Question

A door 1.00 m wide and 2.00 m high weighs 330 N and is supported by two hinges, one 0.50 m from the top and the other 0.50 m from the bottom. Each hinge supports half the total weight of the door. Assuming that the door’s center of gravity is at its center, find the horizontal components of force exerted on the door by each hinge.

Answer 1

Answer:

R2x = 165 N

R1x = -165 N

Explanation:

How the door is in equilibrium, the next conditions must be complied:

$\sum F_{x} = 0$                 $\sum F_{y} = 0$

and   $\sum t_{th} = 0$

        Where $t_{th}$ are the torque moment with respect to the top hinge

The horizontal forces exert over the door are:

R1x: Horizontal reaction in the top hinge

R2x: Horizontal reaction in the bottom hinge

how $\sum F_{x} = 0$, then:

R1x + R2x = 0    R1x = -R2x

How each hinge supports half the total weight of the door

R1y = 165 N   Vertical reaction in the top hinge

R2y = 165 N  Vertical reaction in the bottom hinge

$\sum t_{th} = 0$

-(330)(0.5) + R2x(1) = 0

                  R2x = 165 N

and then    R1x = -165 N

Answer 2

The hori-zon-tal components of force exer-ted on the door by each h-i-n-g-e should be likely R-2-x = 165 N and R-1-x = -165 N.

Calculation of the horizontal components:

Since

A door 1.00 m wide and 2.00 m high weighs 330 N and is supported by two h-i-n-g-e-s, one 0.50 m from the top and the other 0.50 m from the bottom.

So here we can write as the

-(330)(0.5) + R-2-x(1) = 0

                 R-2-x = 165 N

and After this   R-1-x = -165 N

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