A uniform 19 kg door that is 2.2 m high by 1.10 m wide is hung from two hinges t

Question

A uniform 19 kg door that is 2.2 m high by 1.10 m wide is hung from two hinges that are 20 cm from the top and 20 cm from the bottom. If each hinge supports half the weight of the door, find the magnitude and direction of the horizontal components of the forces exerted by the two hinges on the door. (Take the direction of forcing the door away from the hinges to be positive and the direction of forcing the door toward the hinges to be negative.)

____?____ N (Upper hinge)

____?____ N (lower hinge)

Explanation / Answer

if we consider the bottom hinge as a rotatingpoint the su of torque is zero

let the upper Normal force is = Nu
       lower force is = NL
now taking the ratatinal about mid point of the door then :
-19kg*g*.5m+Nu*2.1=0
Nu=44.33

where Nu is the reaction at the upper hinge and this reaction pull the door.

if the upper hing is the rotating point we will the sum of torque there=0
19kg*g*.5-NL*2.1=0

NL=44.33

there NL and Nu is same but opposite direction Nu pull the door NLpush the door

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