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Grade: 11
        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 
7 years ago

Answers : (4)

Fawz Naim
37 Points
							

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

7 years ago
vikas askiitian expert
509 Points
							

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

 

7 years ago
vikas askiitian expert
509 Points
							

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

 

 

 

7 years ago
Mukul Shukla
40 Points
							

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)

7 years ago
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