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## Homework Statement

This is problem 2.17 from Kleppner and Kolenkow, 2nd edition. It is the first problem involving friction, and I want to check my reasoning.

A block rests on a wedge on a horizontal surface. The coefficient of friction of the block on the wedge is ##\mu##. Gravity is directed down. The wedge angle ##\theta## obeys ##\tan \theta < \mu##. The wedge is accelerated horizontally at rate ##a##. Find the maximum and minimum values of ##a## for the block to stay fixed on the plane. [I assume this means "fixed on the wedge".]

## Homework Equations

##F = ma##, friction force ##f## is at most ##\mu N## in magnitude, and if less than this, the block does not slide.

## The Attempt at a Solution

No diagram was provided with this problem, so I drew my own. Please see the attached thumbnail. I chose coordinates such that ##y## points parallel to ##N##, normal to the wedge's hypotenuse, and ##x## points down the slope of the wedge. Also, ##x## and ##y## are inertial, so the origin is at some fixed point on the ground, and does not move with the wedge or block.

We let ##f## denote the friction force, which points up the slope. Then we have the following equations:

$$\begin{align}

mg \sin \theta - f &= m \ddot x \\

N - mg \cos \theta &= m \ddot y\\

\end{align}$$

We are given that the wedge is being accelerated by a rate ##a## horizontally. I oriented this such that positive ##a## points to the right, and negative ##a## points to the left. We presume that the block stays on the wedge (more on that in a moment), so its acceleration in the ##y## direction matches that of the wedge. This means that ##\ddot y = a \sin \theta##. If we also presume that the block does not move up or down the slope, then its acceleration in the ##x## direction must also match that of the wedge: ##\ddot x = a \cos \theta##. Thus the equations become

$$\begin{align}

mg \sin \theta - f &= m a \cos\theta\\

N - mg \cos \theta &= m a \sin\theta\\

\end{align}$$

I'll continue this in a reply to keep the message size under control.

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