Forces Help

  • Thread starter Prim3
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  • #1
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Hello, I have this Physics problem that's due tomorrow in class and I'm having some troubles with it.

Homework Statement



PROBLEM: A racing car has a mass of 1500kg is accelerating at 5.0m/s^2, is experiencing a life force of 600N up (due to its streamlined shape) and group effects of 1000N down (due to air dams and spoilers). Find the driving force needed to keep the car going given that "u" (micro) = 1.0 for the car.

Homework Equations



This is where the problem is. I always get confused and have no idea what formula to use where. To be honest, I'm not sure what I am supposed to find. Fn, Ffr, Fg or Fa. Are there any tips that people have or suggestions on how to determine what formula I need? Also, I get confused on what is what in a question. Like, if it's Fa or Fg or what Ffr.

Some formulas that I do know which I used:

W = mg
Fa - Ffr = ma
Ff = "u"W

The Attempt at a Solution



Regardless of that, I still tried the question. I did get an answer but I would like it if you guys could explain the whole thing. And, not just numbers. I actually want to know why I subtracted 1000N - 600N = 400N and added it to the 22000N.

Heres my work:

http://img262.imageshack.us/img262/3125/scanhe3.jpg [Broken]

Also, I wanted to know if there were any forces tutorials anyone knows about. It'd be really nice if they were video lectures but anything is appreciated since I have a test coming up on it soon.

Thanks for all your assistance.
Prim3.
 
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Answers and Replies

  • #2
PhanthomJay
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You have made a couple of errors. Your force diagram looks good. But the friction force is not uW, it is uN. You should calculate N by looking in the vertical direction only and using Newton's first law. Then look in the horizontal direction only and calculate Fa using Newton's 2nd law.
 
  • #3
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What's uN? We haven't been taught that yet. I'll check the vertical and horizontal thing. Thanks.
 
  • #4
nrqed
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What's uN? We haven't been taught that yet. I'll check the vertical and horizontal thing. Thanks.
"N" is the normal force, it's the force exterted by the ground on the car (so that the car does not "sink" in the ground!).

One thing is that your free body diagram is missing forces. That should always be your starting point: having all the forces shown clearly in your FBD.

By "life force" I guess you meant a "lift force"? This should be another force on your diagram pointing up, in addition to the normal force "N"
(I am not sure what you mean by F_N in your diagram).

I am not sure either about what a "group effect" is, I have never heard that. I am assuming it's a force pushing down due to the flow of air around the car. This shoul dbe yet another force on your car, pointing downward. So you should have 6 forces in your free body diagram.

Now, all you have to do is impose that the net force along y is zero. This will allow you to solve for the normal force "N".

Then you go along the x axis and impose [itex] \sum F_x = m a_x [/itex]. The friction force is [itex] \mu N [/itex]. That will allow you to find th edriving force.


Hope this helps

Patrick
 
  • #5
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Ahh, we refer to every force as this:

Fn would be Normal Force
Ff would be Force of friction
Fg would be Force of gravity
Fa would be Applied Force

Those Fn, Ff, Fg and Fa are on the diagram. Those are my forces.
 
  • #6
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well nrqed is absolutely correct in suggesting that you need to include two more forces, the lift and ground effects which will cause the normal force to be 400N more than the weight of the car as you have mentioned in the original post. As nrqed said, the normal force is the reactive force to the sum of all vertical forces, including the vertical component of forces directed at an angle. Once you have N, then Ff=mu*N. I believe the rest of your analysis is fine, eg Fa-Ff=m*5 m/s^2
 
  • #7
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What's uN? We haven't been taught that yet. I'll check the vertical and horizontal thing. Thanks.
[tex]F_f=(\mu_k)(F_N)[/tex] is your Friction force isn't it? I think that's what he ment by "uN"
 

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