The door of an almirah is $6$ ft high, $1.5$ ft wide and weighs $8$ kg. The door is supported by two hinges situated at a distance of $1$ ft from the ends. If the magnitudes of the forces exerted by the hinges on the door are equal, find this magnitude.

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Solution

For vertical equilibrium,
$v + v = 80$ , $v = 40$
Taking moment about lower hinge,
$H X 4$ = $80 X 0.75 H$ = $15$ .
F by hinge = $s q r t (H^{2} + v^{2})$ = $43 N$

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The door of an almirah is $6 f t$ high, $1.5 f t$ wide and weights $8 k g$ . The door is supported by two hinges situated at a distance of 1 ft from the ends. Assuming force exerted on the hinges are equal, the magnitude of the force is :

Q. The door of an almirah is

6 ft

high,

1.5 ft

wide and weighs

8 kg

. The door is supported by two hinges situated at a distance of

1 ft

from the top and bottom ends. Assuming force exerted on the hinges are equal, the magnitude of the force is
[Take

g = 10 {m/s}^{2}

]