Free Body Diagram of Car on Mountain

  • #1
TimeInquirer
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If a car is on top of a mountain moving with an initial velocity, will it only forces acting on it be the normal force counteracted by the force due to gravity?
 

Answers and Replies

  • #2
brainpushups
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186
No. Is this in relation to a homework problem? Often times you are intended to make assumptions like to ignore certain types of friction, but the intent may be to have you ignore these forces.
 
  • #3
TimeInquirer
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It was an open based discussion in my physics class after various demos. We often neglected reaction forces such as forces between the tires and the ground. I am mostly interested in how to represent the initial velocity and acceleration of such a vehicle.
 
  • #4
22,145
5,078
If a car is on top of a mountain moving with an initial velocity, will it only forces acting on it be the normal force counteracted by the force due to gravity?
The normal force will be less than the gravitational force. Part of the gravitational force will be used to change the direction of the velocity vector as the car travels in the arc over the crest of the mountain.
 
  • #5
brainpushups
437
186
It depends on what you mean by 'top of a mountain.' I interpret that to mean that the road is flat which means the velocity would be horizontal, and there would be no acceleration (assuming the person is not using the controls to do so). If, on the other hand, the car is going up or downhill then there will be a component of gravity that points along the slope.

If you have experience with vectors you may be able to draw the components of the gravitational force that are parallel and perpendicular to the road without much difficulty. You can use this to find an expression for the acceleration of the car under ideal conditions. If you look up 'block on a frictionless inclined plane' I bet you'll find some more detail (though you should try the diagram yourself first!).
 
  • #6
TimeInquirer
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side question related to a different problem involving a spring. I calculate the frequency to be 2.65 (which is f= 2pi/w) but the answer is 0.265? What can be the reason for this? My angular velocity w is correct because it was given. Thank you.

I have noticed it is because of the way I put it into the calculator sqrt(4/1.4)/2pi vs 1.69/2pi
 
  • #7
brainpushups
437
186
If you are calculating frequency your formula is incorrect. Check your units (angular velocity has units of inverse seconds).
 
  • #8
22,145
5,078
It depends on what you mean by 'top of a mountain.' I interpret that to mean that the road is flat which means the velocity would be horizontal, and there would be no acceleration (assuming the person is not using the controls to do so).
We clearly have a difference in interpretation here. In the problem statement, they didn't have to mention that it was on the top of a mountain if they didn't expect you to take into account the curvature (and, instead, assume the car was on a flat horizontal surface).

Chet
 
  • #9
brainpushups
437
186
I agree with you 100%, but why not say 'on a hill' or something? 'Top of a hill' to me says something different. I think you're right that the author of the question intends to discuss the component of gravity on a sloped surface.
 
  • #10
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I agree with you 100%, but why not say 'on a hill' or something? 'Top of a hill' to me says something different. I think you're right that the author of the question intends to discuss the component of gravity on a sloped surface.
I never said that. I said the author intended us to consider a curved mountain surface when the car is at the peak of the mountain.
Chet
 
  • #11
brainpushups
437
186
Yes, after reading your initial post again I see that is what you meant. I think we can probably agree that the question could have been phrased more clearly and leave it at that.
 
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