# Particle Resting On An Accelerating Inclined Plane Problem

1. Sep 26, 2004

### NoPhysicsGenius

I am having difficulties finding an acceptable rationale for my solution to Problem 41 of Chapter 5 from Physics for Scientists and Engineers, 4th edition, by Paul A. Tipler:

Lou has set up a kiddie ride at the Winter Ice Fair. He builds a right-angle triangular wedge, which he intends to push along the ice with a child sitting on the hypotenuse. If he pushes too hard, the kid will slide up and over the top, and Lou could be looking at a lawsuit. If he doesn't push hard enough, the kid will slide down the wedge, and the parents will want their money back. If the angle of inclination of the wedge is $40^\circ$, what are the minimum and maximum values for the acceleration that Lou must achieve? Use $m$ for the child's mass, and $\mu_s$ for the coefficient of static friction between the child and the wedge.

The accelerating reference frame attached to the child of mass $m$ will be used to describe the motion of mass $m$. (Though note that in this reference frame, the child of mass $m$ appears to be at rest.) We take the x-axis as being along the incline of the wedge (with the positive direction taken as down the incline), and the y-axis to be perpendicular to the incline of the wedge (with the positive direction taken as away from the wedge).

The relevant forces are the following ...

The weight $mg$ of the mass $m$ will then have an x-component directed down the incline, and a y-component directed toward the wedge, as follows:

$$\sin 40^\circ = \frac{opposite}{hypotenuse} = \frac{w_x}{mg}$$
$$\Rightarrow w_x = mg \sin 40^\circ$$
$$\cos 40^\circ = \frac{adjacent}{hypotenuse} = \frac{w_y}{mg}$$
$$\Rightarrow w_y = mg \cos 40^\circ$$

There will also be a normal force, $F_n$, exerted by the wedge against the child, directed away from the wedge.

Since the wedge and the child of mass $m$ are at rest relative to each other in the accelerating reference frame of the child of mass $m$, we must introduce a fictitious force $\Vec{F}$ applied to the mass $m$ which possesses the same magnitude as the force applied to the wedge by Lou, but is opposite in direction. This means that the fictitious force $\Vec{F}$ will have an x-component directed up the incline, and a y-component directed toward the wedge, as follows:

$$\cos 40^\circ = \frac{adjacent}{hypotenuse} = \frac{F_x}{F}$$
$$\Rightarrow F_x = F \cos 40^\circ$$
$$\sin 40^\circ = \frac{opposite}{hypotenuse} = \frac{F_y}{F}$$
$$\Rightarrow F_y = F \sin 40^\circ$$

The equation of motion along the y-axis is then:

$$F_{ym,total} = ma_{ym} = m(0) = 0 = F_n - w_y - F_y$$
$$\Rightarrow F_n = w_y + F_y$$

Substituting for $w_y$ and $F_y$ yields:

$$F_n = mg \cos 40^\circ + F \sin 40^\circ$$

In calculating $a_{wedge,min}$, there will also be a force of static friction, $f_s$, directed up the incline (opposite the tendency of the child to slide down the incline).

Therefore:

$$F_{xm,total} = ma_{xm} = m(0) = 0 = -F_x + w_x - f_s$$
$$\Rightarrow f_s = w_x - F_x$$

Substituting the values of $w_x$ and $F_x$ yields:

$$f_s = mg \sin 40^\circ - F \cos 40^\circ$$

In general, $f_s \leq \mu_s F_n$.

But here's my problem ... I can only get the answer that the book gives if I assume that $f_s = \mu_s F_n$. So:

Question #1: What is the rationale for $f_s$ possessing its maximum possible value in order to yield the minimum possible value for the wedge's acceleration, $a_{wedge,min}$?

Continuing right along, using $f_s = \mu_s F_n$ ...

$$f_s = mg \sin 40^\circ - F \cos 40^\circ = \mu_s F_n = \mu_s(mg \cos 40^\circ + F \sin 40^\circ)$$
$$\Rightarrow F \cos 40^\circ + \mu_s F \sin 40^\circ = mg \sin 40^\circ - \mu_s mg \cos 40^\circ$$
$$\Rightarrow F = \frac{mg \sin 40^\circ - \mu_s mg \cos 40^\circ}{\cos 40^\circ + \mu_s \sin 40^\circ}$$
$$\Rightarrow F = m \frac{g \sin 40^\circ - \mu_s g \cos 40^\circ}{\cos 40^\circ + \mu_s \sin 40^\circ}$$

First of all, we assume that there is no friction between the wedge and the ice that the wedge is being pushed along. For the wedge, we will assume that the x-axis is parallel to the motion (and the acceleration) of the wedge, and the y-axis is perpendicular to the motion of the wedge.

The fictitious force $\Vec{F}$ in the accelerating reference frame of the child is now the actual force being applied by Lou to the wedge.

