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A car is pushed forward by a force F(t). The air resistenance acting on the car is \({F}_{air}\)= \(dv^2\)

 

Vertically the car is supported by a spring and damper system at each wheel. Each spring has a stiffness k and each damper has a coeficient of c.

 

The car can only move horizontally and/or vertically, so rotational
effets can be ignored.

 

1. Obtain an ODE in standard form of the car's horizontal velocity by applying Newton's Second Law
in the x-direction.

2.If the cars top speed is 100m/s and the force the car provides at this speed is 2000N, solve for the
drag coefficient d.

3.Obtain an ODE (in standard form) of the cars vertical displacement by applying Newton's Second
Law in the y-direction.

4.When Matt (80kg) gets into the car, the car body lowers by 0.02m, and then remains static. Given this,
solve the spring stiffness k acting at each wheel.

vest4R  Mar 13, 2018
 #1
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having trouble solving this, If anyone could even give me a tip to get started.

 

What I don't understand is,

Q1 how do you get the v^2 into an ode? how do I get rid of the square? If i derive it, I'll loose the drag coefficient d and thus can't solve Q2. If I was to then get v on it's own into an ODE. I'm not sure how to find d as i'll still have m as an unknown.

 

Q5 How do I find the spring stiffness k as I have the other unknowns of c and m.

 

Any help will be very greatly appreciated!

vest4R  Mar 13, 2018
 #2
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Here are parts (1) and (2)

 

.

Alan  Mar 13, 2018
edited by Alan  Mar 13, 2018
 #3
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Here are parts (3) and (4):

 

I'll leave you to put in a value for g and hence work out k (make sure your units are consistent).

Alan  Mar 13, 2018
 #4
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Thanks so much alan!

So I was making it much too complicated!

I just wanted to confirm with you, for question 4. once inputing the units m/s^2 for g,  k would = kg/s^2 ?

also as we're looking for the spring stiffness at each wheel, we're looking for k/4?

vest4R  Mar 13, 2018
edited by vest4R  Mar 13, 2018
 #5
avatar+26971 
+2

Yes and yes (I confess I didn’t notice the “per wheel” bit!).

Alan  Mar 13, 2018

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