Funny. :D:D
Why not?
Ignoring the weight of the rope, ignoring the fact that gravity is minutely less as altitude increases… on both sides, F=Ma=Mg
Same mass, same downward force, height is irrelevant unless the monkey lets go.
But yes, to climb the monkey exerts a force to go upward. Unless, he uses a jet-pack, he’s pulling on the rope. If it was a monkey and a bucket on frictionless ice or sitting on giant air-hockey pucks, the same would apply. He pulls on the rope, he and the bucket move together. since they weight the same, they move together at the same speed.
The only difference is, on air hockey pucks, they would keep moving until they collide. In the pulley example, gravity stops both motions upward once acceleration upward (climbing) force stops.
If the monkey yanked real hard, harder than twice gravity he and the bucket would zoom upward until gravity slowed them to zero and then they both fell to take up any slack in the rope. The difference is the hook and the solid mass anchors your motion, not a movable bucket.
What would happen if you were hanging off a rope tied to a hook, and you yanked yourself up real hard? you’d go up and then gravity would slow you down and you’d fall back to the length of the rope. Same thing if monkey and bucket did that, except both have same mass, same force on them thanks to rope, so identical trajectories.
The monkey must counteract the force of gravity the entire way up. It’s not just a single transition, it’s a sequence of motions that each exert a force.
If the machine in the video smoothly reeled in the rope would the effect be the same?
Three answers:
- Frictionless pulley, light rope.
As the monkey climbs both the monkey and the bucket rise together and both meet at the top.
- Frictionless pulley, rope has mass.
As soon as the monkey has climbed a bit the extra rope makes his side heavier so the monkey falls and the bucket rises to the top where the bucket tips up and pours the water on the monkey. Wet monkey climbs the rope to the top.
- Pulley with friction, smart monkey.
The monkey climbs slowly and steadily so that at no time does the extra force due to his acceleration exceed the friction in the pulley. The bucket stays where it is and the monkey only has to climb half as far.
Quite. The monkey is gaining potential energy all the way up, and this ain’t coming out of thin air.
Rather than assume the rope is weightless, it would be more realistic to assume that it loops like on a flagpole. Simply tie both ends to the bucket handle and there you go. It’s a little more difficult to come up with a real world equivalent of a frictionless pulley though.
Yes, that’s why I specified additional force. As in, in addition to the force to counteract gravity, because that’s already balanced out by the bucket which must also counteract gravity.
Hm, yes, this would work: That’d make the system neutrally-stable.
In this case, then, even with a near-ideal pulley (the bearings are arbitrarily good), the monkey should be able to climb. In such a case, there would be no external torques on the system, so the angular momentum of the system is conserved. The only way to satisfy this is for the monkey and the bucket to both rise at the same rate, until such time as one of them reaches the top. And of course it would be even easier for the monkey if there’s nonzero friction in the pulley.
Ok, but the way you worded that post made it sound like the initial transition to climbing was the only time an additional force was required, once the monkey was in motion that no additional force was required.
Yes, once the monkey is in motion no additional force is required, only the base, balanced amount of force needed to counteract gravity. This is incidentally the same amount of force the bucket must exert, for the same reason.
But for the monkey to stay in motion it must continue to exert an upward force. Each time it grabs the rope with it’s other hand, it must then exert a force to pull itself up the next distance.
yes. But if the monkey is just hanging there stationary, it still needs to exert a force, or else it will fall. If it’s moving at a constant rate, it needs to exert that same force, whether that constant rate is zero or not.
Yep.
Yes, but it needs to continue to exert that force all the way up, each increase in height requires a force (that exceeds the force of gravity) to achieve it, right? The monkey doesn’t just exert a force initially to overcome gravity (as Chronos seemed to say) and then just match gravity the rest of the way up, it needs to exceed it.
It must continue to exert the same force that it did while hanging in place. It need not exert any additional force beyond that, except right at the beginning when it starts climbing. I still don’t see what’s confusing about phrasing it that way.
monkey stays in same place, bucket pulled up (in physics land)
No, it only needs to exceed gravity to accelerate. Once it’s moving upward, it only needs to exert a force equal to its own weight to continue moving upward at the same speed.
Think of it this way, if the monkey exerts the same force gravity is exerting but in the opposite direction, it’s basically going to balance out gravity and it’ll be as though there is no gravity. If the monkey was travelling upward in an environment without gravity and no force acted upon it, would it continue to go up or would it stop there? It would continue to go up, of course.
Maybe you should stop speed-dating.
Because when climbing a rope, it’s done in a sequence of movements that each feel distinct and each one feels like it requires more effort than just hanging at that particular point. Maybe biomechanics are confusing the issue for me but I don’t think so.
It seems like you are saying that on the first motion of reaching up the rope and pulling itself up the monkey exerted a force, but on the second motion when the monkey alternated hands and did the motion again, it didn’t exert a force.
But it seems to me that the force the monkey exerted in the first motion was exactly the amount of force required to raise his mass and the counterweight by X feet, which is much less than the entire distance of Y feet that he will raise himself up by the end, thus the initial exerted force was only part of the total.
First, assume a spherical monkey…