Finding a spring constant of a car

In summary, the conversation discusses the effect of a heavy person getting into a car on the spring constant and natural frequency of the car's springs. The car has a total mass of 2000kg and a ground clearance of 40cm, but with the person added, the ground clearance decreases to 30cm. The problem is that there are two unknowns (the weight of the person and the spring constant) and only one equation. The conversation suggests using two equations by considering the clearance and compression of the springs, but the equation for the natural frequency of the spring is still needed.
  • #1
ChickysPusss
13
1

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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5


First of all, I want to commend you for your efforts in trying to solve this problem. As a scientist, it is important to persevere and keep trying even when faced with challenges.

In order to find the spring constant of the car, we need to use the equation F = -kx, where F is the force applied by the spring, k is the spring constant, and x is the displacement from the equilibrium position. In this case, the displacement is the difference between the original ground clearance (40cm) and the new ground clearance (30cm), which is 0.1m.

We also need to consider the weight of the car and the person inside. When the ground clearance was 40cm, the total weight of the car was 2000kg * 9.8m/s^2 = 19,600N. When the ground clearance was reduced to 30cm, the total weight of the car and the person was (2000kg + Xkg) * 9.8m/s^2 = (2000 + X) * 9.8N.

Using the equation F = -kx, we can set up the following equation:

(2000 + X) * 9.8N = -k * 0.1m

Solving for k, we get:

k = (2000 + X) * 9.8N / 0.1m

However, we still have the variable X, which represents the weight of the person. Without knowing the exact weight of the person, we cannot solve for the spring constant. We would need more information in order to find the value of X and ultimately, the spring constant of the car.

I understand that you want to prove your physics teacher wrong, but it is important to remember that science is about finding answers based on evidence and data. If we do not have all the necessary information, we cannot come to a conclusive solution. Keep up the good work and continue to ask questions and seek understanding.
 

1. What is a spring constant?

A spring constant is a measure of the stiffness of a spring, representing the amount of force required to stretch or compress the spring by a certain distance. It is typically denoted by the letter k and has units of Newtons per meter (N/m).

2. How do you find the spring constant of a car?

To find the spring constant of a car, you will need to perform a simple experiment. First, measure the mass of the car and the distance the spring is compressed when the car is placed on it. Then, apply a known force to the spring and measure the distance it is compressed. The spring constant can be calculated using the equation k = F/x, where F is the applied force and x is the distance the spring is compressed.

3. Why is it important to know the spring constant of a car?

The spring constant of a car is important because it affects the car's suspension system and overall ride quality. A higher spring constant means a stiffer suspension, which can provide better handling but may also result in a rougher ride. A lower spring constant means a softer suspension, which can provide a smoother ride but may also affect the car's handling.

4. Can the spring constant of a car change?

Yes, the spring constant of a car can change over time. Factors such as wear and tear, temperature changes, and modifications to the car's suspension system can all affect the spring constant. It is important to regularly check and adjust the spring constant to ensure optimal performance and safety of the car.

5. What are some common methods for measuring the spring constant of a car?

There are several methods for measuring the spring constant of a car, including using a force gauge to apply a known force to the spring and measuring the resulting displacement, using a spring tester machine, or performing a road test to observe the car's handling and ride quality. It is recommended to use multiple methods for accuracy and to confirm the results.

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