Static Equilibrium of weighted door

[Quincy]

Homework Statement

A door of width 1.08 m and height 1.92 m weighs 281 N and is supported by two hinges, one 0.600 m from the top and the other 0.600 m from the bottom. Each hinge supports half the total weight of the door. Assuming that the door's center of gravity is at its center, find the horizontal components of force exerted on the door by each hinge.

Homework Equations

The Attempt at a Solution

I don't really understand the question; shouldn't the force from the hinges be vertical/upwards? To counter the downward force the door's weight?

 

[PhanthomJay]

As stated, each hinge shares the vertical weight equally, with upward forces exeted by each hinge on the door. But that is just part of the force exerted by each hinge on the door. You must sum torques about one of the hinges, and sum forces in the horizontal direction, to find the horizontal components of the force exerted by each hinge on the door, per Newton's 1st law of equilibrium.

 

[Quincy]

As stated, each hinge shares the vertical weight equally, with upward forces exeted by each hinge on the door. But that is just part of the force exerted by each hinge on the door. You must sum torques about one of the hinges, and sum forces in the horizontal direction, to find the horizontal components of the force exerted by each hinge on the door, per Newton's 1st law of equilibrium.

Besides the weight of the door, what other torque is there?

 

[PhanthomJay]

If you choose the lower hinge as the point aboit which you wish to calculate torques, The weight of the door produces a torque, yes, but also, would not the force of the top hinge on the door also produce a torque about that point? Would the force of the bottom hinge on the door produce a torque about that point?