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Alona [7]
3 years ago
10

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
Physics
2 answers:
pickupchik [31]3 years ago
7 0

Answer:

F_f = 2200 N

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 \times 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 \times 9.8 \times sin7

F_f = 2200 N

Masteriza [31]3 years ago
4 0

2,200 N is the ans for the question asked

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