Physics
posted by Mildred on .
uniform door (0.81 m wide and 2.1 m high) weighs 122 N and is hung on two hinges that fasten the long left side of the door to a vertical wall. The hinges are 1.5 m apart. Assume that the lower hinge bears all the weight of the door.
(a) Find the magnitude and direction of the horizontal component of the force applied to the door by the upper hinge.
magnitude N
direction Select to the left to the right upwards downwards
(b) Find the magnitude and direction of the horizontal component of the force applied to the door by the lower hinge.
magnitude N
direction Select to the left to the right upwards downwards
(c) Determine the magnitude and direction of the force applied by the door to the upper hinge.
magnitude N
Direction Select to the left to the right upwards downwards
(d) determine the magnitude and direction (below the horizontal) of the force applied by the door to the lower hinge.
magnitude N
direction °

This is far to difficult to do in ASCII.
STart with a drawing, at each hinge you have a vertical force and a horizontal force. The sum of the vertical forces equals weight of the door. The sum of the horizontal forces is zero, and the sum of moments about any point is zero. Those relations will yield the answer.