Finding a spring constant of a car

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Homework Help Overview

The discussion revolves around determining the spring constant of a car based on its mass and the effect of an additional weight. The context includes the car's ground clearance before and after the weight is added, along with the natural frequency of the springs.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the car's weight, the added weight, and the resulting compression of the springs. There are attempts to derive equations based on the forces involved and the changes in ground clearance.

Discussion Status

Some participants have offered insights into the need for multiple equations to solve for the unknowns, while others question the validity of certain mathematical manipulations. The discussion is ongoing, with various interpretations being explored.

Contextual Notes

There are constraints regarding the unknown weight of the additional person and the spring constant, leading to a situation with multiple unknowns and insufficient equations. The natural frequency of the springs is also noted as a relevant factor that has yet to be fully integrated into the discussion.

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Homework Statement


A car has a total mass of 2000kg, and a ground clearance of 40cm. But some fat dude got in and he made the ground clearance 30cm. What is the spring constant for this bad boy of a car? (The dude was so fat he made the shocks act like springs if you can believe it.) Also, it's a fact that the natural frequency of said springs are .6667 HZ.

Homework Equations



F = -kx

The Attempt at a Solution



So I tried being like: (2000kg)*(9.8m/s^2) = -kx
and then I was like: (2000kg + Xkg) * (9.8m/s^2) = -k(x-.1m)

I tried solving for k as hard as I could cause I really want it, but there are too many variables and I can't figure out what to do. My physics teacher says he wants me to fail and I'd like to show him that I'm not a total turd. Thank you.
 
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The car weight compresses the springs to 40cm, but it is the car+dude that compresses the springs to 30cm. So the extra weight of the dude compresses the springs an extra how much?

Your trouble is that you don't know the weight of the dude or the spring constant ... that's two unknowns: so you need two equations.

You also need the equation for the natural frequency of a mass on a spring.
 
Ok Physics bros... I think I had a stroke of genius, thanks to Simon Bridge, but I'm wondering of the mathematical/physical legality of what I've done.

So...

(2000kg)*(9.8m/s^2) = -kx = (2000kg)*(g) = -kx

(2000kg + Xkg) * (9.8m/s^2) = -k(x-.1m)= (2000kg +Xkg) * (g)

Is it legal to do this...

(-k(x-.1m)= (2000kg +Xkg) * (g)) - (-kx = (2000kg)*(g)) = (X*g = -.1k)

Then say that k = -(X*g)/.1

Cause if that's k, I can plug it into other formula and I'm good to go.
 
Are you confusing the clearance with the compression?

Let the uncompressed clearance be x0 - then the car weight Mg compresses the springs by x, giving a clearance of 40cm ... eg.

40cm = x0-x: kx=Mg ... thus: 40cm = x0-Mg/k

similarly for Car + dude: 30cm = x0-(M+m)g/k

But you don't know x0, m, or k ... so that's two equations and three unknowns.

You are still missing the equation for the natural frequency of the spring.
 

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