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Pickup truck dollying multiple other pickup trucks

  1. Aug 12, 2011 #1
    Assume you have a brand new Diesel or gas pickup truck of any brand and factory specs you choose. Assume there is a dolly connected to that pickup truck. How many of the same pickup/hitch/dolly combination trucks could the first truck pull if you have them lined up like a train (and the trailing truck are in neutral). Assume that 5 mph is sufficient max obtainable speed to be "successful." It's my last curiosity question...I promise. Is it a hundred or even 150 trucks or maybe even more!?
  2. jcsd
  3. Aug 12, 2011 #2


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    Your question is more difficult to answer than you think because axle bearing drag and rolling resistance will play a dominant role in your problem. Initially getting rolling would be the hardest; once rolling accelerating would probably be relatively easy up to 20 mph or so. I would guess that a truck could probably tow a dozen or so pickups exactly like it, maybe more. A hundred sounds like too many though, it would burn it's tires without being able to accelerate.
  4. Aug 12, 2011 #3
    Thanks. Interesting. Would your answer be different if there was a 2,000 lb. block in the bed of the first pickup truck, considering the tires would have a lower propensity to lose traction?

    Also, considering that acceleration requires far more energy than keeping the trucks in motion, what if you assume all pickup trucks in the "train" had automatic transmissions and all had the gear selector in "D" and were idling and the "train" achieved maximum forward/drive velocity that each truck would would otherwise achieve on its own if it were not connected to anything (let's say that's 5mph), and then at some point the trailing trucks' engines were remotely shut off so that the first truck was the locomotive. How many trucks could the train sustain to keep itself in motion now that acceleration has been eliminated from the experiment?
  5. Aug 13, 2011 #4
    Accelerating from very slow to faster isn't an issue because you can do it as slowly as you like. It's the "keeping them in motion" that's hard. As Mech said, that problem is dominated by rolling resistance, and other drivetrain frictions.

    You could solve it with two bits of information:
    The rolling resistance of a tow truck
    The weight of the first tow truck

    The weight determines how much pulling force it can apply. Assume tyres have coefficient of friction of 1, and all wheels are driven, that means it can pull with a force equal to its own weight.

    Example with made-up numbers: Rolling resistance = 1000N per truck. Weight of 1st truck = 50000N. Then you can tow 50 trucks after you've started moving.
  6. Aug 13, 2011 #5
    >Accelerating from very slow to faster isn't an issue because you can do it as slowly as you like. It's the "keeping them in motion" that's hard.

    Actually Mech said exactly the opposite of that. He said accelerating is hard, keep them in motion is easier and obviously he is right. What you wrote defies the laws of Physics. I think they teach that in the first 3 days of any introductory Physics class. Someone correct me if I am wrong about the 3 days.
  7. Aug 13, 2011 #6

    Please apologize for your rude writing. Here's what I understood Mech to have said, and I agree:

    Changing from stopped to just moving takes force X
    Changing from just moving to a reasonable speed (maybe 20pmh) takes force Y
    Keeping moving at a constant speed takes force Z

    X>Y (static friction, if relevant, but it might be closer to X=Y)
    Y=Z in the limit of acceleration approaching zero (very gentle acceleration)

    The important point which I think you overlooked is that acceleration can be done as slowly as you like, so the additional force it requires can approach zero.

