# Help with Forces, Pulleys, and Whether People get pulled

## Homework Statement

A rope of negligible mass passes over a pulley of negligible mass attached to the ceiling, as shown above. One End of the rope is held by Student A of mass 80 kg, who is at rest on the floor. The opposite end ofthe rope is held by Student B of mass 70 kg, who is suspended at rest above the floor.
(I will provide the image referenced to below)
Student b now climbs up the rope at a constant acceleration of .5 meters per second squared with respect to the floor.
a. Calculate the tension in the rope while Student B is accelerating.
b. As student B is accelerating, is student A pulled upward off the floor? Justify your answer.
c. With what minimum acceleration must student B climb up the rope to life student A upward off the floor?

## The Attempt at a Solution

I solved in other problems related to this that the normal force exerted by the floor on student A is 100 N, and I tried to do part A (I will post pictures in a minute). Part B and C I am very unsure of, and I do not know where exactly to begin with these parts. If I could have any/all hints at a direction for part B and C, that would be greatly appreciated 8D.

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T
this is my attempt at part A and below is my image

so, acording to your calculations, there is less tension on the rope when student b is acelerating upward than when he is just hanging on? does that make sense to you?
Imagine you are student A, do you think it would be harder to hold on to the rope when he is just hanging there or when he is climbing?
sure he is acelerating upward, what about the rope? is it acelerating upward with him?

komarxian
!! Since he is climbing the force he is applying will be in the opposite direction of his motion, and that will make tension larger?

yes, the rope besides "feeling" his weight it would also "feel" the force he is applying to manage to climb it, his weight applies a downward force (on the rope) and so does the climbing

Does this look okay?

just to clarify, your answer to "c." (the last one) is "he must climb at 11.42 ms^-2 in order to achieve lift on student A" ?
if yes:
the thought process is good, you just need to remind yourself of the "components" of the tension force generated by student B, yes, Ft must be greater than the Fg of student A, but what is Ft "made of"?

and remember if Ft=FgA the net force on A is 0, meaning.. ?

komarxian
Yes, 11.42 is my answer, but I will take another look.

hint: you know what Ft must be to lift, so calculate Ft again with the new acceleration, i think that will help you out

This is my solution for part C. Is it correct?

yeah, you see, your previous answer was 11.43, because he (student B) needed to "make" the rope accelerate at that speed, what you forgot was that gravity was allready making it accelerate at 10 m/s^2, so he only needed that 1.43.
and don't forget if Ft is exactly 800N the net force will be 0 (on student A) and basically student A will no longer "feel" is own weight, but wont be enough to lift him.

anyway funny stuff; since tension is 800.1N for that acceleration (1.43), student A would accelerate 0.00125 m/s^2.. which means it would take 40 seconds for him to rise one meter :). .. on the other side of the rope, student B would have climbed 1144 meters in those 40 secs :D

Okay yay!! Thanks for the help. Its nice to finally understand it.

haruspex