3
$\begingroup$

I ran across this problem along with an answer. The answer says that a certain force due to friction can be ignored, but I question whether that is correct.

enter image description here

The question is: find the acceleration of block 2. There is friction between block 2 and the floor with coefficient of friction $\mu_{2f}$, and there is friction between block 1 and block 2 with coefficient $\mu_{12}$.

The given answer starts $$ (m_1 + m_2)a = F_a - F_\mathrm{friction\; on\; 2\; due\; to\; floor} - F_\mathrm{friction\;on\; 1\; due\;to\; 2}$$ $$ (m_1+ m_2)a = F_a - \mu_{2f}(m_1 + m_2)g - \mu_{12}m_1g$$ along with an explanation "There is also a friction force on 2 due to 1, but that can be ignored."

I don't see how we can ignore the friction force on 2 due to 1. In order to make things as clear as possible, I might treat each mass separately in order to clarify the forces on each mass.

$$ m_1a_1 = -T +\mu_{12}m_1g$$ $$ m_2 a_2 = F_a -T -\mu_{2f}(m_1+m_2) - \mu_{12}m_1g $$ $$ a_2 = -a_1 = a$$ The last term in the second equation explicitly accounts for the force on 2 due to 1. This is the force that the given answer says can be ignored. I'm wondering if I'm double counting the friction. I saw this problem and answer in print, so I'm inclined to doubt my analysis. Is the friction force on 2 due to 1 ignorable?

$\endgroup$
8
  • 1
    $\begingroup$ i dont think its ignorable. $\endgroup$
    – Lelouch
    Commented Jul 30, 2016 at 14:56
  • $\begingroup$ you have to substitute for T as well in the final expression. So the friction force on 2 due to 1 must be present, as you can see from the last 3 equations. $\endgroup$
    – Lelouch
    Commented Jul 30, 2016 at 15:00
  • $\begingroup$ "There is also a friction force on 2 due to 1, but that can be ignored." I don't see how we can ignore the friction force on 1. I am confused now; is it a friction force on box 1 or on box 2 that should be ignored? $\endgroup$
    – Steeven
    Commented Jul 30, 2016 at 15:30
  • $\begingroup$ It might mean that in this particular problem, that you ignore that component of the friction. $\endgroup$
    – jim
    Commented Jul 30, 2016 at 16:10
  • $\begingroup$ The one that they are suggesting can be ignored is the friction force on box 2, the lower box, due to box 1, the upper box. $\endgroup$
    – garyp
    Commented Jul 30, 2016 at 17:45

4 Answers 4

1
$\begingroup$

Your equation is written incorrectly confusingly for me as:

$$m_1a_1 = -T +\mu_{12}m_1g$$

I would write it as:

$$T = m_1a_1 + \mu_{12}m_1g$$

Then:

$$F_a - T-\mu_{2f}(m_1+m_2)g - \mu_{12}m_1g = m_2 a_2$$

Then replace T as follows:

$$F_a - m_1a_1 - \mu_{12}m_1g -\mu_{2f}(m_1+m_2)g - \mu_{12}m_1g = m_2 a_2$$

This yields:

$$F_a -\mu_{2f}(m_1+m_2)g - 2\mu_{12}m_1g = (m_1 + m_2)a$$

Where:

$$a = a_1 = a_2$$

You can't ignore the friction between 2 and 1 because that friction is working on each of the masses in such a way as to resist the force, $F_a$. Instead, that frictional force should actually be accounted for TWICE.

$\endgroup$
2
  • $\begingroup$ My first equation is not incorrect. I just prefer to have the positive axis point to the right in all cases. Then I don't have to remember which coordinate frame points left and which points right. And minus signs always mean "to the left". $\endgroup$
    – garyp
    Commented Jul 30, 2016 at 17:44
  • $\begingroup$ @garyp I understand your point Gary. I can't do it your way because then I end up getting confused. I edited my wording above. $\endgroup$ Commented Jul 30, 2016 at 17:48
1
$\begingroup$

I think your analysis is correct. The 'given answer' provides no reason for ignoring the friction force on block 2 due to block 1. Such 'hand-waving' ought to ring alarm bells.

Suppose blocks 1 & 2 are vertically displaced so that, instead of block 1 resting on block 2, it rests on a fixed ledge. Further suppose that another block of mass $m_1$ rests on block 2, and this block is pushed to the left so that it remains vertically below block 1 as that block moves. The same coefficient of friction $\mu_{12}$ exists between each surface.

The forces on each block remain exactly the same as before, but it is now clear that the friction force $\mu_{12}mg$ acts on each block and must be counted twice, once for each.

However, this way of looking at the problem is no better than (and no different from) drawing separate Free Body Diagrams for each block.

$\endgroup$
0
$\begingroup$

I think it refers to the friction on the wheel axis, which actually will be absent if the m1 is absent, and can be ignored because it should be very small if compared with the other frictions. I admit it is a weird interpretation, but I don't see any other friction

$\endgroup$
1
  • $\begingroup$ The complete wording of the problem makes it clear that they are ignoring the friction force on the lower box due to the upper box. $\endgroup$
    – garyp
    Commented Jul 30, 2016 at 17:47
0
$\begingroup$

if we specify sign to be positive when directing to the right,

$$m_1(-a) = -T +\mu_{12}m_1g$$

$$m_2 a = F_a -T -\mu_{2f}(m_1+m_2)g - \mu_{12}m_1g $$

subtract the first equation from the second equation,

$$(m_1+m_2) a = F_a -\mu_{2f}(m_1+m_2)g - 2\mu_{12}m_1g $$

that is your answer.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.