Finding the force of Tractive resistance on a vehicle

[marine345]

Homework Statement

I working on this for a high school thesis. I am trying to find the power required to move a 7100 lb vehicle over a variety of scenarios, but am having trouble calculating the force of tractive resistance.

Homework Equations

I found what i thought was a pretty exhaustive equation for the force of tractive resistance:
F_TR=mg[sin$\alpha$+C₀sgn(V)]+sgn(V)[mgC₁+($\rho$/2)C_DA_F]V²+ma
where:
C₀=dimensionless coefficient of rolling resistance
$\alpha$=the angle of the surface the vehicle is navigating
C₁=Coefficient of rolling resistance while in motion, calculated by (when C₀=0.01) C₁=C₀(V²/100)
$\rho$=air density (lb/ft³
C_D=drag coefficient
A_F=frontal area of the vehicle (ft²)
V=velocity (fps)
g=32fps²

The Attempt at a Solution

For my first scenario, finding F_TR when the vehicle is going 20mph in sand, I plugged in these numbers and got:
F_TR=(7100)(32)[sin(0)+0.35]+[(7100)(32)(0.35(29.3/35))+(0.0718/2)(0.75)(22)](29.3)²=5.72299(10^7)
I used 32fps² as the acceleration of gravity, 29.3fps as velocity because that would give me the answer in ft/lbs, and calculated C₁=0.35(29.3/35) because I figured '100' would change based on C₀. I also thought this number was astronomically large, especially since it would be multiplied by velocity to find the required power to meet the scenario requirements. I have also donr the same with several other scenarios, and gotten really big numbers

 

[rude man]

Homework Statement

I working on this for a high school thesis. I am trying to find the power required to move a 7100 lb vehicle over a variety of scenarios, but am having trouble calculating the force of tractive resistance.

Homework Equations

I found what i thought was a pretty exhaustive equation for the force of tractive resistance:
F_TR=mg[sin$\alpha$+C₀sgn(V)]+sgn(V)[mgC₁+($\rho$/2)C_DA_F]V²+ma
where:
C₀=dimensionless coefficient of rolling resistance
$\alpha$=the angle of the surface the vehicle is navigating
C₁=Coefficient of rolling resistance while in motion, calculated by (when C₀=0.01) C₁=C₀(V²/100)
$\rho$=air density (lb/ft³
C_D=drag coefficient
A_F=frontal area of the vehicle (ft²)
V=velocity (fps)
g=32fps²

The Attempt at a Solution

For my first scenario, finding F_TR when the vehicle is going 20mph in sand, I plugged in these numbers and got:
F_TR=(7100)(32)[sin(0)+0.35]+[(7100)(32)(0.35(29.3/35))+(0.0718/2)(0.75)(22)](29.3)²=5.72299(10^7)
I used 32fps² as the acceleration of gravity, 29.3fps as velocity because that would give me the answer in ft/lbs, and calculated C₁=0.35(29.3/35) because I figured '100' would change based on C₀. I also thought this number was astronomically large, especially since it would be multiplied by velocity to find the required power to meet the scenario requirements. I have also donr the same with several other scenarios, and gotten really big numbers

You say it's a 7100 lb. vehicle. So that's not its mass, that's its weight, and you should not be multiplying by 32.2 ft/sec2.

Then there is the problem of C1, which is apparently some incremental amount to C0 to account for motion - BUT mg*C1*V^2 should be << mg*C0, seems to me. To first order, C1 = 0 as taught in introductory 'serious' college-level physics. I personally have never encountered such a coefficient.

With these two changes you would reduce your calculation of force by around 64:1.

 

[marine345]

Ok so 7100 lbs is mg. that helps. And the C1 term has to be in there because rolling resistance increases with velocity. I guess C0 is what you have to overcome to begin moving

 

[rude man]

No, that's static friction, and it plays no part in a rolling scenario. I would investigate C0 vs. C1 more, were I you.

 

[marine345]

Then there is the problem of C1, which is apparently some incremental amount to C0 to account for motion - BUT mg*C1*V^2 should be << mg*C0, seems to me. To first order, C1 = 0 as taught in introductory 'serious' college-level physics. I personally have never encountered such a coefficient.

Using C1 reduces the magnitude of the answer pretty seriously. Are you saying that I should sub in mg*C0 for C1?

 

[rude man]

How does it reduce the magnitude of F? Or is C1 < 0?

 

[marine345]

Yes, C1 < 0. For the above scenario, it would be 0.35(29.3/35)=0.293

 

[rude man]

But you considered C1 > 0 in your calculation:

, I plugged in these numbers and got:
F_TR=(7100)(32)[sin(0)+0.35[B]]+[(7100)(32)(0.35(29.3/35)[/B])+(0.0718/2)(0.75)(22)](29.3)²=5.72299(10^7)

I personally doubt that rolling friction is much of a function of velocity. As I said, in elementary courses it's treated as the same. And I'm talking physics-major level.

I would just omit the C1 term, leaving of course the air-resistance term. A riolling coefficient of 0.35 sounds very reasonable to me.

 

[marine345]

Alright, so recalculating without C1 and using 7100 as mg gives me 7570.27 ft/lbs, assuming my units are correct. Multiplied by the velocity, 29.3fps, gives me 221,809 ft/lbs/s, which converts to 403.289 hp. Does this sound correct?

 

[rude man]

Alright, so recalculating without C1 and using 7100 as mg gives me 7570.27 ft/lbs, assuming my units are correct. Multiplied by the velocity, 29.3fps, gives me 221,809 ft/lbs/s, which converts to 403.289 hp. Does this sound correct?

Why 7570.27 ft/lbs? Your equation is a force equation so the result is force, which is lbs, right? But you're closing in on it.

Ok, I'll tell - your answer should be numerically correct if you converted ft-lbs/s to hp.

So it's really 7570.27 lbs, times velocity which is ft/s, and the answer is thus in ft-lbs/s which converts directly to horsepower.

Power = force*distance / time = force*velocity.

I leave you with one urgent note: pay attention to units! Checking units in an equation is about the best way to avoid mistakes in math. Check that every term has the same units. Keep track of every parameter, constant, whatever, in terms of units.