Question

A door 1.00 m wide and 2.30 m high weighs 330 N and is supported by two hinges, one 0.70 m from the top and the other 0.70 m from the bottom. Each hinge supports half the total weight of the door.

Part A:

Assuming that the door’s center of gravity is at its center, find the horizontal component of force exerted on the door by hinge on the top.

Ftop h=..... N

Part B:

Find the horizontal component of force exerted on the door by hinge on the bottom.

Fbottom h = ......N

(I got 127 N and -127 N but it still incorrect)

Answer 1

Let's call the force exerted by the upper hinge Upper, and the force exerted by the lower hinge Lower.

For force equilibrium, we know that the horizontal components that the hinges exert must cancel out, as they are the only players acting horizontally. Let's guess that the upper hinge is exerting a force to the right (+), and the lower is exerting a force to the left (-) on the door, our equilibrium looks like:

Upper_x - Lower_x = 0

Vertically, there is nothing too exciting going on either. The weight of the door 330N down (-), and the two hinges lifting up(+) on the door:

Upper_y + Lower_y - 330 N = 0
Now, let's talk about TORQUE!
We need to pick a spot to calculate the torque about, so let's pick the lower right corner of the door. (it doesn't matter what point you pick)
so many forces, so big a door. Well, the vertical forces exerted by the hinges do not contribute to the torque about the lower right corner of the door, as they are directly away from the pivot point (t= Frsinq, and q = 0^o).

The horizontal components of force exerted by the hinges (Upper_x , Lower_x ) create torque.

the Lower hinge is pushing to the left, and the upper is to the right, then we have the upper hinge causing a clockwise (+) torque with a force of Upper_x, acting at a distance of

2.30 m - .70 m = 1.60 m

from the lower right corner to create a torque of +(Upper_x)(1.60 m), and the lower hinge is causing an anticlockwise (-) torque with a force of Lower_x, acting .70 m away from the lower right corner to create a torque of -(Lower_x)(.70 m).

all we have left is the weight.  

The weight acts at the center of mass, which is the center of the door, and so the arrow straight down is the weight of the door (330 N)  
since t=  Frsinq, and since rsinq is just half of the width of the door, we don't even need to know r which would have been the distance from the center of the door to the lower right hand corner, as it is only the horizontal part of r that matters. (Only the horizontal location of the center of mass matters). The weight, of course, exerts an anti clockwise torque about the lower right corner, and acts at half the width, or 1m / 2 = 0.5 m, so we have a torque of -(330N)(0.5 m) due to the weight.  

So our TOTAL TORQUE EQUILIBRIUM looks like:
+(Upper_x)(1.60 m) - (Lower_x)(.70 m) -(330 N)(0.5 m) = 0


Time for math

Here are our equations:
Upper_x - Lower_x = 0
Upper_y + Lower_y - 330N = 0
+(Upper_x)(1.60 m) - (Lower_x)(.70 m) -(330N)(.5 m) = 0

Now, we have too many unknowns to be allowed. (4, with only three equations). I think it is a fair assumption, that if
Upper_x - Lower_x = 0, then Upper_x = Lower_x = Hinge_x (as we will now call it), and then we can assume that the door is well built, and that the vertical components exerted by the hinges are identical.

Upper_y + Lower_y - 330N = 0
So I will assume that Upper_y = Lower_y = Hinge_y

now our equations are more manageable:

Upper_x - Lower_x = 0
Upper_y + Lower_y - 330 N = 0
+(Upper_x)(1.60 m) - (Lower_x)(.70 m) -(330N)(.5 m) = 0

Becomes

2(Hinge_y) - 330N = 0
+(Hinge_x)(1.60 m) - (Hinge_x)(.70 m) -(330N)(.5 m) = 0

Which makes me smile.
2(Hinge_y) - 330 N = 0, so Hinge_y = 165 N

and
+(Hinge_x)(1.60 m) - (Hinge_x)(.70 m) -(330N)(.5 m) = 0
+(Hinge_x)(1.60 m - .70 m) = 165N
Hinge_x = 183.3 N

So both hinges exert an upward force of 165 N (assuming they are mounted in the right place) and the top hinge exerts a force to the right of 183.3 N, and the lower hinge exerts a force to the left on the door of 183.3 N