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Man on boat rows

  1. Dec 12, 2016 #1
    1. The problem statement, all variables and given/known data

    The man pushes the weightless stick, and so exerts a force with a magnitude of 240 N to the bottom of the lake. Hypothesize that the stick is in the same vertical as the boat's keel. At some point in time, the stick creates a 35 degree angle with the vertical and the water exerts an horizontal force of 47.5 N to the boat, with a direction opossite to the velocity's, which faces towards the front of boat, and has a magnitude of 0.857 m/s. The mass of the boat, the cargo and the man is 370 kg.

    a) The water exerts an upthurst force, which faces vertically upwards. Find the magnitude.

    b) Consider that these forces stay constant for very short time spans. With that info, find the velocity of the boat 0.450 s later.

    2. Relevant equations

    ΣF = m*a
    Vf = Vi + a*t

    3. The attempt at a solution

    The thing is, I'm not sure how the stick, the boat and the bottom of the lake connect. When he moves the stick, he exerts a force directly at the bottom, correct? So, since he's moving forwards (let's say forwards is east), he exerts a 240 N force at the bottom, which faces to the east. Then, since the stick is weightless, according to Newton's 3rd Law, where does the bottom exert the opposite force to?

    When the stick is at an angle, then this new force crates a net force, with both horizontal and vertical vectors. At first, I figured I'd find the vertical force, I'd subtract it from the weight, and I'd be done, but that's not the correct result. I'm obviously missing something (I figure the whole "370 kg is the mass of the boat, the man and the cargo" play some role later?), so I'd appreciate the help.
     
  2. jcsd
  3. Dec 12, 2016 #2

    BvU

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    Then the thing to do is: make a drawing.
    The stick can be weightless and exert a force nevertheless; in this case along its length. Man pushes down and backward, stick pushes up and forward. Up you need for a), forward for b).
     
  4. Dec 12, 2016 #3
    So, the 240 N force goes upwards then? I made a drawing, but I'm not sure how the bottom, the stick and the boat are connected. For example, I know that the center of the earth exerts a force on the boat which is what we call the weight, and that's downwards. Then, because the boat is on top of the water, the water/bottom causes an upper thurst to the boat, that faces upwards, and it's what I'm trying to find. But when the stick creates a 35 degree angle with the vertical, and thus the bottom exerts a 47.5 N force to the boat (that faces west while it moves east), then where is the stick facing? How do the bottom, the stick and the boat connect?
     
  5. Dec 12, 2016 #4

    BvU

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    No, it's at an angle of 35 degrees with the vertical. Part of it is upward, part is forward. Upward you need for a), forward you need for b) (Didn't I already write that ? :smile:)

    Would you like some constructive comments ? Then post it.

    No 'thus' . The forward part of the 240 N is not 47.5 N. The 47.5 N is the (viscosity) friction force the water exerts against the moving boat, doing its best to slow it down, while the man does his best to push it forward.
    If you find the water complicating, imagine a skateboard, or a cart: :rolleyes:

    artizani_streetship2_0.jpg

    No and yes: it's not the center but the entire earth. Magnitude ##mg##.

    The stick is facing forward at an angle of 35 degrees with the vertical.
     
  6. Dec 12, 2016 #5
    The point in time where the stick exerts a force of 240 N to the bottom (when it's completely vertical, it doesn't create an angle) is different than the point in time when it creates an angle with the vertical. My problem isn't picturing the net force, it's how the stick affects the bottom. Should I just imagine it that it touches the bottom, and so because the stick is weightless, the 3rd LoN is between the bottom of the sea and the boat as a whole?
     
  7. Dec 12, 2016 #6

    jbriggs444

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    You are expected to assume that the stick is under compression only. It conveys no twisting or other force perpendicular to its axis. You are expected to assume that the magnitude of this force is 240 N regardless of how the stick is angled.

    Edit: Even if those assumptions are not the expected ones, the rule of thumb is to clearly state your clarifying assumptions about the problem up front and then solve. Doing so will assure at least partial credit.
     
  8. Dec 12, 2016 #7

    BvU

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    It's not completely vertical: It is only vertical in the plane of the keel (the 'left-right vertical' ) : it makes an angle of 35 degrees with the 'forward-backward vertical'.
    punt.jpg
     
  9. Dec 12, 2016 #8

    BvU

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    We haven't reached the sea yet. And yes, the boat as a whole will do: actually it's via the man who distributes his ##mg## over the vertical component of the 240 N he exerts on the stick (and therefore the stick on him) and the boat. On with part a) which is now almost solved !
     
  10. Dec 12, 2016 #9

    BvU

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    Is actually not such a good title. Punting is the word.
    Students do it at Oxford and Cambridge. The english learned it from the dutch in the sixteenth or seventeenth century. 'Punteren' was the means of local transportation in Giethoorn well into the twentieth century.
     
  11. Dec 12, 2016 #10
    Yeah, I get it now. I forgot all about friction force, and assumed that the stick is at some point completely vertical, and "pushes" the bottom (like when you try to stick a flag in the ground). So technically, the stick is in the same vertical space, but can still move left and right. The 240 N that it exerts is in a net force, with both an x and y vector. The 47.5 N is just a horizontal force that plays a role in (b).

    So, by that logic, vertically, on the boat I have the following forces:

    |Fg| = 370 kg * 9.8 m/s^2 = 3626 N (which faces downwards)

    |Fsy| = 240 N * cos(35) = 196,6 N (which faces upwards) - The y vector of reaction of the bottom to the force the stick exerts to it

    And so: ΣFy = 0 < = > Fwy + 196,6 N = 3626 N <=> Fwy = 3429,4 N ~ 3,43 kN (the book's answer)

    Thanks for all the tips! I got caught up with the stick and misssed that the 47.5 N and the 240 N were totally different forces and not just the same one at different points in time.
     
  12. Dec 12, 2016 #11

    BvU

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    You're in business! On to part b) and no need to think overly complicated.


    (A vertical 'space'? o0) -- at each pointthere are two vertical planes perpendicular to each other. The boat moves in an east-west plane which is where the stick is. The stick is NOT in a north-south vertical plane)
     
  13. Dec 12, 2016 #12
    Well, I'm still having a bit of trouble with the english definitions, as it's not my native language, and the book I'm using has been translated. I'm mostly going by what I know and online dictionaries. Thanks for the tips though, they're very useful, since at some point I'll have to be able to fluently explain and/or translate problems/thoughts in english, complete with definitions (I don't have a problem with the speaking part, but such advanced and specialized words are a tad new to me).
     
  14. Dec 12, 2016 #13

    BvU

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    Same here. It's a difficult, contorted, ambiguous, class-oriented disaster of a language for science, but I suppose it beats chinese. And it has beaten esperanto ([edit] and latin).
     
  15. Dec 12, 2016 #14
    My native is greek so there is some overlap, but yeah, it is hard to get the hang of it when it comes scientific terms and concepts. Well, try it 'till you make it as they say.
     
  16. Dec 12, 2016 #15

    BvU

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    As far as I can tell, you're doing extremely well: not everyone uses terms like 'a tad new' :smile:
     
  17. Dec 12, 2016 #16
    It's mostly because of my teacher. He was from South Africa and lived in England for a long while, so I got to know the various phrases and British pronunciation early on. It's the various terms that I'm having teeny bit of trouble with, but I'll get the hang of it eventually...I hope...
     
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