Calculating Deceleration Forces: Is It Easier to Use g's or Mass?

  • Thread starter Thread starter wirenut67
  • Start date Start date
  • Tags Tags
    Deceleration Force
AI Thread Summary
To calculate the force experienced by a 150 lb person during deceleration from 12 mph to 8 mph in 0.5 seconds, the correct approach involves converting speeds to feet per second and using standard units. The deceleration rate is found to be 11.8 ft/s², derived from the change in velocity over time. The mass must be converted to slugs by dividing the weight in pounds by the acceleration due to gravity (32 ft/s²), resulting in approximately 4.6875 slugs. The force is then calculated using the formula F = ma, yielding a force of about 55.31 lbs. This force represents a noticeable reaction for the vehicle occupants, and some suggest expressing deceleration in g's for simplicity.
wirenut67
Messages
15
Reaction score
0
Greetings,

First post here.

I am trying to calculate the force applied to a 150 lb. person riding in a vehicle that decelerates from 12mph to 8 mph in .5 seconds.

If I am understanding correctly the formula is:

Initial velocity-Final Velocity/.5

So
12-8/.5 = 8

The to calculate the force,

m x d = f

so mass is 150 lbs

Deceleration is 8

so force is 1200 Lbs?

This does not seem correct to me, any help would be appreciated.

Thanks!

Mike
 
Physics news on Phys.org
You are trying to apply Newton's 2nd law (F = ma), but you are messing up the units.

Use standard units for velocity: ft/sec, not mph.

Mass is 150/32 slugs, not 150 lbs.
 
Doc Al said:
You are trying to apply Newton's 2nd law (F = ma), but you are messing up the units.

Use standard units for velocity: ft/sec, not mph.

Mass is 150/32 slugs, not 150 lbs.

Are you telling me that mass will be 150 LBS divided by 32 slugs?

This is the part I am not understanding.
 
This is why I always convert to metric before I do these calculations. I try to avoid any equation with the word "slug" in it like the plague.
 
wirenut67 said:
Are you telling me that mass will be 150 LBS divided by 32 slugs?
No. The mass is measured in slugs. (32 is the acceleration due to gravity in ft/s^2.)

In the British system of units, mass is measured in slugs. The weight (force) is related to the mass via weight = mg, where g is the acceleration due to gravity. In that system of units, g = 32 ft/s^2 (approximately).
 
Ok so for the first part I would use:
Initial Velocity= 17.6 ft/s (12mph)
Final Velocity = 11.7 ft/s (8mph)
Deceleration time = .5 seconds
So then

17.6-11.7= 5.9 ft/s

5.9ft/s /.5 seconds = 11.8 ft/s

Is this correct for the first part of the calculation?
 
wirenut67 said:
Ok so for the first part I would use:
Initial Velocity= 17.6 ft/s (12mph)
Final Velocity = 11.7 ft/s (8mph)
Deceleration time = .5 seconds
So then

17.6-11.7= 5.9 ft/s

5.9ft/s /.5 seconds = 11.8 ft/s

Is this correct for the first part of the calculation?
Yes. (But the units for acceleration are ft/s², not ft/s.)
 
Doc Al said:
Yes. (But the units for acceleration are ft/s², not ft/s.)

So the final result would be 11.8 Ft/s²?

Al I really appreciate your patience here. This is just something I am beginning to understand.


Mike
 
So written correctly it should be:

17.6 ft/s²-11.7ft/s²= 5.9ft/s²

and then

5.9 ft/s² /.5= 11.8 ft/s²

Is this correct Sir?
 
  • #10
wirenut67 said:
So the final result would be 11.8 Ft/s²?
Yes.

wirenut67 said:
So written correctly it should be:

17.6 ft/s²-11.7ft/s²= 5.9ft/s²
The units for speed would be ft/s, not ft/s².

and then

5.9 ft/s² /.5= 11.8 ft/s²
5.9 ft/s /.5 s = 11.8 ft/s²
 
  • #11
Doc Al said:
Yes.