Denoting the mass of the wedge as $M$ yields:

$$F = (m + M)a_{wedge,min} = m \frac{g \sin 40^\circ - \mu_s g \cos 40^\circ}{\cos 40^\circ + \mu_s \sin 40^\circ}$$
$$\Rightarrow a_{wedge,min} = \frac{m}{m + M}\frac{g \sin 40^\circ - \mu_s g \cos 40^\circ}{\cos 40^\circ + \mu_s \sin 40^\circ}$$

However, this is not the answer given in the back of the book. To get the book's answer, I must then assume that $M = 0$, which yields the book's answer of:

$$a_{wedge,min} = \frac{m}{m + 0}\frac{g \sin 40^\circ - \mu_s g \cos 40^\circ}{\cos 40^\circ + \mu_s \sin 40^\circ}$$
$$\Rightarrow a_{wedge,min} = \frac{g \sin 40^\circ - \mu_s g \cos 40^\circ}{\cos 40^\circ + \mu_s \sin 40^\circ}$$

Question #2: What is the rationale for neglecting the mass of the wedge?

To be continued in the next post ... (When I tried putting all of the text into a single post, I kept receiving a "Cannot Find Server" error, presumably because the post was too lengthy. :grumpy: Oh well ...)

2. Sep 26, 2004

### NoPhysicsGenius

... Continued from my previous post ...

In calculating $a_{wedge,max}$, there will be a force of static friction, $f_s$, directed down the incline (opposite the tendency of the child to slide up the incline).

Therefore:

$$F_{xm,total} = ma_{xm} = m(0) = 0 = -F_x + w_x + f_s$$
$$\Rightarrow f_s = F_x - w_x$$

Substituting the values of $w_x$ and $F_x$ yields:

$$f_s = F \cos 40^\circ - mg \sin 40^\circ$$

In general, $f_s \leq \mu_s F_n$.

Here, it is quite logical that the maximum value of the wedge's acceleration for which the child does not slide corresponds to $f_s$ possessing its maximum value. Therefore, $f_s = f_{s,max} = \mu_s F_n$:

$$f_s = F \cos 40^\circ - mg \sin 40^\circ = \mu_s F_n = \mu_s(mg \cos 40^\circ + F \sin 40^\circ)$$
$$\Rightarrow F \cos 40^\circ - \mu_s F \sin 40^\circ = mg \sin 40^\circ + \mu_s mg \cos 40^\circ$$
$$\Rightarrow F = \frac{mg \sin 40^\circ + \mu_s mg \cos 40^\circ}{\cos 40^\circ - \mu_s \sin 40^\circ}$$
$$\Rightarrow F = m \frac{g \sin 40^\circ + \mu_s g \cos 40^\circ}{\cos 40^\circ - \mu_s \sin 40^\circ}$$

$$F = (m + M)a_{wedge,max} = m \frac{g \sin 40^\circ + \mu_s g \cos 40^\circ}{\cos 40^\circ - \mu_s \sin 40^\circ}$$
$$\Rightarrow a_{wedge,max} = \frac{m}{m + M}\frac{g \sin 40^\circ + \mu_s g \cos 40^\circ}{\cos 40^\circ - \mu_s \sin 40^\circ}$$

However, this is not the answer given in the back of the book. To get the book's answer, I must once again assume that $M = 0$, which yields the book's answer of:

$$a_{wedge,max} = \frac{m}{m + 0}\frac{g \sin 40^\circ + \mu_s g \cos 40^\circ}{\cos 40^\circ - \mu_s \sin 40^\circ}$$
$$\Rightarrow a_{wedge,max} = \frac{g \sin 40^\circ + \mu_s g \cos 40^\circ}{\cos 40^\circ - \mu_s \sin 40^\circ}$$

I must say that I am not at all confident that my approach to solving this problem is valid. Can anyone offer any help?

3. Sep 27, 2004

### maverick280857

My God! This is a damn big post. It'll take me quite a while to finish reading it. But I believe you have "some" trouble analyzing a mass sliding down on an inclined plane which is free to move as well.

The analysis can be done either using "pseudo forces" (in a noninertial frame of reference) or by using simple, faithful, effective real forces ;-). I believe you have finished the problem(?) and have some doubts regarding the solution or the analysis.

Just read a bit of your post. In your case, the mass (boy) is sliding up the incline so the friction force acts opposite to his motion, i.e. down the incline. The problem involves two cases:

1. Max $$a_{wedge}$$ for which the boy slides upwards.
2. Min $$a_{wedge}$$ for which the boy slides downwards.

Supposing the force on the wedge is resolved along and perpendicular the wedge, you can do the analysis simply with two equations.

For Question 1 you must understand that corresponding to each of the abovementioned possibilites (governed really by the man pushing the wedge or more generally, the agency causing the wedge to accelerate in the first place) there is a tendency of the boy (or the mass) to slip. This tendency is different for each case. It is possible (and this analysis at present is indeed intuitive) that the force on the wedge is large enough to PREVENT the boy from sliding down--and perhaps large enough to cause him to rise up (how? well, thats not something that you can accept unless you draw the fbd). On the other hand, it could be small enough to fail to prevent him from sliding down.

The external force in either case has an associated direction of frictional force (which is automatically adjusted) which governs the boy's motion.

At this point in time, the only reason I see for the mass of the wedge to be made zero is that it could be large enough compared to the boy's mass. I don't see why the wedge's mass is zero though. This appears to be more of a mathematical approximation (perhaps the authors solved it taking the wedge's mass to be zero first off). Interestingly, if the wedge were static (of course then this problem would be drastically different) then the question of its mass wouldn't arise at all :-).

However, your analysis is quite correct because this is how you would solve more complex problems involving wedges and masses.

Cheers
Vivek