FRICTION APPLICATIONS
Things to Remember: 1. W = mg (kg does NOT measure forces so NEVER put them in a force diagram!)
2. f = mn
3. not moving = static equilibrium therefore SF = 0
4. constant speed = dynamic equilibrium therefore SF = 0
(inertia keeps the system moving)
5. friction depends on the normal force and NOT on weight
6. start with the y-dimension
1D: A 75 kg skydiver jumps from a plane and obtains terminal velocity. (A constant falling speed.) Calculate the strength of air resistance acting against him.
First convert kg à N W = mg = (75)(9.8) = 735 N SF = 0 (constant speed) f – W = 0 f = W f = 735 N
(Notice that air resistance is equal to the diver’s weight. Inertia keeps them moving.)
1D: If air resistance is 1200 N, then how strong do the thrusters on a 200 kg space probe have to be in order to move up through earth’s atmosphere traveling at a constant speed?
W = mg = (200)(9.8) = 1960 N SF = 0 (constant) F – f – W = 0 F = f + W = 1200 + 1960 = 3160 N
(The thrusters have to fight off the weight of the probe AND air resistance.)
2D: Betsy is pushing her 20 kg table across the floor (no angle). If the coefficient of friction between the table and the floor is 0.24, then how hard does she have to push to slide the table forward at a constant speed?
W = mg = (20)(9.8) = 196 N S F y = 0 (static) f = mn S F x = 0 (constant)
n – W = 0 = (0.24)(196) F – f = 0
n = W = 47 N F = f
n = 196 N F = 47 N
2D: Mulan just fell off the edge of an icy cliff of death with her horse and Mushu. (Watch the movie. You’ll see what I mean.) Their combined mass is 720 kg. Thinking quickly she shoots an arrow and rope back up to her comrades. They have a combined mass of 500 kg. Calculate the coefficient of static friction needed in order to prevent her friends from slipping forward. Assume the tension in the rope is the same as the weight of Mulan and horse. Does this sound like a reasonable coefficient for ice and snow?
W = mg = (500)(9.8) = 4900 N S F y = 0 (static) S F x = 0 (static) f = mn That is WAY too high
T = W = mg = (720)(9.8) = 7056 N n – W = 0 T – f = 0 f / n = m for snow. Mulan is
n = W T = f 7056 / 4900 = 1.44 going to die.
n = 4900 N 7056 = f
Angled: Betty is pushing her 24 kg couch across the living room with 320 N of force at a downward angle of 40 degrees. The couch slides at a constant speed. What is the coefficient of friction between the couch and carpet?
W = mg = (24)(9.8) = 235 N S F y = 0 (static) S F x = 0 (constant) f = mn
n – W – Fsin40 = 0 Fcos40 – f = 0 f / n = m
n = W + Fsin40 = 235 + 320sin40 = 441 N 320cos40 = f = 245 N 245 / 441 = 0.556 = m
Angled: Bob is pulling his 72 kg trampoline across the yard at an upward angle of 30 degrees. The coefficient of friction between the tramp and grass is 0.57. How hard does he need to pull in order to slide the tramp at a constant speed?
(Use simultaneous equations.)
W = mg = (72)(9.8) = 706 N S F y = 0 (static) S F x = 0 (constant) F cos30 = mn 1.151 F = 402
n + Fsin30 – W = 0 Fcos30 – f = 0 F cos40 = 0.57 ( 706 – 0.5F ) F = 349 N
n = W – Fsin30 = 706 – 0.5F Fcos30 = f 0.866F = 402 – 0.285F
Hill: Betty (62 kg) is standing on the side of a wet grassy hill. The angle of elevation is 32 degrees. What coefficient of static friction is needed in order to prevent her from slipping down the hill?
W = mg = (62)(9.8) = 608 N S F y = 0 (static) S F x = 0 (static) f = mn
Angle = 90 – 32 = 58 n – Wsin58 = 0 f – Wcos58 = 0 f / n = m
n = Wsin58 f = Wcos58 322/516 = 0.624 = m
n = 608sin58 = 516 N f = 608cos58 = 322 N
Hill: John (71 kg) is sliding down a giant 48 degree sand hill at Lake Powell on his knee-board. (I have done this. . . . It actually works.) He obtains his maximum cruising speed of 13 m/s and then continues at a constant pace. What is the coefficient of kinetic friction between the board and the sand?
W = 696 N n = 466 N f = 517 N m = 1.11