enter image description here

I think this is better posted here than in Physics.se because it is purely a geometry/vector problem.

A particle (purple dot in diagram) with mass $m$ kg rests in equilibrium due to coefficient of friction $\mu > tan(\alpha)$, where $0^\circ<\alpha<45^\circ$ and $0<\mu<1.$

I was wondering if a force P at a positive clockwise angle $\theta ^\circ > 0$ with the vertical could result in an acceleration of the particle down the slope. I think the answer is no.

$\underline{R}$ is the reaction force on the particle from the slope.

As P increases from $0N$, the particle remains in equilibrium until the resultant force $\underline{Res_{lim}}$ on the particle down the slope, $\underline{Res_{lim}} = m \underline{g} + \underline{R_{lim}}+ \underline{P_{lim}}$ is equal to limiting friction $\underline{F_{lim}}$.

I think we can then show that for $\underline{P}> \underline{P_{lim}}$ (and assuming the particle accelerates down the slope), the resultant force $\underline{Res} = m \underline{g}+ \underline{R}+ \underline{P}$ is strictly less than the friction force $\underline{F}$, using the fact that $\underline{F} = \mu \underline{R},\ \underline{F_{lim}} = \mu \underline{R_{lim}}$ and the fact that $\theta$ is positive clockwise angle to the vertical.

Then the resultant force would be up the slope, which would contradict the assumption that the particle accelerates down the slope, answering my original question.

I'm interested how to formalise my proof with simple geometry and vectors. I've had a go and it shouldn't be that difficult, but haven't managed to formalise it.

My idea for a proof would be: $\underline {P}$ removes a greater proportion of $\underline{Res_{lim}}$ than it removes of $\underline{R_{lim}}$, but like I say, I find formalising this difficult.

$\underline{Res_{lim}} = \mu \underline{R_{lim}}$. Now what?

  • $\begingroup$ To make this a pure vector problem we need a mathematical understanding of the frictional force vector, which I'm not sure has been established here. In particular, if we let $\underline{P}=0$, then the magnitude of $m \underline{g}+ \underline{R}+ \underline{P}$ is greater than that of the vector $\underline{Res_{lim}}$, that is, greater than the limiting frictional force, which would cause the particle to accelerate down the slope in the absence of $\underline{P}$ (unless I misread what you mean by a "limiting frictional force"). $\endgroup$
    – David K
    Oct 11 '20 at 15:52
  • $\begingroup$ Let me put this another way that might be clearer. The diagram shows a force $m\underline g$. Decompose this force into components perpendicular to the slope and parallel to it. The parallel component then is greater in magnitude than the vector $\underline{Res_{lim}}$ shown in the diagram. $\endgroup$
    – David K
    Oct 11 '20 at 16:40
  • $\begingroup$ There is a constant $0<\mu<1$ such that $F = \mu R$ where $\underline{F}$ acts parallel to the slope; $ \underline{R}$ acts perpendicular to the slope. I’m not sure the perpendicularity of \underline{R} and $\underline{F}$ is necessary to solve the problem. Also, you say that $P=0 \implies |m \underline{g} + \underline{R} + \underline{P}|>Res_{lim}$, but as the diagram shows, this is clearly false: they are equal. Limiting friction $F_{lim}= \mu R_{lim}$ is the magnitude of friction when the box is on the verge of moving, and $F=\mu R $ is also assumed true when the box is moving. $\endgroup$ Oct 11 '20 at 16:41
  • $\begingroup$ I didn't change my comment, I was still editing it to correct mistakes. I'm a slow typer it seems. If the box is stationary and not in limiting equilibrium, $F < \mu R$. Friction by definition opposes the direction of motion or the motion that the object has a tendency to move in, i.e. the direction the object would move in if there were no friction. In response to your second comment, when $P = 0, Res \neq Res_{lim}$, if that's what you were saying. When $P=0, Res > Res_{lim}$. So I'm not sure what you're getting at. Yes when $P=0$ we get the right-angled triangle. Your point is what? $\endgroup$ Oct 11 '20 at 16:54
  • 1
    $\begingroup$ Yes, I was already aware of some (though not all) of what you wrote in your answer, and was going to add on some of the information in your answer to my own when I could be bothered at a later date, but it looks like you've done my work (and more) for me so thanks. Plus, your diagrams are much nicer than mine. $\endgroup$ Oct 11 '20 at 21:17

This may be overly detailed in light of the answer already posted, but I had already made the figures so I suppose I might as well try to explain them.

I will interpret the figure in the question as showing a particle in contact with a surface that is immovable and cannot be deformed (so the particle cannot move to any point "under" the inclined line) and rough (so that there is friction).

It may help to work out everything in a coordinate system in which one axis is parallel to the surface and the other is perpendicular to it. Then every force vector decomposes neatly into components parallel to the coordinate axes as illustrated in the figure below:

enter image description here

Now for a little excursion into the physics of the situation in order to define things so that we can eventually get back to the mathematics:

We will suppose that $\vec S$ in this figure represents the vector sum of all forces on the particle except for the forces exerted by the surface, that the component of $\vec S$ perpendicular to the surface is $\vec S_\perp$ and that the component of $\vec S$ parallel to the surface is $\vec S_\parallel.$ If the direction of $\vec S_\perp$ is "into" the surface (as it is in the figure) then the surface exerts a normal force $\vec R$ equal and opposite to $\vec S_\perp$ and a frictional force $\vec F$ opposite to the direction of $\vec S_\parallel.$

If the particle is initially at rest relative to the surface and $\lvert\vec S_\parallel\rvert < \mu_s\lvert\vec R\rvert,$ where $\mu_s$ (the static coefficient of friction) is a constant determined by the quality of the surfaces in contact, $\vec F$ is equal in magnitude to $\vec S_\parallel$ and the particle does not move. Otherwise $\vec F$ is smaller than $\vec S_\parallel$ and the particle slides in the direction of $\vec S_\parallel.$

But if $\vec S_\perp$ is zero or is "away from" the surface then the surface does not interfere with the motion of the particle, which is simply accelerated in the direction of $\vec S.$

I think that ends the excursion into physics.

