How Do You Calculate Tractive Force for Accelerating a Car?

In summary: Yes, this is correct. The average acceleration is calculated as the net force (tractive force + wind resistance) required to achieve the desired velocity.
  • #1
Gixer1127
28
12
Hello all,
This is my first post so please be gentle with me..
I'm trying to work my way through some tricky questions so will post them below and see if anyone can give me some pointers.
The questions are:
A car weighs 1305kgs and accelerates from 0 to 160km/h in 4.2s. The first question asked what the average acceleration would be. I converted the terminal speed to m/s which came out at 44.44m/s.
Then I did a calculation for the average acceleration as shown
ā = (44.44m/s – 0m/s) / 4.2s = 10.58m/s²
Is this correct?
Next question: The tractive force produced by the car to provide this acceleration. Note you must also consider the additional 1000N resistive force.
This is where I am totaly stuck and any help in pointing me in the right direction would be most appreciated.
Cheers,
Gixer1127
 
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  • #2
Welcome, @Gixer1127 !
Where those 1000 Newtons come from?
 
  • #3
Hi Lnewqban,
Sorry, I should have posted the complete description: Average frictional and wind resistance forces can be assumed to be 1000N during the acceleration.
 
  • #4
Hello @Gixer1127 ,
:welcome: ##\qquad ##!​

Gixer1127 said:
Is this correct?
So far, so good.

##\ ##
 
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  • #5
Gixer1127 said:
Hi Lnewqban,
Sorry, I should have posted the complete description: Average frictional and wind resistance forces can be assumed to be 1000N during the acceleration.
What is the NET force required to achieve this acceleration ?
 
  • #6
BvU said:
What is the NET force required to achieve this acceleration ?
That's all the info that's in the question. There is a third part to it but I was hoping that if I could get a help with this one I'd manage the last part myself
 
  • #7
You have enough information to calculate the net force required to acelerate a mass of 1350 kg to the tune of 10.58 m/s2 ...

[edit:]Our homework fora have a nice template with a very useful part: relevant equations (required) !

##\ ##
 
  • #8
What equation do you know that involves forces, mass, and acceleration?
 
  • #9
jack action said:
What equation do you know that involves forces, mass, and acceleration?
Hi Jack Action
Newtons 2nd law. So if I use this where do I input the -1000N? F=ma =ans-1000N? This is the bit I'm stuck on.
 
  • #11
Ok, so does this mean net force is tractive force? I'm doing the F=ma equation and getting 13806.9N as the answer. This seems excessive but since I've zero experience in these types of formulas and equations maybe not? I've not taken out the 1000N resistance so guessing i just subtract this from my answer?
 
  • #12
So the car is pushing as much as it weights, more or less (1 G).
Then, the tires-pavement must have a coefficient of friction of around 1.0, which is higher than street tires have (around 0.8 for dry conditions).

As that push can't be steady, due to engine curve, increasing air drag, and transmission changes, it should have a greater value at times.
 
  • #13
Gixer1127 said:
Ok, so does this mean net force is tractive force? I'm doing the F=ma equation and getting 13806.9N as the answer. This seems excessive but since I've zero experience in these types of formulas and equations maybe not? I've not taken out the 1000N resistance so guessing i just subtract this from my answer?
No, this is wrong.

To find the answer, do a free body diagram (FBD). You have the tractive force, the wind resistance, and the inertia (ma). All of these are vectors. Direction matters. Do the FBD, write the equations, and show them to us in your next post.
 
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  • #14
Lnewqban said:
So the car is pushing as much as it weights, more or less (1 G).
Then, the tires-pavement must have a coefficient of friction of around 1.0, which is higher than street tires have (around 0.8 for dry conditions).

As that push can't be steady, due to engine curve, increasing air drag, and transmission changes, it should have a greater value at times.
Right, so my very inexperienced brain is thinking that to have an average acceleration you need a constant force. The question gave the resistance (total) at 1000N. If the answer to the question is equal, within a very small measurement of the mass of the car, then is this not correct? I did miss a bit from the original question. The car is fully electric so no pick up through the gears, just foot to the floor. Would this make a difference given that I am told the resistance is 1000N?
I'm not at my laptop so don't have the 3rd question but it might influence this answer. I'll post tomorrow and see what you folks can hint at for me to head in the right direction.
 
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  • #15
jack action said:
No, this is wrong.

To find the answer, do a free body diagram (FBD). You have the tractive force, the wind resistance, and the inertia (ma). All of these are vectors. Direction matters. Do the FBD, write the equations, and show them to us in your next post.
Sorry Jack Action missed this post and jumped to the next one. I'll look at this tomorrow and go down the route you have suggested.
Whisky in the back garden time. A beautiful evening in the Highlands of Scotland.
 
