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				   A boy stands on a weighing machine inside a lift.When the lift is going down with an acceleration of g/4 the machine shows a reading of 30kg.when the lift goesupward at acceleration g/4the reading will be............pls solve 
				   

6 years ago

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Answers : (4)

										

When the lift is going in downward direction with an acc. of g/4, net acc. of the lift


g`=g-g/4=3g/4. Weight of the body=mg`=m3g/4


Now the weight of the body is due to the normal reaction of the floor if the lift which is


N=mg`=3mg/4


3mg/4=30


m=12


Now when the lift is going up with an acc. of g/4 then net acc. is g+g/4=5g/4


N=m5g/4


N=12*5*10/4=150kg

6 years ago
										

when lift goes downwards then seudo force acts on boy in upward direction...


if a is the accleration of lift then


 mg - ma = net weight=30g         ...........1         (given)        g is accleration due to gravity


  m(g-a) = 30g


   m=40kg


when lift goes upward then seudo force acts in downward direction so now net weight


 mg + ma = W             


 40(a+g) = W


  50kg = W                 ans


 

6 years ago
										

when lift goes downwards then seudo force acts on boy in upward direction...


if a is the accleration of lift then


 mg - ma = net weight=30g         ...........1         (given)        g is accleration due to gravity


   m(g-a) = 30g


   m=40kg


when lift goes upward then seudo force acts in downward direction so now net weight


 mg + ma = W             


 40(a+g) = W


  50kg = W                 ans


 


 


 

6 years ago
										

While goin down:


mg-N=mg/4


N=3mg/4=30g


Therefore, m=40


While going up:


N-mg=mg/4


N=5/4mg


=50Kg(m being 40)

6 years ago

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