Multi-surface Frictional force

[kftheuidfnaks]

Homework Statement

Does the force of friction on an object depend on the number of surfaces in contact with the object?

Relevant Equations

I attached an image describing my question.

 

[haruspex]

Homework Statement: Does the force of friction on an object depend on the number of surfaces in contact with the object?
Homework Equations: I attached an image describing my question.

View attachment 253025

The safe way is to consider each surface separately, making sure to use only the normal force that applies on that surface in each case.
It gets awkward if the angle between the surfaces is less than 90 degrees, since the frictional force on one can interact with the normal force on the other, leading to indeterminate forces.

Edit: that last part wasn't quite right... I think the indeterminacy arises when it is static friction.

 

[kftheuidfnaks]

The safe way is to consider each surface separately, making sure to use only the normal force that applies on that surface in each case.
It gets awkward if the angle between the surfaces is less than 90 degrees, since the normal forces partly oppose each other, leading to indeterminate forces.

If the surfaces are of equal material, you essentially end up with twice the frictional force?

 

[haruspex]

If the surfaces are of equal material, you essentially end up with twice the frictional force?

No, that's not what I wrote. Consider e.g. when the angle is opened to 180 degrees. It is now one surface, so not doubled.
Do the algebra. What is the normal force on each?

 

[kftheuidfnaks]

No, that's not what I wrote. Consider e.g. when the angle is opened to 180 degrees. It is now one surface, so not doubled.
Do the algebra. What is the normal force on each?

Am I thinking of your proposal wrong if I split the block in two, one left and one right of the center when its opened to 180 degrees?

Normal Force = (m/2)g*cos(Theta) + (m/2)g*cos(Theta)
N = mg*cos(Theta)

 

[haruspex]

Am I thinking of your proposal wrong if I split the block in two, one left and one right of the center when its opened to 180 degrees?

Normal Force = (m/2)g*cos(Theta) + (m/2)g*cos(Theta)
N = mg*cos(Theta)

Unfortunately it is more complicated than that. The normal forces, by definition, act at right angles to the surfaces. So to find the contribution of each to supporting the weight you need to find the angles the surfaces make to the horizontal . This is a tricky bit of 3D geometry.

 

[kftheuidfnaks]

Well I'm probably over-complicating this but I am trying to visualize the problem so I drew this.

Since each dimension has a contribution to the total Normal force, their magnitudes and directions need to get combined, right?

 

[haruspex]

View attachment 253028
Well I'm probably over-complicating this but I am trying to visualize the problem so I drew this.

Since each dimension has a contribution to the total Normal force, their magnitudes and directions need to get combined, right?

I'm not sure what the set-up is in the above diagrams. None of them match the diagram in post #1. In particular, none seem to involve two surfaces. Is that intentional?
Or to put it another way, the two "angle is 90" diagrams do not agree.

 

[haruspex]

In case my point is not clear, look at the diagram in post #1. The two surfaces are not at 45 degrees to the horizontal; they are at a steeper angle. What angle?