Static friction 2 boxes and force question

In summary, the maximum tension Tmax for which the small box rides on top of the large box without slipping can be found using the formula FMax = m2 x g x Us (m1 + m2)/m2, where m1 represents the mass of the bottom box and m2 represents the mass of the top box. However, in the solution attempt, there was an error in isolating the acceleration for the top block, which led to incorrect substitution of variables in the final step.
  • #1
weinbergshaun
12
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
 
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  • #2
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:
 
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  • #3
What's m1 and m2?
 
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  • #4
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!
 
  • #5
Mastermind01 said:
What's m1 and m2?
M1 is top box M2 is bottom box
 
  • #6
Go to bed ! :sleep:
 
  • #7
Your answers wrong. You isolated the acceleration incorrectly for the top block. This is what happens if you change notation.
 
  • #8
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:
 
  • #9
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.
 
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  • #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 !
 
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  • #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!
 
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  • #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:
 
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  • #19
Mastermind01 said:
Bvu is correct, that's where you messed up. Go to sleep now!
Thanks! Yes off to bed!:wink:
 
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  • #20
BvU said:
Then it's really time to go :sleep: !
:biggrin:
 

What is static friction?

Static friction is a type of frictional force that exists between two objects that are not moving relative to each other. It occurs when there is no relative motion between the two surfaces in contact.

What is the difference between static and kinetic friction?

The main difference between static and kinetic friction is that static friction acts to prevent two surfaces from sliding against each other, while kinetic friction acts to slow down or stop an object that is already in motion.

How is static friction calculated?

The magnitude of static friction is determined by the coefficient of static friction (μs) and the normal force (N) between the two surfaces. The formula for calculating static friction is F = μsN, where F is the force of static friction.

What factors affect static friction?

The coefficient of static friction, the normal force, and the roughness of the surfaces in contact are the main factors that affect static friction. Other factors include the type of material, the temperature, and the surface area of contact.

Can the force of static friction be greater than the applied force?

Yes, the force of static friction can be greater than the applied force. This is because static friction always acts in the direction opposite to the applied force, in order to prevent the two surfaces from sliding against each other. The force of static friction will increase until it reaches its maximum value, which is equal to the product of the coefficient of static friction and the normal force.

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