Question

A uniform door (0.79 m wide and 2.0 m high) weighs 113 N and is hung...

A uniform door (0.79 m wide and 2.0 m high) weighs 113 N and is hung on two hinges that fasten the long left side of the door to a vertical wall. The hinges are 2.0 m a part. Assume that the lower hinge bears all the weight of the door. Find the horizontal component of the forces, including sign, applied to the door by (a) the upper hinge and (b) the lower hinge, taking the direction to the right to be the positive direction. Determine the (c) magnitude and (d)direction (as a positive angle counterclockwise from horizontal) of the force applied by the door to the upper hinge. Determine the (e) magnitude and (f) direction (as a positive angle counterclockwise from horizontal) of the force applied by the door to the lower hinge. Answer to 3 significant figures.

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Answer #1

The upper hinge only exerts horizontal force as mentioned in the question. And the lower hinge exerts both horizontal and vertical force.

So we have three unknowns and we need three equations to solve for the unknowns.

Translational Equilibrium equation along horizontal.

Translational Equilibrium equation along vertical.

Rotational equilibrium equation about any of the hinge.

For the detailed solution please find the attached image below

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