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Very simple question.

  1. Apr 5, 2008 #1
    Hello everybody.

    Me and my (stupid) friend had a little debate.
    I was wondering how the forces should work in a case as follow:
    I'm pulling a box using a rope with a certain force. Obviously, the force that the box "feels" is the same as the force im putting in(considering the rope is perfect or whatever).
    Now lets say I add behind this box another rope connect to a box(same mass as the first box) and pull this whole thing with the same force as before. Obviously the first box will "feel" the same force that I put in, and obviously(for me) the second box would "feel" less than that.
    My guess was that considering the whole thing is moving at a constant speed(there is some friction for that matter) the force that "gets" to the second box should be half of what I put in.
    My friend said that both boxes would "feel" the same force which is equal to what I put in. he also said(contradicts himself a bit) that I should look at the whole system as one object and that I cannot know how much force the second box "feels"

    I hope that you guys would be able to understand what im trying to say here(kinda hard with my limited English and Physics knowledge).

    all that I want to know basically is what force each of two boxes connect to a rope "feels" in case that I put in a certain force which makes the whole thing move at a constant speed.

    Thanks in advanced!
  2. jcsd
  3. Apr 5, 2008 #2


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    Welcome to PF,

    The tension in the second rope will certainly be less than the tension in the first rope (the one your holding). To determine the tension in each rope one would have to have knowledge of the masses of both boxes as well as the acceleration (which in your example above is zero).

    The way one would approach this question is that assuming that both ropes are taught, one knows that both boxes have the same acceleration and velocity. Using this information one can then determine the forces acting on each box using Newton's second law.
    Last edited: Apr 5, 2008
  4. Apr 5, 2008 #3


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    Welcome to PF!

    Hi Eshez! Welcome to PF! :smile:

    Yes … if you have lots of boxes, the force gets less for each box.

    It's best to think of the tension in each rope.

    At constant speed, the friction is the same for each box, so the tension in each rope must be the tension in the previous rope minus the friction.

    Each box "feels" the tension in the ropes on either side of it.

    It's the same if you just hang the boxes vertically … with the extra force on each box being its weight rather than friction! :smile:
  5. Apr 5, 2008 #4
    Thank you!

    Thank you :)

    Let me first say that although I don't know much about physics I find it really interesting, I think I'll stay around even after I get the answer to my question.

    anyhow, thank you for that answer, that's what I've expected.
    Now lets take it that the masses are equal(shouldnt matter how much each one is as long as they are the same, I think).

    With one box I've put in a certain force, lets say its 10(force is measured in nuetons isnt it?) this all force would act fully on the one box, causing it to move(as I said, im talking about a case in which the box gets a constant speed due to friction or whatever).
    now when I add another box, the whole thing would move in a constant speed but a slower one(considering the force is enough to move the new mass).
    If the force im putting in in box cases is the same, shouldn't the force that reaches the second box be exactly half of the force that I put inside this system?
    If the answer is not, how would you calculate how much it is?

  6. Apr 5, 2008 #5
    Very nice analogy, thank you.
    So basically the force that reaches the end is the origianal force minus all the force that work on the opposing direction?

    That makes a lot of sense, thank you. I think I understand the physics of this system much more accuratly now!
  7. Apr 5, 2008 #6


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    Welcome aboard, don't forget to stop by General Discussion and introduce yourself (bring a fish :wink:).
    Okay, I'm going to answer your question using Newton's second law (forces are measured in Newton's (N), not nuetons) so hopefully you'll understand where the answer comes from.

    So, lets set up our system. We have two boxes, both of mass m, on a rough , horizontal surface therefore both boxes experience a frictional force of magnitude [itex]\mu mg[/itex]. Let's call the tension in the rope your holding T1 and the tension in the other rope T2, furthermore, let's assume that the tension is applied horizontally, i.e. the rope is horizontal. We further postulate that the boxes travel at a constant velocity so the acceleration is zero.

    Now lets apply Newton's second law to each box individually:

    Box One

    Since the acceleration is zero, we know the net force acting on the box must be zero hence we may write,

    [tex]\sum F = ma = 0[/tex]

    In this case there is three forces acting on the first box: T1 from the first rope, -T2 from the second rope and [itex]-\mu mg[/itex] from the frictional force. Notice that the latter two quantities are negative since the act in the opposite direction to the motion. Hence,

    [tex]\sum F = T_1 - T_2 - \mu mg = 0[/tex]

    [tex]T_1 = T_2 + \mu mg \hspace{2cm}(1)[/tex]

    Box Two

    Again the acceleration of the second box is zero (constant acceleration), however, unlike the first box there are only two forces acting on the second box: T2 from the second rope and [itex]-\mu mg[/itex] from the frictional force. Notice this time that T2 is positive since it acts in the same direction of motion. Hence, using the same method as above,

    [tex]\sum F = ma = 0[/tex]

    [tex]\sum F = T_2 - \mu mg = 0[/tex]

    [tex]T_2 = \mu mg \hspace{2cm}(2)[/tex]

    Hence, if we now substitute equation (2) into (1) we obtain,

    [tex]T_1 = 2T_2[/tex]

    As you correctly postulated. Do you understand qualitatively why this is the case now?

