tension in a string tied to a body falling freely under gravity is
Answers
Answer:
A free body diagram on the 2m mass would have 2mg down and T up. This would give a Newton's 2nd Law equation, assuming up to be the positive vertical direction, of
T−2mg=2ma2v
. The m mass free-body diagram would yield two downward forces, T and mg with a Newton's 2nd Law equation of
−T−mg=ma1v,
assuming the tension magnitude in the rope is the same throughout the rope.
Your statement of constant velocity means that both accelerations must be zero. With that we have
T=2mg
from the first equation
T=mg
from the second. This is clearly an impossible situation unless there are some forces on the masses which are not accounted for.
If they fall with equal and constant acceleration, then we can write
T−2mg=2(−T−mg)
3T=0
and there is no tension in the rope.
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Answer:
A free body diagram on the 2m mass would have 2mg down and T up. This would give a Newton's 2nd Law equation, assuming up to be the positive vertical direction, of
T−2mg=2ma2v
. The m mass free-body diagram would yield two downward forces, T and mg with a Newton's 2nd Law equation of
−T−mg=ma1v,
assuming the tension magnitude in the rope is the same throughout the rope.
Your statement of constant velocity means that both accelerations must be zero. With that we have
T=2mg
from the first equation
T=mg
from the second. This is clearly an impossible situation unless there are some forces on the masses which are not accounted for.
If they fall with equal and constant acceleration, then we can write
T−2mg=2(−T−mg)
3T=0
and there is no tension in the rope.