Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 5 (Q.No. 50)
50.
A body of weight 14 g appears to weight 13 g when weighed by a spring balance in a moving lift. The acceleration of the lift at that moment was
Discussion:
2 comments Page 1 of 1.
Ihsan khan Dir U said:
5 years ago
T = m1a+mg
In an initial case when mass is it\'s original form such as 14g then there was no acceleration i.e a=o.
So, T=0+mg or T=.0014*9.8 = 0.1372 kgm/sec^2.
Now in the second case when there is downward acceleration. The mass appears to be 13g
So, T=m2a+m2g.
Putting values = .1372=.0013(a)+.0013*9.81.
Solving for a.= a=0.75m/sec^2.
In an initial case when mass is it\'s original form such as 14g then there was no acceleration i.e a=o.
So, T=0+mg or T=.0014*9.8 = 0.1372 kgm/sec^2.
Now in the second case when there is downward acceleration. The mass appears to be 13g
So, T=m2a+m2g.
Putting values = .1372=.0013(a)+.0013*9.81.
Solving for a.= a=0.75m/sec^2.
(1)
Aarthi said:
8 years ago
Please explain.
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