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A uniform 26 kg door that is 2.5 m high by 0.80 m wide is hung from two hinges that are 20 cm from the top and 20 cm from the bottom. If each hinge supports half the weight of the door, find the magnitude and direction of the horizontal components of the forces exerted by the two hinges on the door.

(Take the direction of forcing the door away from the hinges to be positive and the direction of forcing the door toward the hinges to be negative.)
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Added Mon, 08 Jun '15
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What are our Givens?
m = 26-kg
H = 2.5-m
W = 0.80-m
d = 0.20-m = distance of a hinge from its closest edge
What's it asking?
What is the horizontal force the two hinges exert.
H_t = ? (top hinge horizontal force)
H_b = ? (bottom hinge horizontal force)

Use the fact the Net Torque = 0 and one of the hinges as a reference point to find the Horizontal force of one of the hinges.

We're going to use the bottom hinge as the reference point.
Tq = H_t * (H - 2d) - m * g * W / 2 = 0
H_t = m * g * W / 2 * / (H-2d)

Plug in the variables to solve for H_t

H_t = 26-kg * 9.81-m/s^2 * 0.8-m / 2 / (2.5-m - 2*0.2-m)
H_t = 48.59-kg*m/s^2 = 48.59-N

Because the directions specified that the force going towards the hinges (to the left) is negative, H_t is a negative value.

H_t = -48.95-N

Now we have to find H_b.
Because the system is in static equilibrium and we know one of the two forces acting in the y direction, expressing the forces in the y direction will enable us to solve for H_b.

Fy = H_t + H_b = 0
>> H_b = -H_t
H_b = 48.95-N
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Added Mon, 08 Jun '15
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