Kleppner - Block on wedge with friction

[jbunniii]

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.

 

[jbunniii]

OK, more on that constraint that the block stays on the wedge. This means we must have $N > 0$. I will also assume that $0 < \theta < \pi/2$ to avoid a degenerate wedge. Now $N > 0$ means
$$\begin{align}
mg \cos \theta + ma \sin \theta & > 0 \\
a \sin \theta &> -g \cos \theta \\
a &> -\frac{g}{\tan \theta}\\
\end{align}$$
We also require $$|f| < \mu N$$ so that the block doesn't slide up or down the wedge. This gives us
$$-\mu (g \cos \theta + a \sin \theta) < a \cos \theta - g \sin \theta < \mu (g \cos \theta + a \sin \theta)$$
The first inequality can be rearranged as
$$a (\cos \theta + \mu \sin \theta) > g (\sin \theta - \mu \cos \theta)$$
Since the coefficient of $a$ is positive, I can divide by it without flipping the inequality to get
$$a > \frac{g (\sin \theta - \mu \cos \theta)}{\cos \theta + \mu \sin \theta}$$
So now we have two candidates for the minimum acceleration: the one we just obtained, and the earlier one derived from $N > 0$, namely $a > -g/\tan \theta$. A bit of algebra shows that the larger (less negative) of these two bounds is $\frac{g (\sin \theta - \mu \cos \theta)}{\cos \theta + \mu \sin \theta}$, so this is our minimum acceleration.

Note that this is a negative number because the numerator is negative, due to the constraint $\tan \theta < \mu$. This makes sense: we expect the minimum acceleration to be negative and the maximum to be positive, because we know if the acceleration is zero, the block does not slide, again because $\tan \theta < \mu$.

In the next and final post I will find the maximum acceleration.

 

[jbunniii]

To find the maximum acceleration, we use the second inequality in the chain in the previous post:
$$a \cos \theta - g \sin \theta < \mu (g \cos \theta + a \sin \theta)$$
This can be rearranged to
$$a(\cos \theta - \mu \sin \theta) < g (\sin \theta + \mu \cos \theta)$$
Notice that the coefficient of $a$ can be either positive or negative. The condition $\cos \theta - \mu \sin \theta > 0$ is equivalent to $\tan \theta < 1/\mu$. This is automatically satisfied if $\mu < 1$, because then $\mu < 1/\mu$ and we are given that $\tan \theta < \mu$. In this case we will have
$$a < \frac{g (\sin \theta + \mu \cos \theta)}{\cos \theta - \mu \sin \theta}$$
so the right hand side is the maximum acceleration.

On the other hand, if $\mu > 1$ then we may have $1/\mu < \tan \theta < \mu$ (for example, if $\theta = \pi / 4$). If this is the case, then $\cos \theta - \mu \sin \theta < 0$. But then the inequality
$$a(\cos \theta - \mu \sin \theta) < g (\sin \theta + \mu \cos \theta)$$
is satisfied for all positive $a$ since the right hand side is positive. Thus if $1/\mu < \tan \theta < \mu$ then there is no maximum acceleration. Please check my reasoning, as I feel I'm on shaky ground here.

 

[AlephZero]

I would suggest this is easier if you use a non-inertial coordinate system fixed to the wedge.

Then you have fictitious force $-ma$ acting horizontally on the block. If the block does not move, the reaction force between the block and the wedge must lie within the "cone of friction" with angle $\tan^{-1}\mu$ from the normal to the wedge, and it must be a compressive force between the block and the wedge. The third force on the block is its weight.

So you can draw the triangle of forces for the block, and get the result by geometry...

From the triangle of forces, it is clear there are two different situations, depending on whether $\tan^{-1}\mu$ is greater or less than the angle of the wedge. In one case, the block does not slide for zero acceleration and the extreme accelerations are positive or negative. In the other case, the block does slide for zero acceleration, so the max and min accelerations both have the same sign.

