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Equilibrium Force on Floating Object

  1. Apr 10, 2017 #1
    1. The problem statement, all variables and given/known data
    This problem is found in The Exercises for the Feynman Lectures on Physics:
    A flat steel plate floating on mercury is acted upon by three forces at three corners of a square of side 0.100 m. Find a single fourth force which will hold the plate in equilibrium. Give the magnitude, direction, and point of application along the line AB.
    upload_2017-4-10_5-13-18.png

    2. Relevant equations
    Trig.

    3. The attempt at a solution
    What I did to find the force vector was to imagine the forces at a centralized location then add them together. Breaking FO down into its components, I found the following:

    Fx=50N-35.6N=14.4N
    Fy=50N-35.6N=14.4N

    Meaning there is a net force of 20.1N at 45° from the X axis. To counteract this, the fourth force must be equal in magnitude and opposite in direction. So F4=20.1N at 225°.

    However, I have no idea how to figure out the point of application. I think this is related to the translation of the forces but I also had the idea of finding the moments about the center of the square due to the forces at P and Q. This yielded nothing fruitful, unfortunately. Any help would be great.
     
  2. jcsd
  3. Apr 10, 2017 #2

    BvU

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    Hi,
    How so ?
     
  4. Apr 10, 2017 #3

    kuruman

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    More than just "Trig" is relevant when you have static equilibrium situations. What are the appropriate equations?
    How exactly did you pick this centralized location?
     
  5. Apr 10, 2017 #4
    Well, I found the torque of these two forces:

    τP=rxF or τ=r*Fsinθ, where r=0.71, F=50N, and θ=135°
    τP=25.1Nm so τtotal=50.2Nm clockwise

    However, this is where I get lost. Isn't the angle of F4 and the lever arm 0? And if it isn't, the equation has two variables, the distance and the angle. Is this the correct approach with botched analysis or is there another way that I haven't thought of?
     
  6. Apr 10, 2017 #5
    I guess I'm not entirely sure of the relevant equations in this case. Perhaps the torque equation? Could the Principle of Virtual Work come into play?
    Again, I wasn't sure if it mattered so I redrew them all at a single location without any bearing to the drawing. After working on the application point for a while, I thought maybe there is some way to properly translate forces, but I was unable to find any reading material on the subject. If there is, then I would only move the force at P to Q, and work from there.
     
  7. Apr 10, 2017 #6

    kuruman

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    The relevant equations are
    1. Sum of all the forces = zero. You already implemented that.
    2. Sum of all the torques = zero. To implement this, remember that it matters where a force are applied; you just can't pick any place. So put F4 at x on line AB, then find x.
     
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