- #1

Jazz

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## Homework Statement

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:

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:

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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