Derivation of angle of repose - Friction

[k.udhay]

Hi,

Recently I came across this image in deriving an equation for angle of repose of an object in an inclined plane:

"W" is the weight of the component which acts vertically down.
"Rn" is the normal reaction by the inclined plane
"W cos alpha" is the component of the weight of the object parallel to the normal reaction.
"alpha" is the angle of inclination of the plane.

My question:
In all calculations, I have observed the force component perpendicular to the tangent of contact point to be the maximum. Yes, I mean the resultant force. Whereas, in the above picturem such resultant force (W cos alpha) is smaller than the vertical component (W). How can this be right?

Even if it is right, if I have to calculate the horizontal reaction by the plane, how can I calculate that? Pl. help. Thanks.

 

[cnh1995]

in the above picturem such resultant force (W cos alpha) is smaller than the vertical component (W). How can this be right?

Wcos alpha is a component of W perpendicular to the incline. It should be smaller than W.

 

[k.udhay]

Wcos alpha is a component of W perpendicular to the incline. It should be smaller than W.

What is the horizontal component of W in that case?

 

[cnh1995]

What is the horizontal component of W in that case?

Zero, since W is perpendicular to the horizontal direction.

 

[k.udhay]

Zero, since W is perpendicular to the horizontal direction.

If I assume the inclined plane to be on rollers, wouldn't it start moving horizontally towards east? So, there is a horizontal component, isn't it?

 

[cnh1995]

If I assume the inclined plane to be on rollers, wouldn't it start moving horizontally towards east? So, there is a horizontal component, isn't it?

I can't visualize how adding the rollers will make the incline move forward. There will be no change in the orientation of the forces after rollers are added, so you'll still have no force on the incline in the horizontal direction. But it's been very long since I passed my mechanics course, so I'm not fluent in such problems.
I believe @haruspex can guide you way better.

 

[k.udhay]

This is what I mean. At this condition, wouldn't the inclined plane start moving east? Thanks.

 

[haruspex]

If I assume the inclined plane to be on rollers, wouldn't it start moving horizontally towards east? So, there is a horizontal component, isn't it?

No, W is the force due to gravity. It has no horizontal component. That force pushes the mass down onto the wedge. The wedge resists being penetrated or moved (this is what normal force is about), resulting in an action and reaction between the mass and wedge.
The mass can accelerate down the slope, but its inertia generates a resistive force up the slope. That has a horizontal component, and leads to the tendency for the wedge to move sideways.

In all calculations, I have observed the force component perpendicular to the tangent of contact point to be the maximum. Yes, I mean the resultant force.

I need an example to understand that.

 

[k.udhay]

Let us take friction out of this case for a minute:

Wouldn't the block start moving horizontally? This means there is a horizontal component generated, right?

 

[haruspex]

Let us take friction out of this case for a minute:
View attachment 108432

Wouldn't the block start moving horizontally? This means there is a horizontal component generated, right?

It can be hard to get your head around what is going on in such theoretical arrangements. If you were to try pushing down on the wedge, you would not be able to help exerting a sideways force. So instead, consider gluing a small mass to the surface of the wedge. That would exert a purely downward force, but the wedge would not move.

 

[Cutter Ketch]

I think the picture provided is a little off and it might have contributed to the confusion. First, in calling the vertical vector "R" as in "the resultant" it must be understood that that is the resultant of only two of the three real forces in the problem: the normal force from the incline and the force of friction up the incline. So far no problem so long as that is understood. Second, in labeling the normal restoring force Rn makes it seem that R is the real force and Rn is just a component of R. Ok, technically it is a component of R, but I'm talking about a "chicken or the egg" problem with that labeling. The real forces are the normal restoring force and the friction and the resultant of those two forces is R, not the other way around. Third, this appears to be from a problem in that the angle between R and Rn is labeled phi rather than alpha. Presumably you are directed to prove that phi is alpha. Let's be aware that it is indeed alpha. Finally, I think this picture tries to do to much showing two ways to think about the problem. You can think of the normal force and friction force each being equal and opposite to the respective components of weight, or you can think of the resultant of the normal force and friction being equal and opposite to the weight. This diagram covers both, but invites the uninitiated to mix apples and oranges in their thinking. Just to be picky I'll throw in that w sine(alpha) and F are the wrong lengths.

Now to your questions. You say "the force perpendicular to ..." and then immediately say "yes I mean the resultant". This confusing because those aren't the same thing. Ok, forget that.

You say that you always find Rn is the biggest vector but w cos alpha is smaller than W. how can this be? Short answer: it can't. You are doing something wrong. I suspect you drew one of your triangles wrong (perpendicular to the wrong vector?)

In any case, all of the vectors in the diagram which appear to oppose each other should be equal and opposite. R and W are the same length and both are longer than any of their components.

Regarding the plane on wheels you are confusing the static case and the dynamic case. In the static case of the drawing where the box isn't moving there is no net force on the plane trying to push it to the right. What you are thinking of is the case where the box slips and accelerates down the plane. In that case there IS a net force on the box towards the left (that's why it is accelerating) and there is an equal and opposite force pushing the ramp to the right.

 

[k.udhay]

It can be hard to get your head around what is going on in such theoretical arrangements. If you were to try pushing down on the wedge, you would not be able to help exerting a sideways force. So instead, consider gluing a small mass to the surface of the wedge. That would exert a purely downward force, but the wedge would not move.

Brilliant! I can now start imagining from a different perspective! I understand the gluing force is caused by friction.

 

[sophiecentaur]

Let us take friction out of this case for a minute:
View attachment 108432

Wouldn't the block start moving horizontally? This means there is a horizontal component generated, right?

There will be two horizontal forces - because of Momentum Conservation. One accelerates the block and the other accelerates the ramp in the other direction. Without the wheels and with the ramp stuck to the ground, there is the same force and the same 'rightwards' gain in momentum but, the v in Mv is nearly zero.