Static friction 2 boxes and force question

AI Thread Summary
The discussion revolves around determining the maximum tension (Tmax) that allows a small box to remain on top of a larger box without slipping, given their masses and the coefficients of static and kinetic friction. Participants analyze the equations governing the forces acting on both boxes, emphasizing the importance of consistent notation for clarity. A mistake is identified in the calculation of acceleration, particularly in the final steps where the friction force is incorrectly substituted. The conversation highlights the need for careful attention to detail in physics problems, especially when dealing with multiple variables. Overall, the thread underscores the collaborative effort to solve a complex physics question.
weinbergshaun
Messages
12
Reaction score
1
Question:
A large box of mass M is pulled across a horizontal, friction-less surface by a horizontal rope with tension T. A small box mass m sits on top of the box. The coefficients of static and kinetic frciton between the two boxes are Us and Uk. find an expreeion for the maximum tension Tmax for which the small box rides on top of the large box without slipping. Relevant equations
None given

The Attempt at a Solution


Know formulas:
FFriction (or FF) = m2 x a (the friction of the top on lower box)
FTension (or FT) - FFriction = m1 x a
the two forces must have the same acceleration in the second formula, will get both acc. alone on both formulasand equal them.

Top block acc. is:
FT = m2 x a (get a alone)
a = FF/m2

Bottom block acc. is:
FT - FF = = m1 x a
FT – (m2 x a) = m1 x a (substituting Ff for the m2 x a value in order to have one force value only)
FT = m1 x a + m2 x a
FT = a(m1 + m2)
a = FT/(m1 + m2)

a = a, getting Tension or FT alone:
FF/m2 = FT/(m1 + m2)
FT = FF(m1 + m2)/m2
FMax = m1 x g x Us (m1 + m2)/m2 (substituting FT with max as it will represent max).

Is this correct, thank
 
Physics news on Phys.org
weinbergshaun said:
Relevant equations
None given
The idea is that here you list the equations you're using in the solution attempt. Like ##F = ma ## and ##F_{\rm fric, max} = \mu_s N##...
In particular the subscript max is important: that's the friction force that determines the maximum T. But you understand that already.

And don't change notation on the way: if your problem statement uses Tmax, stay with it and don't change to Fmax.

weinbergshaun said:
FT = m2 x a (get a alone)
You mean FF

Other than that, it looks flawless to me :smile:
 
  • Like
Likes weinbergshaun
What's m1 and m2?
 
  • Like
Likes weinbergshaun and BvU
BvU said:
The idea is that here you list the equations you're using in the solution attempt. Like ##F = ma ## and ##F_{\rm fric, max} = \mu_s N##...
In particular the subscript max is important: that's the friction force that determines the maximum T. But you understand that already.

And don't change notation on the way: if your problem statement uses Tmax, stay with it and don't change to Fmax.

You mean FF

Other than that, it looks flawless to me :smile:
Thanks! Late nights for me, i sometimes makes mistakes lol!
 
Mastermind01 said:
What's m1 and m2?
M1 is top box M2 is bottom box
 
Go to bed ! :sleep:
 
Your answers wrong. You isolated the acceleration incorrectly for the top block. This is what happens if you change notation.
 
I rest my case
BvU said:
And don't change notation on the way: if your problem statement uses
in this case M and m

How about m1 = bottom box and m2 = top box ? :smile:
 
BvU said:
I rest my case
in this case M and m

How about m1 = bottom box and m2 = top box ? :smile:

I think it's still wrong, if I remember the general solution correctly.
 
  • #10
Yep, I worked it out, it's wrong. Though I can't find out where he went wrong through all the m1 and m2's. :sorry:

EDIT: Found it, you messed up the last step.
 
  • Like
Likes weinbergshaun
  • #11
BvU said:
I rest my case
in this case M and m

How about m1 = bottom box and m2 = top box ? :smile:
Sorry lol, m1 is bottom, m2 is top! Spot on
 
  • #12
Mastermind01 said:
Yep, I worked it out, it's wrong. Though I can't find out where he went wrong through all the m1 and m2's. :sorry:

EDIT: Found it, you messed up the last step.
sorry m2 is top box, m1 is bottom box.
 
  • #13
weinbergshaun said:
sorry m2 is top box, m1 is bottom box.

I get it, you still messed up the last step
 
  • #14
You are absolutely correct, MM, through all the m1 and m2 he missed FF = m2 μ g and filled in m1 in the last step.

When wbh wakes up he'll see it more clearly. I should go :sleep: too for overlooking that !
 
  • Like
Likes weinbergshaun and Mastermind01
  • #15
weinbergshaun said:
sorry m2 is top box, m1 is bottom box.
Where did i mess up last step, sorry been spending a week on this question lol
 
  • #16
Bvu is correct, that's where you messed up. Go to sleep now!
 
  • Like
Likes weinbergshaun
  • #17
BvU said:
You are absolutely correct, MM, through all the m1 and m2 he missed FF = m2 μ g and filled in m1 in the last step.

When wbh wakes up he'll see it more clearly. I should go :sleep: too for overlooking that !
Ah ok lol i see, thanks for your help!
 
  • #18
weinbergshaun said:
been spending a week on this question
Then it's really time to go :sleep: !

And: you're welcome. Well done, MM, wbh and me :smile:
 
  • Like
Likes Mastermind01
  • #19
Mastermind01 said:
Bvu is correct, that's where you messed up. Go to sleep now!
Thanks! Yes off to bed!:wink:
 
  • Like
Likes Mastermind01
  • #20
BvU said:
Then it's really time to go :sleep: !
:biggrin:
 
Back
Top