    Of course if the masses were not equal, the relationship would only be a little more complicated,

    [tex]T_1 = T_2 + \mu g\left(m_1+m_2\right)[/tex]
    Last edited: Apr 5, 2008
  8. Apr 5, 2008 #7
    Thank you.
    This makes a lot of sense. I assumed this how the thing works but didnt understand it all too clearly.
    I just sent my friend a link to this page so hoepfully he will soon read this and understand how wrong he was :D
    He even dared to mock me without checking out his assumptions.

    Is there any written guide I can find online to learn more about some basic Physics principles?
    I used to study Physics in high school but it seems like I dont remember anything and didn't really absorbd any of the material.
    How sad it is that one can get an A without having a clue about the subject.

    Good day.
  9. Apr 5, 2008 #8


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    It was a pleasure :smile:
    Ahh the sweet sweet taste of victory! :biggrin: :tongue:
    It is very sad indeed, I will refrain from a rant regarding High School Physics here :grumpy:

    There are a few decent online Physics resources. Check out our own https://www.physicsforums.com/forumdisplay.php?f=151".
    Last edited by a moderator: Apr 23, 2017
  10. Apr 5, 2008 #9


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    I think you and your friend may both be right, but misscommunicating:

    So far, it seems that everyone is talking about the forces on the ropes, but the original question, and the discussion you had with your friend, was about the forces on the boxes. This is partly a matter of how you figure the forces. If you consider your pull on the rope to be possitive force, and the resistance of friction to be a negative force (because it acts in the opposite direction), then you can see that the two will reach equilibrium at a point where the box is no longer accelerating, but travelling at a constant speed.

    As has been mentioned, once a box is no longer accelerating, the total force on that box must be zero. When you have two equal forces and they must come out to zero, you obviously must be subtracting one from the other. So, when figuring the total force on a box, you must subtract the negative forces from the possitive.

    Now, if you add the second box (with its attendant friction), then that force must be subtracted, if we're subtracting forces. In this case, the total force of your pull is felt by the first rope, but not by the first box, which feels the force of your pull, minus the force of its own resistance, minus the resistance of the second box. The second box will feel the force of the pull on the second rope (which is much less than the force on the first rope) minus its own resistance only (which is much less than the negative force being subtracted from the first box). This will total out to be the same as the force felt by the first box.

    So your friend is right if you add forces that way. Although the forces felt by the two ropes would be very different, the forces felt by the two boxes would be the same.
  11. Apr 5, 2008 #10
    Thank you for that clarification.
    I totally understand what you've just said and it did help me see things ever better.
    The debate was about what force each box will feel, as in, how hard will it feel the rope is pulling it. I can understand that the sum of forces is zero(even that wasnt clear to me before :)) but that wasnt what my friend claimed nor the point I was trying to make to him.

    btw, How will this system behave in a friction-less enviroment?
    I think I might understand how it will work, though it can surely be more clear than what it is.
    The whole thing will keep accelerating. The question is, what force will each box feel that pulls it? Surely the amount of mass should affect the acceleration, therefor the second box should "feel" less force than the one being put in the system. It just hard for me to see where this force "goes". Is it just being translated into the accel of the first box?
  12. Apr 5, 2008 #11


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    In a frictionless environment, the tension in the ropes would depend on both the masses and the acceleration of the system. Lets repeat the previous analysis but this time without friction. If the ropes remain taught we know that the acceleration of both boxes must be equal, a = a0. Furthermore, let us consider the cases where the masses are not equal. Hence, for each box:

    Box One

    [tex]\sum F = ma[/tex]

    This time the acceleration is non-zero, but there is no frictional force. Therefore, there is now only two forces acting on the first box, T1 and -T2, again note the negative sign. Let us also denote the mass of the first box m = m1. Hence,

    [tex]\sum F = T_1 - T_2 = m_1a_0[/tex]

    [tex]T_1 = m_1a_0 + T_2 \hspace{2cm}(3)[/tex]

    Box Two

    Let this box have mass m = m2. Again, in this case there is no frictional force, the only force acting on this box is T2. Hence,

    [tex]\sum F = ma[/tex]

    [tex]\sum F = T_2 = m_2a_0 \hspace{2cm}(4)[/tex]

    Again, substituting (4) into (3) we obtain,

    [tex]T_1 = \left(m_1+m_2)a_0\hspace{2cm}(5)[/tex]

    Hence, we can deduce that the tension in the first rope (5) must account for the acceleration of the first box and the second box combined; whereas the tension in the second rope (4) merely accounts for the acceleration of the second box.

    Does that make sense?
  13. Apr 5, 2008 #12
    Very much so. Thank you :)
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