A meter stick is pivoted at the 0.50-m line. A 3.0-kg object is hung from the 0.15-m line. Where should a 5.0-kg object be hung
to achieve equilibrium (the meter stick oriented horizontal and motionless)?
1 answer:
Answer:
0.71 m
Explanation:
Let the 5 kg is hang at a distance d from the midpoint.
use the concept of moments.
take moments about the mid point.
Anticlockwise moments = clockwise moments
3 x (0.5 - 0.15) = 5 x d
1.05 = 5 d
d = 0.21 m
So, the distance of 5 kg from 0 mark on the stick is 0.5 + 0.21 = 0.71 m
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Weight, (W)
Force due to air resistance, (F)
Upthrust force, (U)
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In this case, the hot-air balloon descends vertically at constant speed.
so,
and
so, ....................(1)
when it is ascending let the weight that it is releasing is R, so
..........(2)
solving equation (1) and (2)
2F is the weight of material that must be released from the balloon so that it ascends vertically at the same constant speed.