What is the normal force on the driver?

  • Thread starter Shatzkinator
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In summary, the normal force on the driver from the seat when the car passes through the bottom of the valley is twice the weight of the driver. This can be determined by considering the values of centripetal acceleration at the top and bottom of the hill, where the normal force is 0 and equal to the weight, respectively. The equation for centripetal acceleration (F = mv^2/r) can be used to find the magnitude of the normal force in each case, and it is found that at the bottom of the hill, the normal force is twice the weight of the driver.
  • #1
Shatzkinator
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Homework Statement


A car is driven at constant speed over a circular hill and then into a circular valley with the same radius. At the top of the hill the normal force on the driver from the car seat is 0. The driver's mass is 70.0 kg. What is the magnitude of the normal force on the driver from the seat when the car passes through the bottom of the valley??


Homework Equations


Fg - Fn = 0 ?
F = mv^2/r


The Attempt at a Solution


Stumped. I looked at this question and was like... wtf?

Only thing I can think of is Fg - Fn cannot equal zero because there is always Fg and if Fn is 0, then that equation doesn't make sense. But I don't know any better than this equation. I'm so confused! =S
 
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  • #2
At the top of the hill if his normal force is 0, then what must his centripetal acceleration be?
 
  • #3
LowlyPion said:
At the top of the hill if his normal force is 0, then what must his centripetal acceleration be?

Fg = 9.8 * 70 = 686
 
  • #4
How is the direction of the centripetal acceleration different at the top as opposed to the valley?
 
  • #5
LowlyPion said:
How is the direction of the centripetal acceleration different at the top as opposed to the valley?

At the top, it is directed downwards toward the centre, at the bottom is directed up toward the centre...
 
  • #6
Shatzkinator said:
At the top, it is directed downwards toward the centre, at the bottom is directed up toward the centre...

... and you still can't answer the original question?
 
  • #7
Fn = Fg .. which would be al;so 686 at the bottom of the hill.

The answer should be 1.37 x 10^3 N

=P

No I still can't answer the original question.

Edit: and yes I see the answer is 2* 686... I have no idea why.
 
  • #8
In the first case Fn = 0 = mg - mv2/r

That means that mv2/r = mg.

In the valley

Fn = mg + mv2/r

But you just figured out that mg = mv2/r so

Fn = 2*mg
 

1. How does a car drive over a hill?

A car drives over a hill by using its engine to generate enough power to overcome the force of gravity pulling it downhill. The car's wheels gain momentum as they reach the bottom of the hill, allowing it to climb up the other side.

2. What factors affect a car's ability to drive over a hill?

The weight of the car, the power of the engine, the angle of the hill, and the road conditions all play a role in a car's ability to drive over a hill. A heavier car or a steeper hill may require more power from the engine to successfully make it over.

3. Is it more efficient for a car to drive over a hill or around it?

It depends on the specific circumstances. In general, driving around a hill may be more efficient because it requires less energy from the car's engine. However, if the hill is not too steep and the car has enough momentum, driving over it may also be a viable option.

4. What safety precautions should be taken when driving over a hill?

It is important to maintain a safe speed and to be aware of any potential hazards on the road, such as sharp turns or steep drops on the other side of the hill. It is also recommended to use low gears and avoid sudden braking or acceleration while driving over a hill.

5. How does driving over a hill affect a car's fuel efficiency?

Driving over a hill can decrease a car's fuel efficiency as it requires more power from the engine to overcome the force of gravity. However, the impact on fuel efficiency will vary depending on the car's weight, engine size, and the angle of the hill.

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