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:
2,200 N
9,300 N
11,350 N
18,000 N
I got A, but I was told I was wrong! :(
11,350 N. i got the same answer at first and then my teacher helped me.
Can you please explain??
The answer is actually 2,200. I think the test has been updated, but that's definitely correct.
Does anyone know how the answer is 11,350??????????
To find the force of static friction between the tires and the road, we can use the equation: Force of friction = coefficient of friction * normal force.
The normal force can be calculated by finding the component of the weight of the car that acts perpendicular to the road. Since the road has an elevation angle of 7°, the normal force can be found by calculating the weight of the car and then taking the component of that weight that is perpendicular to the road.
The weight of the car can be calculated using the equation: weight = mass * gravity, where mass is given as 1,800 kg and gravity is approximately 9.8 m/s^2.
weight = 1,800 kg * 9.8 m/s^2 = 17,640 N
The component of the weight that acts perpendicular to the road can be found using trigonometry. By using the angle of elevation (7°) and the weight of the car (17,640 N), we can calculate the normal force.
normal force = weight * cosine(angle of elevation)
normal force = 17,640 N * cos(7°) = 17,640 N * 0.99254615164 ≈ 17,503 N
Now that we have the normal force, we can find the force of static friction using the given coefficient of friction (0.65).
Force of friction = coefficient of friction * normal force
Force of friction = 0.65 * 17,503 N ≈ 11,378 N
Therefore, the approximate force of static friction between the tires and the road is 11,378 N.
Now let's check the answer options provided:
- 2,200 N is incorrect.
- 9,300 N is incorrect.
- 11,350 N is incorrect.
- 18,000 N is incorrect.
None of the options match the calculated force of static friction, which is approximately 11,378 N. It's possible that there was a mistake in the answer choices or in the calculation.