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Combining Truck Power

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data

    A loaded truck (truck 1) has a maximum engine power P and is able to attain a maximum speed v. Another truck (truck 2) has a maximum engine power 2P and can attain a maximum speed of 1.5v . The two trucks are then connected by a long cable, as shown.

    To solve this problem, assume that each truck, when not attached to another truck, has a speed that is limited only by wind resistance. Also assume (not very realistically)
    A) That the wind resistance is a constant force (a different constant for each truck though). i.e. It is independent of the speed at which the truck is going.
    B) That the wind resistance force on each truck is the same before and after the cable is connected, and,
    C) That the power that each truck's engine can generate is independent of the truck's speed.

    Find v_max, the maximum speed of the two trucks when they are connected, assuming both engines are running at maximum power.
    Express the maximum speed in terms of v.


    2. Relevant equations

    [​IMG]



    3. The attempt at a solution

    I don't have the slightest clue. Am i using Work-Energy theorem in one form or another? Some guidance would be great.
     
    Last edited: Oct 14, 2007
  2. jcsd
  3. Oct 14, 2007 #2

    dynamicsolo

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    Power is the rate at which work is done, so you will need to look at work. Since you need to evaluate the work, it would be good to start by showing all of the forces acting on each truck.
     
  4. Oct 14, 2007 #3
    well, if they are moving at a maximum velocity there is no force acting on either truck in that respect since the velocity is constant.

    The thing i'm thinking of right now is the Tension of the cable. But even at that point, i'm having a hard time understanding this problem.
     
  5. Oct 14, 2007 #4

    dynamicsolo

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    When the velocity is constant, there is no net force acting on each truck; there are still forces present, but they add up to zero. Because the forces add to zero, the total work on each truck is zero. But the power from each engine is the rate at which it must do work in order to overcome forces opposing it.

    Besides the tension force, what other forces in the horizontal direction do we need to consider?
     
  6. Oct 14, 2007 #5
    We have the drag force, which is independent of the velocity, and we also have the tension. I'm assuming the power is a force that needs to be considered as well?

    I'm getting confused with the Power in this problem. I haven't worked much with power.

    Here's a shot in the dark:

    [tex]\sum[/tex]_truck 2 = -T - F_d + 2P = 0

    [tex]\sum[/tex]_truck 1 = T + P - F_d = 0

    ???
     
  7. Oct 14, 2007 #6

    dynamicsolo

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    The power isn't force, it is the rate at which work is done. The force would be the applied force from each engine.

    Keep in mind that the drag forces are different on each truck. Let's also call the applied force from each engine F1 and F2 for the moment; we need to work out what their actual relationship is, and it's not 2:1.

    Otherwise, that's the idea for the force equations. They'd be

    truck 2: F1 - F_d1 - T = 0 and

    truck 1: F2 - F_d2 + T = 0 .

    I see no mention of friction, so I take it that the rolling friction of the trucks can be neglected.

    We need to figure out what the problem is telling us about the forces from the engines. What is the equation for work?
     
  8. Oct 14, 2007 #7
    In this situation, since it is moving horizontally W = F*[tex]\Delta[/tex]x

    But there is no mention of [tex]\Delta[/tex]x in the eq?
     
  9. Oct 14, 2007 #8
    ok, after doing a little research i've read that P = F * v

    I have a gut feeling we're going to have to use this?
     
  10. Oct 14, 2007 #9

    dynamicsolo

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    That's where the rates come in. Power is [tex]\Delta[/tex]W / [tex]\Delta[/tex]t . So if we divide the work equation through
    by [tex]\Delta[/tex]t , we get

    P = [tex]\Delta[/tex][ F·x ] / [tex]\Delta[/tex]t ,

    but when the force is constant, we have

    P = F·[[tex]\Delta[/tex]x / [tex]\Delta[/tex]t ] = F·v .

    We can now apply this to each truck. The maximum power of the engine in truck 1 is

    P = F1 · v . What can we write for truck 2?
     
  11. Oct 14, 2007 #10
    aha! i was right.

    Ok P_max_truck 2 = (F2 * 1.5v)/2 ??
     