The units for speed would be ft/s, not ft/s².


5.9 ft/s /.5 s = 11.8 ft/s²

Understood so far.

So my rate of deceleration is 11.8 ft/s²

So now I move to force.

Force= mass x deceleration.

Is this correct Al?

So my deceleration is 11.8ft/s²

Where I am lost is in figuring the mass based on slugs. Can you show me how to do this?
 
  • #12
wirenut67 said:
Understood so far.

So my rate of deceleration is 11.8 ft/s²

So now I move to force.

Force= mass x deceleration.

Is this correct Al?

So my deceleration is 11.8ft/s²
All good.

Where I am lost is in figuring the mass based on slugs. Can you show me how to do this?

Weight = mass X g(acceleration due to gravity)

so:

mass = Weight/g(acceleration due to gravity)

Here we measure weight in pounds, g is 32 ft/s², and the mass is in slugs. So:

mass = [150 lbs]/[32 ft/s²] =
 
  • #13
mass = [150 lbs]/[32 ft/s²] =4.6875 slugs?
 
  • #14
So then

4.6875 slugs x 11.8 ft/s ^2 is a force of 55.31 slugs or 55.31 lbs of force?
 
  • #15
wirenut67 said:
So then

4.6875 slugs x 11.8 ft/s ^2 is a force of 55.31 [STRIKE]slugs[/STRIKE] or 55.31 lbs of force?
The force will be in pounds, not slugs. (The mass was in slugs.)
 
  • #16
Doc Al said:
The force will be in pounds, not slugs. (The mass was in slugs.)

So the final result is that the force is 55.31 lbs of force?
 
  • #17
wirenut67 said:
So the final result is that the force is 55.31 lbs of force?
Right. That's the average force for the given acceleration.
 
  • #18
Al could I bother you for just a second to review this for me.

I am posting this written like this so please look this over for me.

We can calculate the force created by braking from 12mph to 8 mph in .5 seconds, and its force upon a 150lb human.

To calculate the deceleration rate we take this formula:

Initial Velocity - Final Velocity / time = deceleration rate.

First we convert mph to ft/s:

12mph is 17.6 ft/s
8mph is 11.7 ft/s
Braking time is .5 seconds

So

17.6 ft/s - 11.7 ft/s = 5.9 ft/s

then

5.9 ft/s / .5 seconds (time) = 11.8ft/s²

Now we can figure force.

Force = mass x deceleration

deceleration as defined above = 11.8 ft/s²

Mass has to be converted to slugs
A slug is weight/g (acceleration due to gravity)

So

150lbs/32 ft/s²= 4.6875 slugs. This is the mass in slugs based of deceleration due to gravity.

So

Mass= 4.6875
Deceleration = 11.8ft/s²

11.8ft/s² (deceleration) x 4.6875 (mass)= 55.31 pounds of force.If this is correct than you have done a fantastic job for me AL and it is much appreciated. I thank you for your patience.Mike
 
  • #19
Looks good to me!
 
  • #20
Doc Al said:
Looks good to me!

Again Al many thanks!

In your opinion would this amount of force create a noticeable reaction from the vehicle occupants?
 
  • #21
wirenut67 said:
In your opinion would this amount of force create a noticeable reaction from the vehicle occupants?
It's 1/3 their weight, so it's noticeable. (No big deal, though.)
 
  • #22
Doc Al said:
It's 1/3 their weight, so it's noticeable. (No big deal, though.)
Agreed and Thanks Again Al!
 
  • #23
Wouldn't it be easier to state the decel rate in g's, then multiply the mass by that to get the force?

11.8 ft/sec^2/32 ft/sec^2 = 0.369 g
0.369 g x 150 lbs = 55.3 lbs

Same thing but in a different order; makes more sense to me because I deal with g's rather than slugs.
 
Back
Top