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Finding average horizontal force?

  • #1

Homework Statement


A car is traveling at 60.0mph when it collides with a stonewall. The car comes to rest after the first foot of the car is crushed. What was the average horizontal force acting on a 150lb driver while the car came to rest?

Homework Equations


I'm not a hundred percent sure which equations to use but, I searched through questions I've had in the past and found an example however, it deals with distance.

vf^2=vi^2+2ax

SIGMA F= MA

The Attempt at a Solution


Again, I wasn't quite sure where to even start so, I decided to find acceleration because I thought it might be important with SIGMA F=MA. I've listed the givens which, may or may not be correct.
Vi = 60MPH ---> 88 FT Per Second
Vf = 0 m/s
m = 150lbs
 

Answers and Replies

  • #2
Nathanael
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If I told you how long the collision lasted, would you be able to tell me the average force?
 
  • #3
Do you mean time?
If so, I think I would.
 
  • #4
Nathanael
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Do you mean time?
Yes I mean the time that it took the car to come to rest from when it first touched the wall.

If so, I think I would.
Specifically, if I say the collision-time is T, then tell me what the average force is.


If you can figure out the average force based on the duration of the collision, then the problem becomes a problem of finding "how long did the collision last?"
Any ideas about how to find how long the collision lasted?
 
  • #5
I'm not sure. I do know that Impulse is a force applied over a period of time. J=Ft
And I also know that mΔv = Ft.

I'm so sorry. I honestly don't understand this question. I get that I need to find time in order to help find acceleration to find ∑F. However, this is a collision problem, so wouldn't I have to use momentum? Yet the collision equation needs two masses and two velocities in order to find it. I know this is possibly off topic to your original question and I honestly apologize. Is there something that I'm not seeing? The equations given by my teacher do not help.
 
  • #6
Nathanael
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I also know that mΔv = Ft.
Good.

wouldn't I have to use momentum?
Indeed, the way you would use momentum is in the above equation, m∆v=Ft

Yet the collision equation needs two masses and two velocities in order to find it.
I have a feeling you might have been asking "wouldn't I have to use momentum conservation?"
Conservation of momentum isn't useful here. The car is essentially colliding with the entire Earth (in a sense) because the momentum that it loses is gained by the Earth. You can't hope to measure the insignificant change in velocity of the Earth (and even if you could, it wouldn't be useful for what we're trying to find).


About the problem,
You can pretend the force on the person is always equal to the average force. What this means is that the force is constant, which means that the acceleration is constant.

So you want to find how long the collision lasted. You know 3 things:
You know the acceleration is constant (which is not true, but if we only care about the average force, then that is how we treat the situation)
You know the initial velocity
And you know how far the car travelled in the collision-time
(You were told the car was crushed by 1 foot, so how far did the car travel from the moment it first touched the wall to the moment it stopped moving?)

You'll have to use those 3 pieces of information to find the collision-time.
 
  • #7
Vi = 88 ft per second
a= 0 m/s ^2
x = 1 ft

Would I have to use the kinematics equation, x=vit + (1/2)at^2, in order to find time?

Also, are any of my givens incorrect?
 
  • #8
Would I end up using quadratic or is that too far?
 
  • #9
Nathanael
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Vi = 88 ft per second
a= 0 m/s ^2
x = 1 ft
a = 0? That doesn't make sense. There is an average force acting on the person Favg, so the acceleration would be Favg/m
but anyway the value of a is not important
Would I end up using quadratic or is that too far?
It isn't necessary. Do you know an equation for the average velocity given that acceleration is constant? Use that to find [itex]V_{avg}[/itex] and then find the time by [itex]V_{avg}[/itex]*time=distance
 
  • #10
haruspex
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given that acceleration is constant
That is the key question here.
If the acceleration is constant then the force is constant, and you can use ##E = F\Delta s##. And since you are given no other way to solve the question, it seems you are expected to assume constant force. Yet the question asks for average force. If the force varies then you do not have enough information. Finding the duration is crucial.
At this point you have my permission to sack the problem setter - he or she has inadequate expertise to be teaching physics.
In a car crash, the force obviously starts low but rapidly increases. In a well designed car body, the force should rise to its maximum quite quickly and stay at about the same value as long as possible. This minimises the maximum force exerted on passengers. So taking the force to be constant might be reasonable, but the question should tell you to assume that it is.
 
  • #11
Hello haruspex,
The formula you suggested; I've never seen it before. What does it mean?
 
  • #12
Do you know an equation for the average velocity given that acceleration is constant? Use that to find [itex]V_{avg}[/itex] and then find the time by [itex]V_{avg}[/itex]*time=distance
So, I need to find an equation that equals V avg?
If so, I have Vi+Vf/2.
I'm not sure if acceleration is constant in that equation.

And if that is correct, I can use Vavg / x = t, right?
 
  • #13
haruspex
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Hello haruspex,
The formula you suggested; I've never seen it before. What does it mean?
Work = force times distance.
 
  • #14
haruspex
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So, I need to find an equation that equals V avg?
If so, I have Vi+Vf/2.
I'm not sure if acceleration is constant in that equation.
Yes, it assumes constant acceleration.
And if that is correct, I can use Vavg / x = t, right?
That's dimensionally wrong. Try again.
 

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