  12. Oct 14, 2007 #11

    dynamicsolo

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    Careful: the problem states the maximum powers relative to each other. So you are close, but you need to write

    P_max_truck 2 = 2P = (F2 * 1.5v) , where P = F1·v is the maximum power of truck 1.

    (We need the comparative relationship because the question wants the maximum speed from the combined engines expressed in terms of v.)

    So how does F2 compare to F1?
     
  13. Oct 14, 2007 #12
    Would i be substituting P = F1*v into the other equation?

    2(F1*v) = (F2 * 1.5v) But that would leave us with 2 different variables?

    or would i set F1 = F2 because it is ultimately the same force relative to each other since they are moving at the same velocity?

    P/v = 2P/1.5v But that doesn't make much sense either because v would cancel out....
     
    Last edited: Oct 14, 2007
  14. Oct 14, 2007 #13

    dynamicsolo

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    Exactly: we only have one equation in two unknowns, F1 and F2, so we can only find the relation between them. (That's OK -- we won't need to know the magnitudes of these forces.)

    So v can be divided out, leaving 2 · F1 = F2 · 1.5 ,
    or F2 = (4/3) · F1 .

    This next part isn't essential to getting the final answer, but it's a good idea to look at the physics. If you go back to your force equations for the trucks, solve one of them for T and substitute the result into the other equation, what do you get? (That is, we are eliminating T as a variable, which is also OK, since we don't need to know its value either...)

    It won't be true that the applied forces are equal. Recall that the forces of air resistance to be overcome are different for the two trucks (and we are told nothing about their masses), so there's no real reason to assume F1 should be equal to F2.
     
    Last edited: Oct 14, 2007
  15. Oct 15, 2007 #14
    huh?

    But what about v? The question calls for the answer to be answered in terms of v. How is this possible if we've eliminated it?

    So if i substitue back into force eq. of truck 2 with T solved from the eq from truck 1, i get:

    (4/3)F1 - (Fd1 - F1) - Fd2 = 0

    What is going on with the Fd1 and Fd2 now.....

    Wow, i need to sleep!!
     
  16. Oct 15, 2007 #15

    dynamicsolo

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    We're not quite there yet. This v is the maximum speed of truck 1 alone; we're going to find v_max as a multiple of v.

    We're nearly there. We can re-arrange this as

    Fd1 + Fd2 = F1 + F2 = F1 + (4/3)·F1 = (7/3)·F1 .
    [You flipped a sign before the substitution, so I fixed that, since it's getting late...]

    The point is that the two trucks working together provide a combined applied force of (7/3)·F1 . The combined force equation we created just says that this combined force is equal to the combined drag force, as we'd expect, since the net force is zero because the velocity is constant. This is just the sum of the individual drag forces, as specified by the conditions of the problem: the drag forces do not depend on speed (rather unrealistic) and are not affected by linking the trucks with the cable.

    All of this is by way of justifying what we are about to do to finish the problem. The two trucks linked together provide a combined power of 2P + P and a combined force of F1 + F2 = (7/3)·F1. So if we write a power equation for the combined engines, we have

    P_total = 2P + P = (F1 + F2) · (v_max) .

    Since we stated earlier that P = F1·v , we can now find v_max in terms of v.

    You're on...
     
    Last edited: Oct 15, 2007
  17. Oct 15, 2007 #16

    Wow, like i would've ever gotten this on my own! haha.

    ok so...

    2(F1*v) + (F1*v) = (7/3)F1 * v_max

    v_max = [3(F1*v)]/[(7/3)F1) ---------> F1 cancels and we're just left w/ v =

    v_max = (9/7)v ---------->BAH! tell me that's right, lol
     
  18. Oct 15, 2007 #17
    That is the right answer. Thank you very much dynamicsolo. I have a hard time thinking about setting problems up like this, but i always understand the concepts behind it. Setup is by far the hardest part.
     
  19. Oct 15, 2007 #18

    dynamicsolo

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    That's what I got, too. You're sort of forced to go about this in a somewhat roundabout way because only relative values are given, rather than magnitudes.

    You get a value a bit less than 1.5 v , because the more powerful truck is using some of its power to tow the weaker truck (#1) to greater than #1's own maximum speed.
     
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