A mass $M$ is drawn up a straight incline of given height $h$ by a mass $m$ which is attached to the first mass by a string passing from it over a pulley at the top of the incline (Fig 10-10) and which hangs vertically. Find the angle of the incline in order that the time of ascent be minimum.
I know that the net force acting on $M$ is $32m - 32MsinA$, (assuming that the pull of gravity is 32 ft/s) but I'm not sure how to use this information to minimize the time of ascent.
I've tried applying the work formula as follows:
$$W = Fd = (32m-32MsinA) \frac{h}{sinA}$$ $$W = \frac{32mh}{sinA} - 32Mh$$
I then decided to take the derivative with respect to $h$ and set it to zero. By minimizing $h$, the hypotenuse is minimized which means that the time of ascent will be less since less distance has to be covered for the mass $M$.
The derivative of the above gave me the following:
$$ \frac{32m}{sinA} - 32M = 0$$ $$sin A = \frac{m}{M}$$
However, my book said that the answer is $sin A = \frac{m}{2M}$. I'm not really sure why.
Any help would be appreciated. Thanks in advance.
A little research reveals that the book gives the statement, "The net force acting on $M$ is $32m − 32M\sin A$," as a "suggestion" for solution of the problem.
I think this is actually a mistake by the author.
Gravity pulls the mass $M$ down the incline with a force $Mg \sin A.$ (Dr. Kline would like you to assume $g = 32.$ I prefer to write $g$ for the acceleration due to gravity until we absolutely need a numerical result.) In order for $mg − Mg\sin A$ to be the force on $M$ alone, the string would have to be pulling on $M$ with the force $mg.$
Why would it do that? Apparently we are supposed to assume it is because the mass $m$ pulls down with a force $mg.$ But if $m$ actually pulls the string down with that much force, the string pulls $m$ up with the same force (this is an equal and opposite reaction, one of Newton's three laws), and since gravity only pulls $m$ down with the force $mg,$ the net force on $m$ would be zero. So if we set up the experiment by hanging the two masses by passing their connecting string over the pulley as shown, and then let them go, nothing would move. The mass $M$ would never reach the top of the incline.
What you managed to do by setting your derivative equal to zero was to solve for the case where $mg − Mg\sin A = 0.$ And you get the correct answer to a different problem, which is how steep the incline should be so that there is no tendency for $M$ to slide either up or down the incline.
(To reason like this, by the way, we have to make a bunch of assumptions, such as that friction, air resistance, the mass of the string, and the mass of the pulley are all negligible, and the masses are initially at rest, but the given answer makes no sense unless we make all those assumptions, so let's make them.)
A physical intuition one might apply is that the force $mg − Mg\sin A$ is the force accelerating the object of mass $M + m$ (the two masses joined by the string) along its path up the incline and straight down the vertical face. If $s$ is the distance the combined mass $M + m$ has traveled along this path, then the acceleration of the combined mass is $$ \frac{d^2s}{dt^2} = \frac{32m − 32M\sin A}{M + m},$$ and you can continue as in the other answer.
