Friction question

[Obnoxious]

I was going over the standard block-on-ramp-calculate-the-coefficient-of-friction problem. The coefficient turns out to be tan x, where x is the angle involved. I was just wondering, just what happens to friction when it approaches 90 degrees.

Certainly, when in real life when I have a block next to a wall, friction isn't infinite, as otherwise the block wouldn't fall. But then again, there must be some friction, as stick services (like velcro) wouldn't work either.

[J.C.MacSwell]

I was going over the standard block-on-ramp-calculate-the-coefficient-of-friction problem. The coefficient turns out to be tan x' date=' [/b'] where x is the angle involved. I was just wondering, just what happens to friction when it approaches 90 degrees.

Certainly, when in real life when I have a block next to a wall, friction isn't infinite, as otherwise the block wouldn't fall. But then again, there must be some friction, as stick services (like velcro) wouldn't work either.

 

Why? (the bold part)

 

The block would fall regardless of the coefficient since the normal force approaches zero as the angle approaches 90 degrees. (If it didn't "roll" off first)

[Obnoxious]

But can we just absolutely ignore friction like that?

[DQW]

If you state the problem explicitly, it'll make more sense.

 

The problem states (and this is my guess based on the OP) that the block does not slide down the incline and you are asked to find the minimum coefficient of friction that ensures this.

 

As the incline gets steeper, the component of the weight down the incline increases (with the sine of the angle) and the normal reaction gets smaller and smaller (with the cosine). It will naturally require more stickiness (or a greater coefficient of friction) to keep the block from sliding down (since the frictional force is the product of the normal reaction and the coefficient).

 

When the incline is nearly vertical, the weinght of the block is essentially mg and the normal reaction is tiny. The coefficient of friction must be truly enormous to keep the block from sliding down.

 

When the incline is exactly vertical, the normal reaction is exactly zero. The force down the plane, however, is exactly mg. The frictional force upwards needs to balance this downward force for the block to remain stationary. In other words, you want to find a number, which when multiplied by zero, gives mg. Or :

 

[math]0 \cdot \mu_s = mg [/math]

 

But you know that, by definition

 

[math]0 \cdot x = 0 [/math] for any finite number, x.

 

Thus, [imath]\mu_s[/imath] can not be a finite number number ("and hence, must be infinite").