Question

The hinges of a uniform door which weighs 200N are 2.5m apart. One hinge is a distance d from the top of the door, while the other is a distance d from the bottom. The door is 1.0m wide. The weight of the door is supported by the lower hinge. Determine the forces exerted by the hinges on the door. (Ans: The horizontal force at the upper hinge is 40N. The force at the lower hinge is 200N at 79deg above the horizontal.)

Answer 1

At top hinge

Horizontal force = Rxtop

Vertical force = 0 (because weight balances by lower hinge)

At bottom hinge

Horizontal force = Rxbottom

Vertical force = Rybottom

As weight of the door is supported by lower hinge so Rybottom = 200 N

In horizontal direction

Rxtop = Rxbottom

Torque about bottom hinge = 0

-200×(0.5) + Rxtop ×(2.5) = 0

Rxtop = 40 N

So Rxbottom = 40 N

So force at top hinge or upper hinge = 40 N in horizontal direction

Force at bottom hinge = (402 + 2002)1/2 = 203 N

Angle = tan-1 (200/40) = 79o above the horizontal