Another approach is to look at the vertical height $y$ gained by the mass $M.$ We have $y=0$ at the bottom of the incline and $y=h$ at the top. When the mass $M$ moves a distance $\Delta y$ vertically upward, it moves a linear distance $\frac{1}{\sin A} \Delta y$ along the slope, and the mass $m$ moves $\frac{1}{\sin A} \Delta y$ straight down. Converting this to rates of change, the linear velocities of $M$ and $m$ are both $\frac{1}{\sin A} \frac{dy}{dt},$ and therefore the total kinetic energy is $$ KE = \frac12 M \left(\frac{1}{\sin A} \frac{dy}{dt} \right)^2 + \frac12 m \left(\frac{1}{\sin A} \frac{dy}{dt} \right)^2 = \frac{M+m}{2\sin^2 A} \left(\frac{dy}{dt} \right)^2.$$ The rate of change of kinetic energy is $$\frac{d}{dt} KE = \frac{d}{dt} \left(\frac{M+m}{2\sin^2 A}\left(\frac{dy}{dt}\right)^2\right) = \frac{M+m}{\sin^2 A} \frac{dy}{dt} \frac{d^2y}{dt^2}. $$
On the other hand, potential energy is $$PE = Mgy - \frac{mgy}{\sin A} = \left(M - \frac{m}{\sin A}\right)gy,$$ and its rate of change is $$\frac{d}{dt} PE = \frac{d}{dt} \left(\left(M - \frac{m}{\sin A}\right)gy\right) = \left(M - \frac{m}{\sin A}\right)g \frac{dy}{dt}.$$
Now since any increase in kinetic energy comes from an equal decrease in potential energy, we know that $\frac{d}{dt} KE = -\frac{d}{dt} PE,$ that is, $$ \frac{M+m}{\sin^2 A} \frac{dy}{dt} \frac{d^2y}{dt^2} = -\left(M - \frac{m}{\sin A}\right)g \frac{dy}{dt}.$$
Simplifying and reearranging this to get acceleration in terms of everything else, $$ \frac{d^2y}{dt^2} = \frac{g}{M+m} (m\sin A - M\sin^2 A). \tag1$$
This is constant with respect to the height of $M,$ that is, $M$ starts at $y=0$ and accelerates upward with a constant vertical acceleration $\frac{d^2y}{dt^2}$ until $y = h.$ The fastest way to get from $y=0$ to $y=h$ is to maximize that acceleration.
To maximize this by varying $A,$ for $A$ between $0$ and a right angle, we find a maximum when $A=0,$ when $A$ is a right angle, or when $$ 0 = \frac{d}{dA} \left(\frac{g}{M+m} (m\sin A - M\sin^2 A)\right) = \frac{g}{M+m}\left(m\frac{d\sin A}{dA} - 2M\sin A \frac{d\sin A}{dA}\right) = \frac{g}{M+m}\frac{d\sin A}{dA}\left(m - 2M\sin A \right), \tag2 $$ provided that the derivative exists everywhere in that interval (which it does).
If $A=0$ then Equation $(1)$ tells us that $$\frac{d^2y}{dt^2} = 0.\tag3$$ If $A$ is a right angle then $\sin A = 1$ and Equation $(1)$ says that $$\frac{d^2y}{dt^2} = \frac{g}{M+m} (m - M).\tag4$$ But $A$ is neither zero nor a right angle and Equation $(2)$ is true, then $\frac{d\sin A}{dA} \neq 0$ and therefore the only way the right-hand side of Equation $(2)$ can be zero is if $m - 2M\sin A = 0.$ In that case, Equation $(1)$ says that $$\frac{d^2y}{dt^2} = \frac{g}{M+m} \left(m \frac{m}{2M} - M\left(\frac{m}{2M}\right)^2\right) = \frac{g}{M+m}\left(\frac{m^2}{4M}\right).\tag5$$
Unlike the two possibilities at the ends of the interval (Equations $(3)$ and $(4)$), Equation $(5)$ has the nice property that it is always positive (since we assume $M,$ $m,$ and $g$ are all positive). But if $m > 2M$ then there is no solution for $m - 2M\sin A = 0,$ Equation $(5)$ does not apply, and instead the minimum time occurs when $A$ is a right angle, that is, the "incline" should go straight up. That detail seems to have been neglected in the textbook.
Assuming $m \leq 2M,$ the last thing to check is whether there is any possibility that $m - M$ is greater than $\frac{m^2}{4M},$ because then the time would be minimized when $A$ is a right angle rather than when $m - 2M\sin A = 0.$ This check is another detail that I suspect may not have been considered in the writing of the textbook, but it turns out to be OK, because $$ \frac{m^2}{4M} - (m - M) = \frac{(m-2M)^2}{4M},$$ which is never negative, so we can be sure that $\frac{m^2}{4M} \geq m - M.$ Therefore, provided that it is possible to solve $m - 2M\sin A = 0,$ that solution gives the maximum vertical accelaration and the minimum time.
Taking all of this into consideration, you should not feel badly about having difficulty with this problem. After all, it is a much more complex problem than one might guess at first glance.