# Forces on hindges of a door

1. Aug 13, 2010

### diegocas

1. The problem statement, all variables and given/known data

A door of mass $$M$$, height $$h$$ and width $$b$$ is held by two hindges a distance $$d$$ from the lower and upper edges of the door. What is the force on the hindges?

2. Relevant equations

3. The attempt at a solution

We may assume the weight of the door ($$M\vec{g}$$) is applied in the geometric center of the door. Since the sum of all the forces on the door must be zero, the horizontal components of the forces on the hindges must be opposite and the vertical components must add to $$Mg$$.

If we compute the torques with respect to the lower hindge, we get that the torque of the force on the other hindge is $$F_x (h-2d)$$. On the other hand, the torque of the weight of the door is $$Mg\frac{b}{2}$$. Since these two torques must compensate, we get $$F_x = \frac{1}{2}Mg \frac{b}{h-2d}$$.

Now to the vertical components. They must add to $$Mg$$. However, I don't seem to find any way to determine each one of the forces! Am I overlooking something?

Thanks!

2. Aug 13, 2010

### PhanthomJay

Not really,the problem is statically indeterminate. If the door is installed properly, each hinge should vertically carry half the weight of the door. But if it were installed say with the lower hinge supporting all the weight before the upper hinge was fully installed , then the upper hinge might carry no vertical load. The usual assumption is that the vertical force on each hinge is Mg/2.

3. Aug 13, 2010

### diegocas

Thanks, it is much clearer now.

And sorry for my spelling! "hindges" looked odd to me, but I didn't know why.