Question

A 1,800 kg car is parked on a road that has an elevation angle of 7°. Suppose the coefficient of static friction of the kinds of rubber and asphalt involved is 0.65. Which is approximately the force of static friction between the tires and the road? Question options: 11,350 N 9,300 N 18,000 N 2,200 N

Answer 1

2,200 N is the ans for the question asked

Answer 2

`F_f = 2200 N`

Step-by-step explanation:

As we know that car is parked on inclined plane

So here we have

`F_n = mg cos\theta`

`F_n = (1800)(9.81)cos7`

`F_n = 17526.4 N`

now we know that maximum limiting friction between tyres and road is given as

`F_s = \mu_s F_n`

here we have

`F_s = 0.65 * 17526.4`

`F_s = 11350 N`

Now at static condition of the car net force is balanced on it

so we have

`F_f = mg sin\theta`

`F_f = 1800 * 9.8 * sin7`

`F_f = 2200 N`