A car moves at a constant speed on a straight but hilly road. One section has a crest and dip of the same 250-m radius. (a) As the car passes over the crest, the normal force on the car is one-half the weight of the car $(mg)$. What will be the normal force on the car as it passes through the bottom of the dip? (b) What is the greatest speed at which the car can move without leaving the road at the top of the hill? (c) Moving at the speed found in (b), what will be the normal force on the car as it moves through the bottom of the dip?

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Here is what I have tried so far. I know that an object on a $\theta$ degrees slope will experience a normal force of $F_n = mg \cos(\theta).$ So, if a normal force is being acted upon a body is known, we can compute the angle of inclination using this relation and so theta would be:

$$\frac{1}{2} mg = mg \cos(\theta).$$ $$\theta = \arccos( 1/2 )$$ $$\theta = \frac{\pi}{3}\ \mathrm{radians}$$

But for a dip the angle of inclination will still be the same and so the normal force's magnitude will still be $\frac{1}{2} mg.$

But I am not sure wheather this answer is correct. And I also can't solve the other two parts of the question.

  • $\begingroup$ Have you learned about centripetal acceleration yet? Have you drawn a free body diagram of the car passing over the crest of the hill, and another free body diagram passing through the dip in the road. $\endgroup$ – Chet Miller Feb 15 '18 at 14:49

The Normal force you compute is for an object that has no acceleration in the direction perpendicular to the ground, such that Normal force cancels out gravity.

When an object is following a perfect circular motion at constant speed, it needs an acceleration of $\frac{v^2}{r}$ directed at the center of the circle, to follow that path.

Multiplying this acceleration with the mass tells you what the net force must be on the car, which is the sum of the weight and the Normal force.

Also the Normal force can not be negative. Think what is happening when this force becomes 0.


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