 

[jbunniii]

Edited a bit to fix some typos and to fix the directions of the coordinates in the figure.

I would suggest this is easier if you use a non-inertial coordinate system fixed to the wedge.

Non-inertial coordinates and fictitious forces were mentioned slightly in chapter 2, but won't be covered until a later chapter. (For that matter, friction wasn't discussed until chapter 3, so this problem seems out of place. In the previous edition of the book, chapters 2 and 3 were combined, so it's probably just an editing error.) I'll think about your suggestion and see if I can do it that way as well.

 

[jbunniii]

From the triangle of forces, it is clear there are two different situations, depending on whether $\tan^{-1}\mu$ is greater or less than the angle of the wedge. In one case, the block does not slide for zero acceleration and the extreme accelerations are positive or negative. In the other case, the block does slide for zero acceleration, so the max and min accelerations both have the same sign.

Yes. The problem gives a constraint which rules out the second case: $\tan \theta < \mu$.

But even the first case has subcases, if the reasoning in my third post is right: subcase 1 is $\tan \theta < 1/\mu$ and subcase 2 is $\mu > 1$ and $1/\mu < \tan \theta < \mu$. In the latter subcase (e.g. if $\mu > 1$ and $\theta = \pi /4$), if my work is correct, there is no maximum acceleration, which seems kind of freaky to me.

The $\mu > 1$ case might seem bogus, but according to K&K, some surfaces/materials can have $\mu > 1$. They didn't mention which ones, though.

 

[AlephZero]

OK, since the block doesn't move relative to the wedge, using a non-inertial coordinate system is not very relevant to the question. But I think drawing a triangle of forces is useful. Maybe engineers like drawing pictures instead of doing algebra.

When I started thinking about this, I mis-read the question as meaning "the block can slide on the wedge but it doesn't lose contact with it". In that case, using a non-inertial coord system IS simpler that getting tied in knots about what is the actual acceleration of the block when it may or may not be sliding in one direction or the other.

 

[jbunniii]

When I started thinking about this, I mis-read the question as meaning "the block can slide on the wedge but it doesn't lose contact with it". In that case, using a non-inertial coord system IS simpler that getting tied in knots about what is the actual acceleration of the block when it may or may not be sliding in one direction or the other.

I believe that if it's allowed to slide and we just want it to stay in contact (until it slides off one end or the other anyway), that's equivalent to $N > 0$. In that case, we just get the constraint mentioned at the start of my second post:
$$a > -\frac{g}{\tan \theta}$$
So there is a minimum acceleration but, unsurprisingly, no maximum. The coefficient of friction is irrelevant here.

 

[AlephZero]

but according to K&K, some surfaces/materials can have $\mu > 1$. They didn't mention which ones, though.

"Soft" materials like rubber can have $\mu > 1$, in contact with some other materials. For metal on metal, typical values are about 0.1 to 0.3, though I have seen measured coefficients up to 0.6 for steel on steel at high temperatures (of the order of 1000 C).

 

[AlephZero]

So there is a minimum acceleration but, unsurprisingly, no maximum.

Yup, coming to that conclusiion (by thinking about the problem, not doing the math) was what made me re-read the question!

 

[AlephZero]

In that case, we just get the constraint mentioned at the start of my second post:
$$a > -\frac{g}{\tan \theta}$$
So there is a minimum acceleration but, unsurprisingly, no maximum.

Another way to interpret that is, the acceleration of the block downwards must be less than g, otherwise it can't fall fast enough to stay on the the wedge.

At the limiting condition when it just lifts off, there is no normal force, and therefore no friction force, so as you said the coefficient of friction is irrelevant.

 

[jbunniii]

Another way to interpret that is, the acceleration of the block downwards must be less than g, otherwise it can't fall fast enough to stay on the the wedge.

At the limiting condition when it just lifts off, there is no normal force, and therefore no friction force, so as you said the coefficient of friction is irrelevant.

