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A rope with length L and mass density l(kg/m) is held in a pile, and you grab an end that sticks up a tiny bit. The pile is then released, and falls downward. As a function of time, what force must your hand apply to the top of the rope to keep it motionless? Assume there is no friction of the rope with itself, so the pile is always in freefall.

User Adam Fox
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1 Answer

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Answer:

F= l x ( L - 1/2 g t²) g

Step-by-step explanation:

Given that

length =L m

mass density = l kg/m

Distance cover by time t

y = 1/2 g t²

The remaining length of the pile = L - 1/2 g t²

So the total weight of the remaining length of the pile ,m

m= l x ( L - 1/2 g t²)

The force apply to the top of the rope to keep it motionless is F

F= m g

F= l x ( L - 1/2 g t²) g

A rope with length L and mass density l(kg/m) is held in a pile, and you grab an end-example-1
User Thisgeek
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