    What that all boils down to is the problem is dominated by rolling resistance. So you can use my method to estimate the solution.
  8. Aug 13, 2011 #7
    It was civil but tongue-in-cheek because acceleration-vs-constant speed energy consumption is just so fundamental in Physics and I really know almost nothing compared to real physicists like you all must be on here. I am someone without formal Physics training beyond an accelerated-level (no pun intended) Physics course in high school and that was decades ago. I can't use your formula, but rather I am looking for you or one of your brethren on the forum to use a formula of their choosing based on some arbitrarily chosen pickup truck with specs available to them on the internet or elsewhere. That is, not with made-up numbers as you've done to come up with a final number but with basically real numbers taken from specs and best guesses when necessary for various variables that are unknown (like axle bearing resistance). That it's a new pickup is arbitrary. That I chose 5mph is arbitrary. It might as well have been 1 mph. It's just obvious to me that even without knowing advanced physics that acceleration takes more energy than keeping an object at a constant speed. Just look at fuel efficiency of cars...they don't do nearly as well in stop and go traffic as they do cruising along at a constant 45mph. I though most people knew that. Maybe I'm wrong. Please tell me if so. Again, please read that all as civil and sincere and not sarcastic.
  9. Aug 13, 2011 #8
    OK, sorry if I misunderstood your comment, sarcasm doesn't communicate well in writing. You're right, acceleration does take extra energy. But that extra energy can approach zero if the acceleration approaches zero. You didn't specify any minimum acceleration so I assume it's OK if it takes all day to get up to 5mph.

    You can find the numbers easily enough. Just the weight of a tow truck, and the rolling resistance of a typical car/truck.
  10. Aug 13, 2011 #9
    Would you mind doing it for me? I'm really stupid when it comes to this stuff. Take a 2011 Ford F250 Diesel Quad Cab pickup truck for example, or anything else. I am just looking for the total number of trucks that would be movable in the train. Yes, it can take a long time but still reasonable amount of time to get to its fastest speed...like 10 minutes to reach 1mph if you want (all day would not be acceptable).
  11. Aug 13, 2011 #10
    OK, this changes things. Now there's an additional force to accelerate. That needs to be calculated and added to the rolling resistance. It is:
    m = mass of the whole train
    a = 1mph / 10minutes

    I think this forum has a policy of question askers doing some of the work themselves. Here's a start:
    Google: weight of 2011 Ford F250
    Google: rolling resistance
  12. Aug 13, 2011 #11
    Maybe you're just getting back at me for my seemingly snide retort.... ;-). I am just kidding. OK, ugh, I'll Google those numbers....alright, it's 9,900 or 10,000 lbs. depending on whether it's 4wd or 2wd. Let's assume this is a 4x4 vehicle for better traction. It lists the 10k lbs. as GVWR. Does "gross" weight include a full tank of fuel and the maximum number of passengers and cargo allowed? I think it does. Otherwisek, what would net weight be net of? So, it must. I have a feeling I am looking for "curb weight" which Ford doesn't provide. I know this will come as a shocker to you, but Ford doesn't provide the numbers for axle resistance. I don't know what number to plug in there. It's probably close to zero right? I mean these would be brand new axles with brand new roller bearings not filled with dirt and grime in the grease. Hey...I did some work!!! Are you proud? :-)
  13. Aug 13, 2011 #12
    Depends how accurate you want it. Does a 200lb person matter for a 10,000lb truck? Shall we just say 10,000lb?

    It probably is close to zero, but there are other parts of rolling resistance too. Most significantly the tyres on the road. We can assume it's all zero if you want. That simplifies the problem but may make it quite unrealistic. The rolling resistance of tyres won't change much with vehicle model or brand, mostly just with weight.

    Althought you don't like formulas, here are overall formulas:

    Force required = number of trucks * (truck mass * acceleration + truck rolling resistance)
    Force required = weight of one truck [assuming 4x4, coefficient of friction=1, etc]

    We can combine them to one formula:

    weight of one truck = number of trucks * (truck mass * acceleration + truck rolling resistance)

    All the variables can be found except "number of trucks". Solve for that and it's the answer.
  14. Aug 13, 2011 #13
    I don't think that would be to have a totally unrealistic number. If you just assume something other than zero for the resistance that's as close as possible to your best WAG (wild a** guess) what's your best estimate for the number just based on looking at the formula?
  15. Aug 14, 2011 #14
    Solving formulas is boring. I think that's a job for the person who's most interested in this problem! Which might actually be me before long :P

    You'll have to convert everything to consistent units. If I do it I'll use SI.

    I don't have a clue about rolling resistance. But If you find the force required to accelerate, you might see that it's much higher than any possible rolling resistance, and can therefore ignore rolling resistance.
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