(III) A door 2.30 m high and 1.30 m wide has a mass of 13.0 kg. A hinge 0.40 m from the top and another hinge 0.40 m from the bottom each support half the door's weight (Fig. 9-69). Assume that the center of gravity is at the geometrical center of the door, and determine the horizontal and vertical force components exerted by each hinge on the door.

Step-by-step solution

Step 1 of 5 ^

FAx; x components of the force at hinge A

FAy; y components of the force at hinge A

FBx; x components of the force at hinge B

FBy; y components of the force at hinge B

M.g: the weight of the directed downward from the center of mass of the door.

H=2.30m the height of the door.

w=1,30m : the width of the door

d=40cm=0.4m the distance from the ends of the door to the hinges.

Step 2 of 5 ^

We are going to write the torques around hinge A.

Step 3 of 5 ^

We can write the first condition of equilibrium on x axis to find FAx