Particle on a slope - forces: vector/geometry-based proof formalisation

Question

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?

Answer 1

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.

Answer 2

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:

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.

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.

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.