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A large gate weighing 205 N is supported by hinges at the top and bottom of the wooden frame, and is further supported by a wire.  
 (Assume the positive x-direction is to the right and the y-direction is upward.)


(a) What must the tension in the wire be for the force on the upper hinge to have no horizontal component?

(b) What is the horizontal force on the lower hinge?
(take the direction of pushing the gate away from the hinges to be positive and the direction of pulling the gate towards the hinges to be negative.)

(c) What are the vertical forces on the hinges?
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Added Mon, 08 Jun '15
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What are our Givens?
L = 3m
H = 1.5-m
W = 205-N

(a) What must the tension in the wire be for the force on the upper hinge to have no horizontal component?
We can do this by realizing that the net torque is 0.

Set the reference point at the bottom hinge.
Tq = T * cos(45) * L/2 + T*sin(45)*H - W * L/2 = 0

Solve for T
T * (cos(45) * L / 2 + sin(45) * H ) = W * L / 2
T = ( W * L / 2 ) /  (cos(45) * L / 2 + sin(45) * H )

T = (205-N * 3-m / 2) / (cos(45) * 3-m / 2 + sin(45) * 1.5-m)
T = 307.5 / (1.06 + 1.06) 
T = 145.05-N

(b) What is the horizontal force on the lower hinge?
For this we will need to express the forces in the x direction.

H_b = bottom hinge horizontal force

Fx = H_b - T*cos(45) = 0
H_b = T*cos(45)
H_b = 145.05-N * cos(45)
H_b = 102.57-N

(c) What are the vertical forces on the hinges?
For this we will need to express the forces in the y direction.

V_b = bottom hinge vertical force
V_t = top hinge vertical force

Fy = V_b + V_t + T*sin(45) - W = 0
V_b + V_t = W - T*sin(45)
V_b + V_t = 205-N - 145.05-N * sin(45)
V_b + V_t = 102.43-N
Edit
Added Mon, 08 Jun '15
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