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  • #16
1681205733867.png

1681205788937.png

This is what I have so far. I've looked at other posts along a similar line and followed their examples with regards FR. Is any of this correct or am I missing something? The posts I saw had the 2 answers added together F + FR to give the force required to travel the object at a constant acceleration. The question I've been asked is what is the tractive force produced to provide this acceleration taking into account the 1000N resistance. Same answer?
 
  • #17
Gixer1127 said:
View attachment 324733
View attachment 324734
This is what I have so far. I've looked at other posts along a similar line and followed their examples with regards FR. Is any of this correct or am I missing something? The posts I saw had the 2 answers added together F + FR to give the force required to travel the object at a constant acceleration. The question I've been asked is what is the tractive force produced to provide this acceleration taking into account the 1000N resistance. Same answer?
Or do I subtract the 2 answers F - FR = 1285.05N as the answer? This, to me, makes more sense as surely the resistance is the force the car needs to overcome to maintain acceleration and then hold a constant speed?
 
  • #18
Gixer1127 said:
...The car is fully electric so no pick up through the gears, just foot to the floor. Would this make a difference given that I am told the resistance is 1000N?...
Being electrically powered, you are correct, the torque on the driving wheels (and the tangential force on the pavement) should be about constant.

Those 1000 N resist the forward movement, not in a linear way, as air drag increases with the square of speed.
Therefore, at the beginning of the run is about zero, and at the end is greater than the estimated value: 1000 N is just an average.

The wheels must push harder to overcome that resistance and still accelerate the mass of the car as it did.
It is like the car is going slightly uphill while accelerating.
Therefore, those 1000 N should be added to the product of mass times actual acceleration.
 
  • #19
Gixer1127 said:
View attachment 324733
View attachment 324734
This is what I have so far. I've looked at other posts along a similar line and followed their examples with regards FR. Is any of this correct or am I missing something? The posts I saw had the 2 answers added together F + FR to give the force required to travel the object at a constant acceleration. The question I've been asked is what is the tractive force produced to provide this acceleration taking into account the 1000N resistance. Same answer?
The friction coefficient ##\mu## is not given in the problem so forget about it. The force ##F## is what you are looking for. You have put the wind resistance vector on your FBD but not the inertia vector (ma). Add those three vectors (considering their correct directions) to find the magnitude of ##F##.
 
  • #20
OK, let's see if I get this correct. On my FBD I should have input the result of the F=ma (13806.9N)? Then add this to the 1000N for wind resistance vector and gravity/mass vector (not including the coefficient) (12802.05N). Considering the wind resistance vector is pushing back and the gravity/mass is on a downward trajectory does this mean they are both in the negative? Adding them together gives me an F of 27608.95N. Am I on the right track with this?
 
  • #21
Gixer1127 said:
OK, let's see if I get this correct. On my FBD I should have input the result of the F=ma (13806.9N)? Then add this to the 1000N for wind resistance vector and gravity/mass vector (not including the coefficient) (12802.05N). Considering the wind resistance vector is pushing back and the gravity/mass is on a downward trajectory does this mean they are both in the negative? Adding them together gives me an F of 27608.95N. Am I on the right track with this?
Or is it add the 1000N to the sum of mass x acceleration to get the F?
 
  • #23
Gixer1127 said:
Or is it add the 1000N to the sum of mass x acceleration to get the F?
This is right but you don't seem sure.

If you had done the FBD as I asked, it should be clear. In the figure you presented in post #16 you show the force ##F## pointing to the right and the wind resistance pointing to the left. This is good. The ##ma## vector should be opposing the ##a## vector. In your case, it will point to the left. Intuitively, you should be able to figure out that a car that is accelerating (acceleration pointing to the right) requires an effort from your engine. Thus the ##ma## vector is pointing to the left. So you get:
$$F = F_{wind} + ma$$
Do you see how your tractive force ##F## is opposing both the wind and the inertia of your vehicle? The greater the wind or the greater the acceleration, the greater the required tractive force. Does it make sense?
 
  • #24
The net horizontal force is what produces the horizontal acceleration that you have calculated.
That net force is the summation of all the horizontal forces acting on the chassis of our car.

The shaft of each tire-motor assembly pushes the chassis forward as the contact patch acts as an instantaneous grounded pivot.

The 1000 N force simultaneously pushes the chassis rearward.

Therefore:
Net force = car mass x acceleration = F shaft - 1000 N

Car mass x acceleration + 1000 N = F shaft

Because those 1000 N of resistive force, the batteries must supply more electricity to the motors, which must produce a higher torque than the one needed only to accelerate the car's mass, in the idealized case of absence of friction and air resistance.

image33.png
 
  • #25
1681293738217.png

Folks, is this any closer to the correct answer?
 
