Asked by Mike
A car weighing 2000 lbs is parked on a hillside in San Francisco at an angle of 55o
(relative to the ground). Approximately what is the minimum coefficient of static friction necessary to keep the car from sliding downhill?
(relative to the ground). Approximately what is the minimum coefficient of static friction necessary to keep the car from sliding downhill?
Answers
Answered by
Henry
Mc = 2000Lbs * 0.454kg/Lb = 908 kg = Mass of car.
Wc = m*g = 908kg * 9.8N/kg = 8898.4 N. = Wt. of car.
Fc = 8898.4 N @ 55o = Force of car.
Fp = 8898.4*sin55 = 7289 N. = Force parallel to hill.
Fv = 8898.4*cos55 = 5104 N. = Force
perpendicular to hill.
Fs = Force of static friction.
Fp-Fs = m*a.
7289 - Fs = m*0 = 0.
Fs = 7289 N.
Wc = m*g = 908kg * 9.8N/kg = 8898.4 N. = Wt. of car.
Fc = 8898.4 N @ 55o = Force of car.
Fp = 8898.4*sin55 = 7289 N. = Force parallel to hill.
Fv = 8898.4*cos55 = 5104 N. = Force
perpendicular to hill.
Fs = Force of static friction.
Fp-Fs = m*a.
7289 - Fs = m*0 = 0.
Fs = 7289 N.
Answered by
Henry
Fs = uFv = 7289 N.
u*5104 = 7289
u = 1.43.
u*5104 = 7289
u = 1.43.