When you take your 1200 kg car out for a spin, you go around a corner of radius 54.6 m with a speed of 15.3 m/s. The coefficient of static friction between the car and the road is 0.91. Assuming your car doesn't skid, what is the force exerted on it by static friction?
2007-10-12 04:48:29 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
If we ignore rolling friction of the tires and assume that the speed is constant so that there is no angular acceleration, than the only acceleration is the centripetal which is v^2/r
so the first step is to see if the car skids
The centripetal force is
1200*15.3^2/54.6
=5145 N
The normal force is
1200*9.81
and max friction is
1200*9.81*0.91
or
10712.52
This is much larger than the actual force of 5145 so the car does not skid and the frictional force is
5145 N
j
2007-10-12 08:38:16 · answer #1 · answered by odu83 7 · 0⤊ 0⤋