Question

32. A body of mass m is placed on the floor of a litt. Find
ed on the floor of a lift. Find its apparent weight when the lift is
(a) moving upwards with uniform acceleration a
(b) moving downwards with uniform acceleration a
(c) at rest or moving with a uniform velocity v.
(d) falling freely. Take acceleration due to gravity g.
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Answer 1

When lift is accelerating upward with acceleration :
It’s net force will be upward so
Net force =T-W
F= T-W
Ma=T-W
T. = W+ma.
Result :
Apparent weight is increased by factor ma
When lift is accelerating downward
Net force=W-T
F=. W-T
Ma=W-T
T. =W-ma
If lift is falling freely
Net force=W-T
Ma= T-W
T= W-ma
T=mg-mg
As W=mg
a=g
So
T=0
When lift is at rest or moving with uniform speed so
T=W
Uniform speed
Net force =T-W
Ma =T-W
As a=0.
So
0. = T-W
T= W
If it is moving with uniform motion in downward direction
Net force = W-T
F= W-T
Ma = W-T
As v=0
a=0
So
W = T