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Stability: Leaning Horse against a Wall

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  1. Dec 15, 2014 #1
    1. The problem statement, all variables and given/known data
    image.jpg

    Data:

    ##m = 500\ kg##
    Distances given in the image.

    2. Relevant equations

    ##\tau = rF\sin(\theta)##

    ##F_{net} = ma##

    3. The attempt at a solution

    It seems this problem is intended to be one where torque applies, but I don't see it in that way. And of course my answer doesn't agree with that given by the textbook :).

    The diagram:

    image.jpg

    According to my understanding, there is a component of the weight acting along the horse's body (the diagonal I've drawn) and another one perpendicular to its body, which is making it rotate to the left. I labeled the latter ##w_{\perp(horse)}##. The component of ##w_{\perp(horse)}## that lies perpendicular to the wall, I think, has the same magnitude that the force exerted by the wall on the horse. This is what I mean:

    image.jpg

    Then:

    ##w_{\perp(horse)} = w\sin(\theta)##

    ##F_{wall} = w_{\perp(horse)}\cos(\theta)##

    ##F_{wall} = w\sin(\theta)\cos(\theta)##

    And by Newton's Third Law, this is the force exerted on the wall.

    The angle is found to be ##14.04º##, so

    ##F_{wall} = 500\ kg \cdot 9.80\ m/s^2 \cdot \sin(14.04º) \cdot \cos(14.04º) = 1.15 \times 10^3 N##

    Maybe it should have a negative sign since it's in the opposite direction, but that is not what I'm worried about. The textbook's answer is ##1.43 \times 10^3 N##. I have no idea what I'm missing /:.

    Thanks !!
     
    Last edited: Dec 15, 2014
  2. jcsd
  3. Dec 15, 2014 #2

    PhanthomJay

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    You are over complicating this problem and getting messed up. Sum torques about the horses feet. Torque of a force is force times perpendicular distance.
     
  4. Dec 15, 2014 #3
    Ah, I see. I confused the perpendicular distance with the perpendicular component of the force (I've been doing so all the time :) ).

    ##\sum \tau = 0 = \tau_{ccw} - \tau_{cw}##

    ##\sum \tau = 0 = \tau_{weight} - \tau_{wall}##

    ##\tau_{wall} = \tau_{weight}##

    ##F_{wall} = \frac{500\ kg \cdot 9.80\ m/s^2 \cdot 0.35\ m}{1.20\ m}##

    ##F_{wall} = 1.43 \times 10^3\ N##

    If I were to solve a similar problem where ##\sum F = 0##, I must then assume ##F_{wall} = friction ## and ##w = F_{normal}## about the horses feet, right? I want to be sure about it, because I also got confused with that and with the fact that the forces are acting along different heights/distances.
     
  5. Dec 15, 2014 #4

    PhanthomJay

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    Yes, that is correct. Normal force from ground acts up on the horses feet equal in magnitude to its weight, and the ground friction force on its feet acts left , equal in magnitude to the normal wall force.
     
  6. Dec 15, 2014 #5
    Nice!

    Thanks for helping me [:
     
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