet started, the driver must push it with a force equal to the maximum static friction.
P = f_{s} = µ_{s}N = µ_{s}mg P = (0.28)(600 kg)(9.8 m/s^{2}) P = 1,646 N Once the pallet starts moving, the coefficient of friction drops from its static value to its kinetic value.
f_{k} = µ_{k}N = µ_{k}mg f_{k} = (0.17)(600 kg)(9.8 m/s^{2}) f_{k} = 1,000 N But the forklift is still pushing with 1,650 N of force. Thus we have a nonzero net force.
Σ F = P − f_{k} Σ F = 1,646 N − 1,000 N Σ F = 646 N A net force causes acceleration.
a = Σ F/m a = (646 N)/(600 kg) a = 1.08 m/s^{2} Acceleration goes with a change in velocity.
v = v_{0} + at v = (1.08 m/s^{2})(0.5 s) v = 0.54 m/s Once the forklift stops pushing, kinetic friction becomes the net force. This net force will cause an acceleration opposite the direction of motion. When one vector is opposite another, one of the two needs to be negative. The convenient thing to do for this problem is to let friction be the negative one.
a = ΣF/m = ƒ_{k}/m a = (−1,000 N)/(600 kg) a = −1.67 m/s^{2} Pick the appropriate equation of motion
v^{2} = v_{0}^{2} + 2a∆s
Eliminate the zero term (final velocity), solve for distance, substitute, and calculate. Watch how the negative signs disappear. This has to happen. An object moving forward should be displaced forward.
∆s = −v_{0}^{2} 2a ∆s = −(0.54 m/s)^{2} 2(−1.67 m/s^{2}) ∆s = 0.087 m
practice problem 2
- Determine the car's maximum starting acceleration with and without "burning rubber". How do these two methods of starting a car compare?
- Determine the car's minimum braking distance with normal brakes and antilock brakes as a function of initial speed. How do these two methods of stopping a car compare?
solution
The net external force propelling a car comes from the friction force between tires and pavement. When a driver starts a car by "flooring it" (pressing the accelerator to the floor) the tires grind on the road producing a smoke of burning rubber and pavement. Since the tires are slipping, the coefficient of kinetic friction determines the maximum acceleration. Under normal circumstances, however, most drivers are not willing to subject their tires to such extreme punishment. Typical car tires rotate over the surface of the road without slipping, thus the coefficient of static friction determines a car's maximum acceleration in most situations.
To solve this problem, set the frictional force on level ground equal to the net force of the second law of motion.
∑F = ma ƒ = μmg = ma a = μg a_{burnout} = ⅔(9.8 m/s^{2}) = 6.54 m/s^{2} a_{normal} = ¾(9.8 m/s^{2}) = 7.35 m/s^{2} a_{burnout} = μ_{k}g = μ_{k} = ⅔ = 8 = 88.9% a_{normal} μ_{s}g μ_{s} ¾ 9 Contrary to popular belief, flooring the accelerator is not an effective method of starting a car. Burning rubber is only about 90% as effective as accelerating a car normally from rest.
The net external force stopping a car comes from the friction force between tires and pavement. Stopping a car with ordinary brakes may result in wheel lock; that is, the wheels lock in position and are not able to rotate. When this happens, the tires skid and the coefficient of kinetic friction determines the braking distance. Cars equipped with an antilock braking system (ABS) have a sensor that releases the brake pads the instant the wheel locks up. After a brief pause the brakes are then quickly re-engaged. If they don't lock up again, all is well. If they do, the ABS releases the brake pads again. This processes can repeat many times a second. In any case, the tires are not allowed to lock for more than a few milliseconds. The car is then stopped using the force of static friction alone.
To solve this problem, determine acceleration using the displacement-velocity formula of kinematics. Set this equation equal to the formula for acceleration due to friction derived above.
v_{0}^{2} = 2aΔs = 2μgΔs
Δs = v^{2} 2μg Δs_{antilock} = v^{02}/2μ_{s}g = μ_{k} = ⅔ = 8 = 88.9% Δs_{normal} v^{02}/2μ_{k}g μ_{s} ¾ 9 Antilock brakes need 90% of the distance of regular brakes to stop a car traveling at the same speed. This decrease in distance is certainly significant, but doesn't really seem all that great given the high cost of an ABS. In addition to reduced braking distance, however, antilock braking systems also increase performance during extreme braking. Locked brakes are useless for steering. ABS ensures that the wheels retain their static frictional grip on the road, which allows for maneuvering while braking in an emergency.
practice problem 3
solution
- Answer it.
practice problem 4
This tab-delimited text file contains the stopping distance data for 123 cars tested by Road & Track magazine in 1998. Two initial speeds were used: 26.82 m/s (60 mph) and 35.76 m/s (80 mph). Use the data in this file and your favorite data analysis software to determine the coefficient of static friction of car tires on pavement.
solution
Start with Newton's second law of motion.
∑F = ma
A stopping car is acted upon by three forces: weight pointing down, normal pointing up (we'll have to assume the test track is level), and (as long as the wheels don't lock and the car doesn't skid) static friction. Of course, there's also aerodynamic drag, but worrying about that force in this problem would be a waste of time. Weight and normal cancel out since the car is neither accelerating up nor down. Static friction is therefore the net force acting on a braking car.
f_{s} = ma
Replace ƒ with its classical formula.
µ_{s}N = ma
Earlier, we assumed (quite sensibly) that the test track would be level, which means that normal equals weight (W = mg).
µ_{s}mg = ma
Work the magic of algebra and solve for the goal of this problem — the coefficient of friction.
μ_{s} = | a |
g |
Great, but what is a? Go back to the good old days when you learned the equations of motion. Pick the one that doesn't involve time and solve it for acceleration.
v^{2} = | v_{0}^{2} + 2aΔs | |
a = | v^{2} | |
2Δs |
Substitute this expression into the previous one.
μ_{s} = | v^{2} |
2gΔs |
Take all the numbers in road-test-summary.txt and run them through this final equation. These results are given in road-test-summary-solution.txt. (Note: I used g = 9.8 m/s^{2}, but one could also use the value of standard gravity g = 9.80665 m/s^{2}.) Using the mean of these 246 trials as the value and the standard deviation as the uncertainty yields the following answer.
μ_{s} = 0.91 ± 0.10