Ranking friction forces between two boxes accelerating upward together

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isukatphysics69
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Diagram of two boxes labeled A and B pressed between a hand on the left and a smooth, frictionless wall on the right, with an upward velocity arrow above them.
1. Homework Statement

in picture

Homework Equations


My equations are
netforceyA = -MG + Friction between A and Hand + Friction between A and B
netforceyB = -MG + Friction between B and A

The Attempt at a Solution


Friction between A and B are equivelant due to Newtons third law since they are both accelerating upwards at the same rate correct?
I am not understanding this problem very much.. i know that the netforce = 0 for both cases since they are accelerating upwards at same rate (shown by velocity vector in middle of boxes) so the acceleration is 0. I don't know how to figure the rankings after solving the equations for the forces..​
[/B]
 
Attachments
  • Diagram of two boxes labeled A and B pressed between a hand on the left and a smooth, frictionless wall on the right, with an upward velocity arrow above them.
    apr4.PNG
    14.7 KB · Views: 863
Online quiz question reading: Suppose the blocks move up the wall with constant speed. Rank the magnitudes of the vertical forces, greatest first. Answer field contains: FGAE > FGBE ? FFAB = FFBA, followed by a Check button.


So i know that the force of gravity on A is greater than the force of gravity on B (problem says A has more mass).. and i believe that Newtons third law will tell us that the frictional forces between the two blocks are equivalent since they are both accelerating upwards at the same rate.. i am not sure how i would rank these two against each other and i believe that the frictional force between the hand and A is greater than the force of gravity..
 
Attachments
  • Online quiz question reading: Suppose the blocks move up the wall with constant speed. Rank the magnitudes of the vertical forces, greatest first. Answer field contains: FGAE > FGBE ? FFAB = FFBA, followed by a Check button.
    APR42.PNG
    4.4 KB · Views: 490
Online quiz screenshot: prompt asks to rank magnitudes of vertical forces when blocks move up a wall at constant speed; answer entered as FFAH > FGAE > FGBE = FFAB = FFBA, marked Correct, 2.00/2.00 marks.


WOW i actually just figured it out using Newton laws
 
Attachments
  • Online quiz screenshot: prompt asks to rank magnitudes of vertical forces when blocks move up a wall at constant speed; answer entered as FFAH > FGAE > FGBE = FFAB = FFBA, marked Correct, 2.00/2.00 marks.
    APR43.PNG
    6.1 KB · Views: 902
haruspex said:
I'm glad you figured it out, but you should not have abandoned the thread at https://www.physicsforums.com/threads/a-problem-on-ranking-forces.943704/ and posted it afresh. I was helping you there. Indeed, it sounds like you solved it by taking my advice.
Hey sorry about that.. I can delete that other thread, i am now on part 2 and not understanding something..
I have
netforceyA = -MG + Friction between A and Hand + Friction between A and B > 0
netforceyB = -MG + Friction between B and A > 0

Friction between B and A > MG
FFAB = FFBA because Newton 3rd law so FFAB = FFBA > FGBE

Friction between A and Hand > MG - Friction between A and B

Since accelerating upwards the frictional force beats gravity so FFAH > FGAE
i think my ranking is wrong after FGAE, But i am unsure why
Screenshot of an online quiz: question asks to rank vertical force magnitudes, answer entered FFAH > FGAE > FFAB = FFBA > FGBE, with feedback Reconsider FGAE and marks 0.50/2.00.
 
Attachments
  • Screenshot of an online quiz: question asks to rank vertical force magnitudes, answer entered FFAH > FGAE > FFAB = FFBA > FGBE, with feedback Reconsider FGAE and marks 0.50/2.00.
    con6.PNG
    7.8 KB · Views: 496
isukatphysics69 said:
netforceyB = -MG +
The masses are different. To avoid confusion, use m for the mass of B. (And it is g everywhere, not G.)
isukatphysics69 said:
Friction between A and B
isukatphysics69 said:
Friction between B and A
isukatphysics69 said:
FFAB = FFBA
Careful about directions. The friction between A and B cannot be pushing both up. One problem is that "between" does not specify which way around. Better to write "frictional force of A on B" etc.
It is not entirely clear, but I would assume the question intends that "FFAB" etc. are only supposed to represent the magnitudes of the forces.
 
haruspex said:
The masses are different. To avoid confusion, use m for the mass of B. (And it is g everywhere, not G.)
Careful about directions. The friction between A and B cannot be pushing both up. One problem is that "between" does not specify which way around. Better to write "frictional force of A on B" etc.
It is not entirely clear, but I would assume the question intends that "FFAB" etc. are only supposed to represent the magnitudes of the forces.
Hi, yes, it is magnitude ranking i should have mentioned that.. The program is giving a hint tho to check the force of gravity on A..Not sure what I'm doing wrong with A's MG
 
Ok i figured out that i had everything correct but i just had to eliminate FGAE because there was not enough info to actually give a ranking for it (i think)
 
isukatphysics69 said:
Ok i figured out that i had everything correct but i just had to eliminate FGAE because there was not enough info to actually give a ranking for it (i think)
it makes sense to me right now because using Newtons laws i was totally confused about how to prove FGAE ranking in this picture given increased acceleration..it seemed like there was no way to prove it. not sure
 
isukatphysics69 said:
it makes sense to me right now because using Newtons laws i was totally confused about how to prove FGAE ranking in this picture given increased acceleration..it seemed like there was no way to prove it. not sure
How do you deduce that FGAE>FFAB?