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  • #26
Gixer1127 said:
View attachment 324767
Folks, is this any closer to the correct answer?
The FBD is wrong. The tractive force ##F## is not on it and the inertia vector is in the wrong direction.

Can't you see that 13806.9 N is not equal to 1000 N? Use this one as an inspiration (source):

accelerating-forces.gif

 
  • #27
jack action said:
The FBD is wrong. The tractive force ##F## is not on it and the inertia vector is in the wrong direction.

Can't you see that 13806.9 N is not equal to 1000 N? Use this one as an inspiration (source):

Damn, back to the drawing board.
So the mass x acceleration is pointing the wrong way? I've got the car driving left to right so I thought that would have been the direction of inertia. I'm really struggling here but will persevere and get the right answer...eventually.
 
  • #28
OK, so is my answer of 14806.9N correct for the tractive force? I've been over all the posts and this is the only answer (so far) that I can come up with. If this is correct then I'll submit the wording only on how I accomplished this with a nod to Newtons 2nd Law of Motion. I do want to know where I've gone wrong with the FBD though and have seen an FBD with inertia (ma) pointing backwards. I'll have a look at this to try and figure out the why's and how's of this.
 

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  • #29
Yes, 14806.9 N is right.

For the FBD, there are two ways to look at it.

The way I explained it: What is pushing onto the car?
  • The ground is pushing the car with the tractive force
  • The wind is pushing the car in the other direction
  • The inertia is "pushing" the car backward (because the car doesn't want to slow down unless compelled to)
fbd-1.png

The opposite way: What is the car pushing against?
  • The car pushes against the ground
  • The car pushes against the wind
  • The car's inertia "pushes" it forward
fbd-2.png

Both analyses will have equal and opposite arrows but will lead to the same conclusions because of Newton's third law: To every action, there is always an opposed and equal reaction.

A third way of looking at way might be this way: A diagram on each side of the equal sign of ##\sum F = ma##. (The second one is actually called a kinetic diagram) This way, all arrow directions might be more intuitive.

fbd-3.png
 
  • #30
Many thanks. I've a few more "interesting" questions to get through on this assessment so you might see another post in the very near future. Thanks again for prompting me in the right direction.
 
  • #31
I think it was Sir Isaac Newton that stated something like this:
"A body at rest tends to stay at rest."
"A body in motion tends to stay in motion."

Since the body you are talking about is a car that is trying to accelerate, its tendency (inertia) is tending to keep it at rest.

I know, we usually use the word "inertia" in the case of something that is moving tends to keep moving; but it works both ways!

So for a FBD, the arrow for Inertia points toward where the object (car) would be if no force were applied to it.

Hope this helps!

Cheers,
Tom
 
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  • #32
Tom.G said:
I think it was Sir Isaac Newton that stated something like this:
"A body at rest tends to stay at rest."
"A body in motion tends to stay in motion."

Since the body you are talking about is a car that is trying to accelerate, its tendency (inertia) is tending to keep it at rest.

I know, we usually use the word "inertia" in the case of something that is moving tends to keep moving; but it works both ways!

So for a FBD, the arrow for Inertia points toward where the object (car) would be if no force were applied to it.

Hope this helps!

Cheers,
Tom
Thanks Tom. Yup, that's what I'm trying to get my head around. I'll get there in the end and have almost completed this particular assessment with the help of the folks on here. Very much obliged to all.
 
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  • #33
1682255666255.png
1682255666255.png

Hey folks. I've just had this come back as wrong from my tutor. He's saying the decimal point in (a) is in the wrong place. This in turn leads onto an incorrect answer for (b) and (c) Any pointers on where I've gone wrong?
 
  • #34
Gixer1127 said:
View attachment 325369View attachment 325369
Hey folks. I've just had this come back as wrong from my tutor. He's saying the decimal point in (a) is in the wrong place. This in turn leads onto an incorrect answer for (b) and (c) Any pointers on where I've gone wrong?
It's OK, I've seen the error of my ways
 
  • #35
Gixer1127 said:
... Hey folks. I've just had this come back as wrong from my tutor. He's saying the decimal point in (a) is in the wrong place. This in turn leads onto an incorrect answer for (b) and (c) Any pointers on where I've gone wrong?
In your calculations, avoid using mm for length, area or volume.
Before you start plugging numbers into equations, transfer all given data to the International System of Units:

https://en.wikipedia.org/wiki/International_System_of_Units

In your case, make those 16 mm of diameter 16/1000 meters first.
One good clue about your error in a) answer was the result of a fraction of 1 mm2 as the cross-section area of a circle of 16 mm diameter: impossible.

Another clue on b) and d) answers: Pascals are N/m2 rather than N/mm2.
 
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