Stability: Leaning Horse against a Wall

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Homework Statement


image.jpg


Data:

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

Homework Equations



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

##F_{net} = ma##

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 !!
 
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Answers and Replies

  • #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.
 
  • #3
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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.
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.
 
  • #4
PhanthomJay
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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.
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.
 
  • #5
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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.
Nice!

Thanks for helping me [:
 

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