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I have done all of the work to this problem, but I am having trouble conceptualizing the answer(s) I am arriving at.

A block is placed on an inclined plane at an angle of $30^\circ$ to the horizontal. A horizontal pushing force of 6N prevents the block sliding down the plane. If $\mu = \frac{1}{3}$ what is the mass of the block?

My book gives just one answer but I found 3.

$$m_1=0.543\ kg$$$$m_2=1.060\ kg$$$$m_3=2.992\ kg$$

For my calculations I used $g=9.800\ m\ s^{-2}$ and rounded to 3 decimal places.

$m_2$ is when friction doesn't come into play. The gravity component equals the pushing component.$$mg\ \sin30^\circ=6\ \cos30^\circ$$ $$m=1.060\ kg$$

$m_1$ is when the pushing force component exceeds the gravity component. Friction acts down the slope.$$mg\ \sin30^\circ +\mu(6\ \sin30^\circ+mg\ \cos30^\circ)=6\ \cos30^\circ$$ $$m=0.543\ kg$$

$m_3$ is when the gravity component exceeds the pushing component. Friction acts up the slope. $$mg\ \sin30^\circ =6\ \cos30^\circ -\mu(6\ \sin30^\circ+mg\ \cos30^\circ) $$ $$m=2.992\ kg$$

Have I overlooked something here?

Are there really 3 solutions to this problem?

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    $\begingroup$ "A horizontal pushing force ... prevents the block sliding down the plane.", and "if $\mu = \frac{1}[3}". The second tells you that you have to take friction into account (and just simple Coulombic friction at that). The first is your key that you need to figure the boundary between the force stopping the mass, and the mass overcoming the combined friction and force. $\endgroup$ – TimWescott Dec 3 '19 at 0:37
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    $\begingroup$ You are waaaay over-thinking it. The question is about friction, not pedantic nit-picking. Obviously, the friction and stopping force are countering the gravitational force. Write out the equation and find $m$. That's all. $\endgroup$ – Oscar Bravo Dec 3 '19 at 14:05
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    $\begingroup$ @OscarBravo You are actually under thinking it. If you think the friction force is equal to $\mu N$ then there is your error. $\endgroup$ – Aaron Stevens Dec 3 '19 at 14:48
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    $\begingroup$ Stictly speaking, the question is insufficiently defined and all three of your answers are valid for the given question, for the circumstances you already set. If the book has answers in the back you can determine what the the writer was intending to ask and therefore correct the wording of the question. $\endgroup$ – Alopex Dec 3 '19 at 15:57
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    $\begingroup$ @OscarBravo You are falling into the usual trap that the force of static friction is always $\mu N$. This is not always the case. Since the problem as written does not give any indication that that static friction force is at its maximum value, you cannot assume this to be true. $\endgroup$ – Aaron Stevens Dec 3 '19 at 17:16
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There are actually an infinite number of solutions. You have just been assuming there is no static friction, or that static friction is equal to its maximum magnitude$^*$. But really all we have for the static friction magnitude is $$0\leq F_\text{fric}\leq \mu N$$ and then the friction force could be acting up or down the ramp, depending on the value of the mass (i.e. you could have in your equations $-\mu N\leq F_\text{fric}\leq \mu N$)

So really your mass should be able to take on any value between $m_1$ and $m_3$ (assuming your work is correct, which I have not checked).

If you want to check for another one of the infinite solutions you can do one of two things. The first thing you can do is just pick some other valid friction force value and direction (for example, $F_\text{fric}=\frac12\mu N$) and determine what the mass needs to be to prevent sliding. The other way is to pick a new mass $m_1<m_4<m_3$ and then determine what $F_\text{fric}$ needs to be in order to prevent sliding. If $F_\text{fric}\leq \mu N$ for that mass $m_4$, then you have found another valid solution.


$^*$Don't fall into the common intro-physics misunderstanding that $F_\text{fric}=\mu N$ for static friction. The only time you can use equality this is if you know that the object is right on the verge of slipping. In general all you can say is $0\leq F_\text{fric}\leq\mu N$, and so it is often incorrect to replace $F_\text{fric}$ with $\mu N$ in certain problems.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Dec 2 '19 at 22:11
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    $\begingroup$ Some other comments deleted. Use the chat, folks. $\endgroup$ – rob Dec 3 '19 at 19:43
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In the way you interpreted the problem, you will not only find 3 solutions, but an infinite number of them. This is because the friction can take any value between 0 and $\mu N$ in both directions.

The way I interpret the problem though, is that the force is the minimal force necessary to prevent the block from moving. The block will move if the component of gravity is larger than $\mu N$, so the solution would be your number 3.

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    $\begingroup$ I also think you need the calculate the minimal force necessary, based on "A horizontal pushing force of 6N prevents the block sliding down the plane." To me that means that if the force were any smaller, the block would start sliding down the plane. The other extreme case seems not compatible with the wording in the question, because in that case the friction + gravity would prevent the block from sliding up the plane. $\endgroup$ – Roel Schroeven Dec 3 '19 at 17:04
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I have settled on an answer.

The wording is important to avoid ambiguity.

If the question were: What values could $m$ have such that the block does not move ? Then the answer would be: $$0.543\ \le m\ \le 2.992\ (kg)$$

It is a range of values.

If the question were: What is the min or max value of $m$ such that the block doesn't move ? Then the answer would be: $m=0.543\ kg$ and $m=2.992\ kg$ respectively.

The book stated ''what mass'' which implies that the author was looking for one answer. But the author did not mention min or max so the question became ambiguous. This was the author's mistake.

The answer the author included was $m=0.543\ kg$ which means that ''min mass'' should have been specified. This is what the author had in mind.

I think it is an interesting problem conceptually as many people (myself included) when first looking at this problem don't realise that friction could act down the slope.

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    $\begingroup$ I finally see what you're getting at... You think the block might be on the point of sliding up the slope? I still think that's a mis-interpretation; the key point in the question is that the force prevents the block sliding down. So we're at the other extreme - the block is about to slide down and the force is only just holding it. The answer is the big number. $\endgroup$ – Oscar Bravo Dec 3 '19 at 17:15
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    $\begingroup$ @OscarBravo The OP has already confirmed that the accepted answer is the smaller one. This problem was obviously not written very carefully. Either way, the OP is interested in exploring the more open-ended version of the problem. They are not interested in getting the single right answer, as they already have it. $\endgroup$ – Aaron Stevens Dec 3 '19 at 17:20

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