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The following question was in a recent exam:

A package P of weight $10$ N is moving up an inclined plane under the action of a horizontal force of magnitude $20$ N. The plane is inclined at angle $\alpha$ to the horizontal, where $\tan{\alpha}=\frac{3}{4}$. Package P is modelled as a particle. Find the range of values $\mu$, the coefficient of friction.

To answer this I found $R=20$N and the force acting up the plane which is $10$ N. From here I feel like there is no way to continue as I do not know acceleration.

The given answer is $\mu \leq \frac{1}{2} $, as they use $20\mu \leq 10. $

This assumes acceleration is either $0$ or positive. I argue there is nothing in the question that eliminates the possibility that acceleration could be negative and that the particle is slowing down, in which case $\mu$ could be greater than $\frac{1}{2}$.

Can someone confirm this, or explain why I am wrong? Many thanks.

Joseph
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    What was your answer and how did you get it? – John Douma Mar 28 '21 at 06:41
  • If that is the exam question it is poorly worded, but trying to be cocky should get you 0 mark (as the "find $x$" joke), because it is quite clear what the examiner really wanted. – user10354138 Mar 28 '21 at 07:01
  • ?? I’m really confused by your reply. I think it is not clear. I genuinely don’t know how to explain this answer, and it’s still not clear to me how to get to the given answer. Instead of insulting me, couldn’t you please explain? How else can I ask a question and not seem “cocky”?? – Joseph Mar 28 '21 at 07:08

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The force perpendicular to the supporting plane is $10\cos\alpha+20\sin\alpha$, therefore frictional force is $\mu(10\cos\alpha+20\sin\alpha)$ acting downhill. Force acting uphill is $20\cos\alpha-10\sin\alpha$. If you want the package to move upwards: $20\cos\alpha-10\sin\alpha-\mu(10\cos\alpha+20\sin\alpha)>0~\implies~\frac{1}{2}>\mu$.

Vítězslav Štembera
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  • No, the horizontal force applied increases the normal reaction force and the friction. – user10354138 Mar 28 '21 at 06:59
  • This is incorrect. – John Douma Mar 28 '21 at 06:59
  • I have corrected myself, thank you for your comments. – Vítězslav Štembera Mar 28 '21 at 07:10
  • I would also have used <, but the mark scheme clearly states $\leq$. I'm still not sure which is correct. – Joseph Mar 28 '21 at 08:18
  • If you use $>$, then it really moves upward, that is clear. In the case $=$ there is no force and it can move upwards only if $v_0>0$ (or if the acting force is $>20$ for some time which causes $v_0>0$). In this case $=$ is sufficient. However in the problem there is nothing about any velocity, so I would personally write >, if I worked on this problem myself. (Remark: in reality, every parameter is given with some error, so this problem is only theoretical). – Vítězslav Štembera Mar 28 '21 at 08:49
  • @vitezslavstembera I could also make the case that a<0. Why is that possibility ruled out? – Joseph Mar 28 '21 at 09:04
  • What is $a$ here? – Vítězslav Štembera Mar 28 '21 at 09:13
  • Sorry I meant the package is slowing down so it’s still moving up the plane but decelerating. – Joseph Mar 28 '21 at 09:17
  • I see. In this case you need nonzero velocity at the beginning $v_0>0$. However in this case, with negative acceleration, you would reach a moment, when it starts moving downwards. Therefore I would not see it as a solution. – Vítězslav Štembera Mar 28 '21 at 09:25
  • But all we are told is that it is moving up the plane. We are not told it will continue to move up the plane indefinitely. I don’t see how that possibility is ruled out. – Joseph Mar 28 '21 at 09:28
  • No. In this example there is no time explicitly given and also you do not solve here a time dependent problem. Therefore you solve it in one time instant, in which the result should support the upward move. – Vítězslav Štembera Mar 28 '21 at 09:32
  • Yes upwards movement could also mean slowing down. – Joseph Mar 28 '21 at 09:44
  • As I said, you do not solve a time dependent problem here. In this case it is not clear about which time instant you talk. Consequently, you need every time instant to be equivalent with respect to the result. – Vítězslav Štembera Mar 28 '21 at 09:50
  • Sorry I don’t understand that. So can acceleration be negative or not? And if not, why not? Sorry it’s still not clear!! I feel thick as a plank for still not understanding. – Joseph Mar 28 '21 at 09:57
  • I try to explain it clearly. You can always ask 1) what the example means, 2) if this meaning is put correctly in words (some assumptions are not always explicitly stated). Here the intention of the example was to find a balance between forces, which would be valid in any time. So no negative acceleration, because then the package would move in one time instant upwards and in some other time instant downwards. On the other hand, you can always try to solve time dependent problem. You could ask for example: "having $v_0<0$, which Force R you need in order to force the package to move up". – Vítězslav Štembera Mar 28 '21 at 10:10
  • Continuation: in this case however, any small positive force would cause that at some time instant the package would start to move upwards. So this would be really trivial. Conclusion: 1) in the example they really mean that it moves upwards in every time instant, 2) time dependent problem, in order to have sense, would have to be completely different formulated. And do not forget then time dependent problems have to be solved via differential equations, which is a different kind of game. Is it clearer now? – Vítězslav Štembera Mar 28 '21 at 10:14
  • I don’t think it’s clear it moves upwards in every moment. If it’s not a time dependent problem how can we assume that? I don’t think I’m being pedantic, I think this question would cause a lot of confusion for students and that it has an ambiguous answer. I think they should have added more information such as “assuming the particle is not accelerating down the slope” or something but as it is the question and given answer is wrong. – Joseph Mar 28 '21 at 10:21
  • From my point of view the example is clearly stated. On the other hand, what is good on your thoughts is that you try to look to the problem as time dependent - and you can always try to solve this kind of problems via differential equations. – Vítězslav Štembera Mar 28 '21 at 10:25
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Supposedly in this case it is to be assumed that the force pushed the particle from rest despite the fact that in the ideal conditions outlined the force would not be sufficient to overcome friction if $\mu=\frac{1}{2}$. So then we can assume acceleration is 0 or positive in the direction of movement, thus getting the desired answer.

Applied math. Yuck.

Joseph
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