A 1200 kg car is traveling at 27 m/s around a level curve of radius 110m....
A 1200 kg car is traveling at 27 m/s around a level curve of radius 110m. What centripetal force must be applied to prevent the car from slipping, and what is the minimum value for the coefficient of friction required to provide the force?
Homework Answers
F = m*v^2 / r = 1200 * (27^2)/110 = 7952.727272727272
Friction = u(fr.cofficient ) * N = 7952.727272
N = mg = 1200*9.8
u (fr. cofficient )= 7952.727272/(1200*9.8) = 0.6765
a)
F = m*v^2/r
= 1200*27^2/110 = 7952.73 N
b)
N = m*g = 1200*9.8 = 11760 N
m*v^2/r = u*N
7952.73 = u*11760
u = 0.676
centripetAL FORCE = mv2 / r
=795207272 newton
now ,
coefficient of friction = centripetal force/ mg
= 0.67625
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