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:(adsbygoogle = window.adsbygoogle || []).push({});

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 [itex]40^\circ[/itex], what are the minimum and maximum values for the acceleration that Lou must achieve? Use [itex]m[/itex] for the child's mass, and [itex]\mu_s[/itex] for the coefficient of static friction between the child and the wedge.

The accelerating reference frame attached to the child of mass [itex]m[/itex] will be used to describe the motion of mass [itex]m[/itex]. (Though note that in this reference frame, the child of mass [itex]m[/itex] 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 [itex]mg[/itex] of the mass [itex]m[/itex] will then have an x-component directed down the incline, and a y-component directed toward the wedge, as follows:

[tex]\sin 40^\circ = \frac{opposite}{hypotenuse} = \frac{w_x}{mg}[/tex]

[tex]\Rightarrow w_x = mg \sin 40^\circ[/tex]

[tex]\cos 40^\circ = \frac{adjacent}{hypotenuse} = \frac{w_y}{mg}[/tex]

[tex]\Rightarrow w_y = mg \cos 40^\circ[/tex]

There will also be a normal force, [itex]F_n[/itex], exerted by the wedge against the child, directed away from the wedge.

Since the wedge and the child of mass [itex]m[/itex] are at rest relative to each other in the accelerating reference frame of the child of mass [itex]m[/itex], we must introduce a fictitious force [itex]\Vec{F}[/itex] applied to the mass [itex]m[/itex] 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 [itex]\Vec{F}[/itex] will have an x-component directed up the incline, and a y-component directed toward the wedge, as follows:

[tex]\cos 40^\circ = \frac{adjacent}{hypotenuse} = \frac{F_x}{F}[/tex]

[tex]\Rightarrow F_x = F \cos 40^\circ[/tex]

[tex]\sin 40^\circ = \frac{opposite}{hypotenuse} = \frac{F_y}{F}[/tex]

[tex]\Rightarrow F_y = F \sin 40^\circ[/tex]

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

[tex]F_{ym,total} = ma_{ym} = m(0) = 0 = F_n - w_y - F_y[/tex]

[tex]\Rightarrow F_n = w_y + F_y[/tex]

Substituting for [itex]w_y[/itex] and [itex]F_y[/itex] yields:

[tex]F_n = mg \cos 40^\circ + F \sin 40^\circ[/tex]

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

Therefore:

[tex]F_{xm,total} = ma_{xm} = m(0) = 0 = -F_x + w_x - f_s[/tex]

[tex]\Rightarrow f_s = w_x - F_x[/tex]

Substituting the values of [itex]w_x[/itex] and [itex]F_x[/itex] yields:

[tex]f_s = mg \sin 40^\circ - F \cos 40^\circ[/tex]

In general, [itex]f_s \leq \mu_s F_n[/itex].

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

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

Continuing right along, using [itex]f_s = \mu_s F_n[/itex] ...

[tex]f_s = mg \sin 40^\circ - F \cos 40^\circ = \mu_s F_n = \mu_s(mg \cos 40^\circ + F \sin 40^\circ)[/tex]

[tex]\Rightarrow F \cos 40^\circ + \mu_s F \sin 40^\circ = mg \sin 40^\circ - \mu_s mg \cos 40^\circ[/tex]

[tex]\Rightarrow F = \frac{mg \sin 40^\circ - \mu_s mg \cos 40^\circ}{\cos 40^\circ + \mu_s \sin 40^\circ}[/tex]

[tex]\Rightarrow F = m \frac{g \sin 40^\circ - \mu_s g \cos 40^\circ}{\cos 40^\circ + \mu_s \sin 40^\circ}[/tex]

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 [itex]\Vec{F}[/itex] in the accelerating reference frame of the child is now theactualforce being applied by Lou to the wedge.

Denoting the mass of the wedge as [itex]M[/itex] yields:

[tex]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}[/tex]

[tex]\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}[/tex]

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

[tex]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}[/tex]

[tex]\Rightarrow a_{wedge,min} = \frac{g \sin 40^\circ - \mu_s g \cos 40^\circ}{\cos 40^\circ + \mu_s \sin 40^\circ}[/tex]

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 ...)

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