Yes, that's a good interpretation. The net acceleration (vector sum of the accelerations due to gravity and the motion of the wedge) cannot point at a slope higher than that of the wedge if the block is to stay on the wedge.

 

[AlephZero]

The "draw a vector triangle" solution to this works out very nicely. Consider $F = ma$ for the block.

Take the position of the block as in your diagram.

For no slipping, the $ma$ vector is horizontal. The $F$ vector has two components, the weight $mg$ downwards, and the reaction force $R$ from the wedge ($R$ is the normal force + the friction force).

So drawing a picture of the three vectors representing $F = ma$, we get a right angled triangle with $R$ as the hypotenuse.

The angle between $R$ and the horizontal is $\tan^{-1}(g/a)$ (positive anticlockwise)

The normal to the wedge is at an angle $\pi/2 - \theta$ to the horizontal, and for no slipping, $R$ must be at an angle between $\pi/2 - \theta \pm \tan^{-1}\mu$.

So the extreme values of $a$ are given by $\tan^{-1}(g/a) = \pi/2 - \theta \pm \tan^{-1}\mu$.

 

[jbunniii]

The "draw a vector triangle" solution to this works out very nicely. Consider $F = ma$ for the block.

Take the position of the block as in your diagram.

For no slipping, the $ma$ vector is horizontal. The $F$ vector has two components, the weight $mg$ downwards, and the reaction force $R$ from the wedge ($R$ is the normal force + the friction force).

So drawing a picture of the three vectors representing $F = ma$, we get a right angled triangle with $R$ as the hypotenuse.

The angle between $R$ and the horizontal is $\tan^{-1}(g/a)$ (positive anticlockwise)

The normal to the wedge is at an angle $\pi/2 - \theta$ to the horizontal, and for no slipping, $R$ must be at an angle between $\pi/2 - \theta \pm \tan^{-1}\mu$.

So the extreme values of $a$ are given by $\tan^{-1}(g/a) = \pi/2 - \theta \pm \tan^{-1}\mu$.

OK, I drew a picture and I agree with all of the above, with the caveat that we define $\tan^{-1}(g/a)$ to assume values in the range $[0, \pi]$ instead of the standard $[-\pi/2,\pi/2]$. This is because $g/a < 0$ means $a < 0$ and $g > 0$, not vice versa.

Note that this caveat imposes the additional constraint that
$$0 \leq \tan^{-1}(g/a) \leq \pi$$
Thus your expressions for the extreme values are valid provided that
$$\begin{align}
\pi/2 - \theta + \tan^{-1}(\mu) &\leq \pi \text{ and} \\
\pi/2 - \theta - \tan^{-1}(\mu) &\geq 0\\
\end{align}$$
or equivalently,
$$-\pi/2 + \tan^{-1}(\mu) \leq \theta \leq \pi/2 - \tan^{-1}(\mu)$$
Since $$\tan^{-1}(\mu) + \tan^{-1}(1/\mu) = \pi/2$$ (draw a right triangle with sides $1$ and $\mu$), the above is equivalent to
$$-\tan^{-1}(1/\mu) \leq \theta \leq \tan^{-1}(1/\mu)$$
The left inequality is always true. The right inequality is the same as one of the two subcases I listed in post #3. For the other subcase, $\theta > \tan^{-1}(1/\mu)$, we have to replace your expression for the smaller extreme value by $0$, yielding
$$\tan^{-1}(g/a) \geq 0$$
which is always satisfied, in other words, there is no maximum acceleration when $\theta > \tan^{-1}(1/\mu)$.

 

[AlephZero]

I agree my sign conventions were a bit sloppy.

The nice thing about the diagram is that you can play with "what if" questions directly from diagram. In particular, what happens if friction and wedge angle are such that the reaction force $R$ can act vertically (the block does not slip with zero acceleration), and/or horizontally (the block can never slip up the wedge).