What is the tension felt when a rock is dropped off a cliff connected to a rope?

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When a rock is dropped from a cliff while attached to a rope, the tension felt in the rope once the slack runs out is influenced by the rock's momentum and the force of gravity. The tension force will exert an upward pull on the hand holding the rope, which can be calculated using the rock's mass and acceleration due to gravity. This tension is not simply the weight of the rock but also accounts for the change in momentum as the rock stops its freefall. The discussion highlights the complexity of understanding the "feel" of the tension, emphasizing the need to consider both the force exerted and the dynamics of motion. Ultimately, the tension felt is a result of the interplay between gravitational force and the rock's momentum at the moment the rope becomes taut.
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This isn't a HW question, just a thought. If I held a rock in my hand and connected it to a rope which I held in my other hand, then dropped it off of a cliff, how heavy would it "feel" when the slack ran out?

I know that the force is the same the entire way down, but when the slack ran out there would be a tension force going upwards towards me. However, it is obvious that it has to do with more than the acceleration, which would assume the tension of the rock just sitting there, not having been dropped.

Is it something to do with momentum? I know p=mv, so If I replaced v with at, I come up with p=m(at). How can I find the tension knowing the momentum?

Thanks for your help.

Alex
 
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apmcavoy said:
This isn't a HW question, just a thought. If I held a rock in my hand and connected it to a rope which I held in my other hand, then dropped it off of a cliff, how heavy would it "feel" when the slack ran out?
I know that the force is the same the entire way down, but when the slack ran out there would be a tension force going upwards towards me. However, it is obvious that it has to do with more than the acceleration, which would assume the tension of the rock just sitting there, not having been dropped.
Is it something to do with momentum? I know p=mv, so If I replaced v with at, I come up with p=m(at). How can I find the tension knowing the momentum?
Thanks for your help.
Alex

I think I know what you're getting at, in which case I'd suggest also seeing that p = (ma)t , now, the question of what would it 'feel' like would kind of be a question of what exactly you'd consider as the 'feel'. Personally, I'd say that what it feels like would be the force it exerts on the hand. .. and then to put it into a perspective for others I'd take it as a weight .. and thus I could do mass based on 9.81 for the a. (which, if its free falling really tells a lot)
 
By feel I mean the tension at the moment the rope runs out of slack and the rock stops its freefall (stopped by the rope).

Thanks,

Alex
 
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