What Forces Act on a Car at the Peak of a Hill?

[pippintook]

A 1000 kg car traveling on a road that runs straight up a hill reaches the rounded crest at 25 m/s. If the hill at that point has a radius of curvature of - in a vertical plane - 75 m, what is the net downward force acting on the car at the instant it is horizontal at the very peak? How fast must the car go in order to leave the ground?


Honestly, I'm not even sure how to get started. I'm looking for F_N and the car's velocity, but I have no idea what the first step should be. Help please?

 

[PhanthomJay]

A 1000 kg car traveling on a road that runs straight up a hill reaches the rounded crest at 25 m/s. If the hill at that point has a radius of curvature of - in a vertical plane - 75 m, what is the net downward force acting on the car at the instant it is horizontal at the very peak? How fast must the car go in order to leave the ground?


Honestly, I'm not even sure how to get started. I'm looking for F_N and the car's velocity, but I have no idea what the first step should be. Help please?

Part 1: Draw a free body diagram of the car when it is at the top of the crest, and identify the forces acting on it (there are just 2 forces acting in the vertical direction). The net sum of these 2 forces is the centripetal force acting on it. you know v, so just calculate F_centripetal. Part 2: When the car leaves the ground, what is the Normal force acting on it? Then solve for v using the centripetal force equation per Newton 2.

 

[thomate1]

I can help you with this problem. The first step would be to understand the forces acting on the car at the instant it is horizontal at the very peak of the hill. These forces include the car's weight (mg), the normal force (FN) from the road, and the centripetal force (Fc) due to the car's circular motion on the curved hill.

To find the net downward force acting on the car, we can use the equation: Fnet = mg - FN. Since the car is at the very peak of the hill, the normal force will be equal to the weight of the car (mg). Therefore, the net downward force will be equal to zero.

To find the car's velocity at the very peak, we can use the equation for centripetal force: Fc = mv^2/r, where m is the mass of the car, v is the velocity, and r is the radius of curvature of the hill. Rearranging this equation, we get v = √(Fc * r / m). Plugging in the given values, we get v = √(1000 kg * 25 m/s * 75 m / 1000 kg) = 15.5 m/s.

In order for the car to leave the ground, the centripetal force must be greater than or equal to the weight of the car. This means that the velocity must be greater than or equal to √(mg * r / m). Substituting the values, we get v ≥ √(1000 kg * 9.8 m/s^2 * 75 m / 1000 kg) = 27.4 m/s. Therefore, the car must be going at least 27.4 m/s in order to leave the ground at the very peak of the hill.

I hope this helps you understand the problem better. Let me know if you have any further questions.