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Introductory Physics Homework Help
Stability: Leaning Horse against a Wall
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[QUOTE="Jazz, post: 4946427, member: 526805"] [h2]Homework Statement [/h2] [ATTACH=full]174827[/ATTACH] Data: ##m = 500\ kg## Distances given in the image. [h2]Homework Equations[/h2] ##\tau = rF\sin(\theta)## ##F_{net} = ma## [h2]The Attempt at a Solution[/h2] 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: [ATTACH=full]174828[/ATTACH] 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: [ATTACH=full]174829[/ATTACH] 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 ! [/QUOTE]
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Stability: Leaning Horse against a Wall
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