Accel. of gravity = 9.8m/s^2
What is the Net centripetal force needed to keep the car from skidding sideways?
If there were no friction between the car's tires and road, what would be the centripetal force provided just by the banking of the road??
Suppose the friction force is sufficient to keep the car from skidding. whats the magnitude of the Normal force exerted on the car by the road's surface???
What is the Magnitude of the Friction Force?
And what could be the lowest possible value of the static friction coeffcient that would prevent the car from skidding?
please help!
2007-10-01 13:13:26 · 1 answers · asked by michael 1 in Science & Mathematics ➔ Physics
sorry i forgot to mention that the weight of the car is 2100kg
and the angle of the bank is 19 deg.
2007-10-01 13:15:13 · update #1
weight of the car is 2100kg (mass?)
and the angle of the bank is 19 deg.
the centripetal force of the turn is
m*v^2/r
the component parallel to the surface of the road is
cos(19)*m*v^2/r
the force down the bank is
m*g*sin(19)
the normal force is
m*g*cos(19)+sin(19)*m*v^2/r
use these equations to answer all the permutations
No skid without friction requires
cos(19)*m*v^2/r-m*g*sin(19)=0
No skid with friction requires
cos(19)*m*v^2/r-
m*g*sin(19)-friction=0
friction is
u*m*(g*cos(19)+sin(19)*v^2/r)
j
2007-10-02 06:14:23 · answer #1 · answered by odu83 7 · 0⤊ 0⤋