- #1

Mr Davis 97

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## Homework Statement

A block rests on a wedge inclined at angle ##\theta##. The coefficient of friction between the block and plane is ##\mu##. The wedge is given a horizontal acceleration ##a##. Assuming that ##\tan \theta > \mu##, find the minimum acceleration from the block to remain on the wedge without sliding

## Homework Equations

Newton's second law

Law of static friction

## The Attempt at a Solution

So essentially, I feel as though I have the correct solution, but I am not sure why it is the solution, physically.

So we have three forces acting on the block. the normal force, friction, and gravity.

If we define our coordinate system so that the y-direction is up and the x-direction is down, then we have the following equations:

y-direction: ##N \sin \theta + f \sin \theta - mg = 0##

x-direction: ##N \sin \theta - f \cos \theta = ma##

then

##N \cos \theta + f \sin \theta = mg##

Since ##\tan \theta > \mu##

we know that ##f = \mu N##

Then

##N (\cos \theta + \mu \sin \theta) = mg##

##N (\sin \theta - \mu \cos \theta) = ma##

##\displaystyle a = \frac{g(\sin \theta - \mu \cos \theta)}{\cos \theta + \mu \sin \theta}##

I think that this is the expression for the minimum, but I am not sure why this is the minimum. What is the condition that makes this expression the minimum, rather than just some acceleration?

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