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Magnitude of force

  1. Feb 22, 2009 #1
    Hello, I am having with this ( actually) kinda easy problem. I don't know how to approach it.



    There's a worker poling a boat. He pushes parallel to the length of the light pole, exerting on the bottom of the lake a force of 240 N. Assume the pole lies in the vertical plane containing the boat's keel.
    At one moment, the pole makes an angle of 35 degrees with the vertical and the water exerts a horizontal drag force of 47.5 N on the boat, opposite to its forward velocity magnitude 0.857 m/s. Total mass of the boat is 370 kg.

    The water exerts a buoyant force vertically upward on the boat. Find the magnitude of this force.



    I've been trying to find an equation to get started, but I have no idea.
     
  2. jcsd
  3. Feb 22, 2009 #2

    LowlyPion

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    Resolve the force provided by the pole into x,y components.

    Then look at the forces on the boat. How will the vertical component of the poleman's force affect the buoyant force provided by the water?
     
  4. Feb 22, 2009 #3
    I thought about that too, but can't find the right answer.

    Vertical would be the 240 N ( cos, I think) and horizontal would be the drag force of 47.N ( sin). Is that what you mean?
     
  5. Feb 22, 2009 #4

    LowlyPion

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    Cos35 is the correct vertical component of the 240 N force. So how many N is the boat? And then what is the net?
     
  6. Feb 23, 2009 #5
    Ok, I think I used a slightly different approach. Now I just need to calculate the buoyant force. I don't know how to do this for this problem. The book deals with the buoyant force several chapters after the one we are doing right now.
     
  7. Feb 23, 2009 #6

    LowlyPion

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    Look they don't give you displacement to figure the weight of the water displaced. It's not needed. No need to get chapters ahead.

    Make a diagram of the boat. There are 2 forces holding the boat up. The water (aka buoyant force) and the pole (the vertical component of the 240 N the guy is pushing down with.)

    Like I said before, just figure the net ... weight - vertical component.
     
  8. Feb 23, 2009 #7
    I still don't get that. Originally I thought I could do it like this...

    Vertical component of the 240 N force (up) - Weight of boat (down) + Buoyancy force (up) = 0

    But this all will not get me the right result. It's supposed to be 3.43 KN...
     
  9. Feb 23, 2009 #8

    LowlyPion

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    Why don't you show your calculation then for that equation?
     
  10. Mar 18, 2009 #9
    I don't understand why is doesn't work for you since it works really well for me...

    But my answer in part b isn't the same as in the book and I'm wondering why.

    The question is: Model the forces as constant over a short interval of time to find the velocity of the boat 0.450 s after the moment described.

    First, I used my x components to find the acceleration.
    After, I used the formula Vf=Vi+at.

    My answer is 0.914 and the real answer is 0.967.

    According to me, my mistake must be in my calculation of the acceleration.

    Here is what I've done:

    F=ma
    Fwater/horizontal=ma

    Is that right?
     
  11. Mar 29, 2009 #10
    What is the equation for total buoyant force for a volume that changes as it ascends to the surface, like an air bubble?
     
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