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3 years ago

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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

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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Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.

azamat

Answer:

$\theta=5.71^{o}$

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

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$\sum F_{x}=0$

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$-\mu_{k}N+mg sen(\theta)=0$

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we can divide both sides into mg so we get:

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According to coulomb’s law ,if a unit

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