What is the Force on the Hindges of a Door?

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SUMMARY

The discussion centers on calculating the forces acting on the hinges of a door with mass M, height h, and width b, supported by two hinges located a distance d from the top and bottom edges. The weight of the door, represented as M*g, acts at its geometric center. The torque analysis reveals that the horizontal force on the upper hinge is F_x = (1/2)Mg(b/(h-2d)). It is established that under normal conditions, each hinge should support half the weight of the door, or Mg/2, although the system is statically indeterminate if the door is improperly installed.

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  • Knowledge of hinge mechanics and their role in door support
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  • Explore the concept of statically indeterminate structures
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Students studying physics, mechanical engineers, and anyone involved in structural design or door installation will benefit from this discussion.

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Homework Statement



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?



Homework Equations





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!
 
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diegocas said:
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!
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.
 
Thanks, it is much clearer now.

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

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