Now if we let $S_\perp = \lvert\vec S_\perp\rvert$ and $S_\parallel = \lvert\vec S_\parallel\rvert$, and if we take the directions of the coordinate system so that $S_\perp$ is positive when the direction of $\vec S_\perp$ is "into" the surface, then $S_\perp = \lvert\vec S\rvert \cos\theta$ and $S_\parallel = \lvert\vec S\rvert \sin\theta$ where $\theta$ is the angle between $\vec S$ and the direction of $\vec S_\perp.$ So the condition $\lvert\vec S_\parallel\rvert < \mu_s\lvert\vec R\rvert$ becomes

$$ \lvert\vec S\rvert \lvert\sin\theta\rvert < \mu_s \lvert\vec S\rvert \lvert \cos\theta \rvert,$$

and under the conditions that $\lvert\vec S\rvert \neq 0$ and $\cos\theta \neq 0,$ this is equivalent to

$$ \lvert\tan\theta \rvert < \mu_s. $$

So this means that for $\lvert\vec S\rvert > 0,$ the equilibrium conditions are obtained when the direction of $\vec S$ is within the angle $\alpha_0$ of the normal vector "into" the surface, where $\tan\alpha_0 = \mu_s$, as shown below.

enter image description here

For an inclined plane in the presence of gravity, we get a force (the weight of the particle) along the "physically vertical line" in the figure below, whose angle from the normal vector is $\alpha,$ the same as the angle of inclination of the plane. In order for the particle to be at rest when influenced only by its weight and the forces exerted by the surface, the angle of the "physically vertical line" from the normal to the surface must be $\alpha_0$ or less.

enter image description here

If we now add a force $\vec P$ directed to the right of the "physically vertical line", we get a sum force $\vec S$ which in the case shown in the figure is in a "no motion" direction. Other possible directions and magnitudes of $\vec P$ could produce $\vec S$ in a "slides rightward" direction or in a "flies away rightward" direction (actually rightward of the "physically vertical line", not just rightward of the normal vector), but never in a "slides leftward" direction.

To make this a more rigorous solution, translate these graphical concepts further into inequations based on angles and magnitudes of vectors.

  • $\begingroup$ Actually I think I get all of this. It's quite nice and informative. I think my answer answers the question I posed better (to give a proof). But your departure into physics was a creative way of looking at the problem. It reminds me of this problem: When a particle is on a slope, what's the range of angles perpendicular to the slope that an *infinite*/very large force could take such that the particle remains in equilibrium. And then you think about it, and the mass of the particle and the direction of the slope doesn't matter: just the angle the infinite force makes to the perpendicular! $\endgroup$ Oct 11 '20 at 21:38
  • $\begingroup$ You might want to wait a bit to see if anyone else chimes in, but other than that I think you're ready to accept your own answer. (No sense in leaving this question in the "unanswered" state for too long!) $\endgroup$
    – David K
    Oct 11 '20 at 22:50
  • $\begingroup$ I agree but I can’t accept my own answer until 2 days time apparently lol. $\endgroup$ Oct 11 '20 at 22:51
  • $\begingroup$ I guess they really want to enforce the "wait a bit" part. :-) In two days, then! $\endgroup$
    – David K
    Oct 11 '20 at 22:53
  • $\begingroup$ Thanks for your help on the question! $\endgroup$ Oct 12 '20 at 9:07

As David K pointed out in the comments, the solution is simple. When $P = 0, R = m g \cos{\alpha} > Res_{lim} = F_{lim} =$ max{ ${F}$ } $\implies $ positive acceleration down the slope when $P=0$, contradicting static equilibrium when $P=0$.

More interesting is when $ \mu = \tan \alpha $, so that when $P=0$, the object starts at rest in static equilibrium. Again, we look at the right-angled triangle formed by $\underline{R_{lim}}, \ \underline{Res_{lim}}$ and $m\underline{g}$ when $P=0$ and compare it to the right-angled triangle formed by $\underline{\Delta R}, \ \underline{\Delta Res}$ and $\underline{P}$.

$$F = \mu R\ \text{and} \ R = R_{lim} - \Delta R, \ \therefore F = \mu(R_{lim} - \Delta R)$$

$$\Delta F = F_{lim} - F = \mu R_{lim} - \mu (R_{lim} - \Delta R) = \mu \Delta R = \Delta R\tan \alpha, \implies \frac{\Delta F}{\Delta R} = \tan \alpha.$$

$$\text{However, for the right-angled triangle formed by} \ \underline{\Delta R}, \ \underline{\Delta Res} \ \text{and} \ \underline{P},\ \frac{\Delta F}{\Delta R} = \tan(\alpha - \theta) \neq \tan \alpha, \text{because} \ \theta > 0 ^\circ. $$

So no, if a particle is at rest in static equilibrium on an inclined plane, then if $P$ is pointing in the same quadrant as up the slope (and also if $\theta = 0$, i.e. $P$ is vertical), then the particle cannot accelerate down the slope.


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