Solve Farm Gate Problem: Horiz Force on Bottom Hinge

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The discussion revolves around calculating the horizontal force exerted on a farm gate by its bottom hinge, given its dimensions, mass, and the tension in a guide wire. Participants highlight the importance of understanding the forces acting on the gate, including tension and gravitational forces. A suggested approach involves taking moments about the point of application of the unknown force to simplify the calculations. The conversation also touches on common mistakes, such as using incorrect values for gravitational acceleration. Ultimately, the focus is on solving the problem using the correct method and values.
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Old MacDonald had a farm, e i , e i , oh ! And
on that farm he had a gate, e i , e i , oh ! And
a squeak, squeak, here. A squeak, squeak,
there. And a squeak, squeak, everywhere . . .
The gate is ℓ = 3 m wide and h = 2.47 m
tall with hinges attached to the top and bot-
tom. The guide wire makes an angle of
α = 39 ◦ with the top of the gate and has
a tension of 221 N. The mass of the gate is
30.9 kg.

Determine the magnitude of the horizon-
tal force exerted on the gate by the bottom
hinge. The acceleration of gravity is 9.8 m/s2 .
Answer in units of N.

Okay the reason why I am having problems with this is because well I am just not sure I know where to start. I think I am just missing something obvious, and my book does not have a whole lot to say about this type of problem. Any Help?
 
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Hi xRandomx210! :smile:

(ooh, that gate is really annoying :frown: … if i help with the question, will you please oil it?)

There are only four external forces on the gate.

You know 2 of them, and you want one of them, but you don't know the fourth one, and and you don't want to know it, so … standard trick … take moments about the point of application of that unknown force. :wink:
 
Sorry for being a noob, but what does oiling it mean?
 
Just tried something, don't know if it is right or not.

Σfx= TCos(Theta) = H1x+H2x

where Hx is the force of the hinge in the x direction

Σfy=TSin(theta) + Hy1 + Hy2 = mg

ΣT = mg(L/2)-Tsin(theta)L= Hx(h)

from there you can slove out, but is the method correct?
 
Never mind the reason i got it wrong the first time was because i used 9.81 instead of 9.8 :C
 
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