Monkey accelerating up and down a rope

[Iamconfused123]

So does that resolve your question in post #26?

It does, lol. Thank you for helping me.

 

[Iamconfused123]

But in c) the monkey wouldn't be accelerating at greater than $g$? Tension in the rope is still positive. If the monkey was still relative to the rope the acceleration of the table mass is $2 \rm{m/s^2} \rightarrow$, with it moving down the rope the acceleration of the table mass is $1.6 \rm{m/s^2} \rightarrow$, giving the monkey an acceleration of $3.6 \rm{m/s^2} \downarrow$ in the inertial frame.

The intutive answer is that the monkey accelerating up the rope with magnitude $a$ is equivalent to the monkey in a gravitational field of $g +a$, as far as the block $M$ is concerned. The acceleration of bock $M$ is, therefore,$$a_M = \frac m{m+M}(g + a)$$

It took me a while to figure this out, but I did eventually. Thank you for helping me.

 

[kuruman]

I can now post this because the question has been answered. I think that a lot of clarity can be achieved if one were to do this symbolically.

From the FBD of the mass we have
$T=MA$ where $A$ is the acceleration of the mass in the lab frame.
Note that $A>0$ and denotes the magnitude of the acceleration of the mass on the table.

From the FBD of the monkey we have ("up" is positive)
$T-mg = ma$ where is $a$ is the acceleration 1d vector of the monkey also in the lab frame.
Note that $a$ can be positive, negative or zero. Now we put the two equations together to get $$MA=mg+ma.$$ Let $a_0$ be the magnitude of the monkey's acceleration relative to the rope. We have two cases

Case 1. The monkey is accelerating up relative to the rope and the rope is accelerating down relative to the lab.
Then $a=-(A-a_0)$ because the relative sign must be $-$ and the acceleration of the rope is "down".

Case 2. The monkey is accelerating down relative to the rope and the rope is accelerating down relative to the lab. Then $a=-(A+a_0)$ because the relative sign must be $+$ and the acceleration of the rope is "down".
Then we have $$\begin{align} & MA=mg+ma=mg-m(A\mp a_0) \nonumber \\
& \implies A=\frac{m(g\pm a_0)}{M+m}. \nonumber\end{align}$$ The upper sign goes with monkey relative acceleration up and the lower sign with monkey relative acceleration down.

Can the monkey accelerate up relative to the rope so that it is at rest in the lab frame? Yes, that happens when $$A=a_0=\frac{m(g+a_0)}{M+m}\implies a_0=\frac{m}{M}g.$$In this special case, the tension in the rope is $$T=MA=Ma_0=mg.$$ This makes sense. Only two forces act on the monkey and they must cancel in order to have the monkey's center of mass at rest in the lab frame.

 

[haruspex]

Can the monkey accelerate up relative to the rope so that it is at rest in the lab frame?

@Iamconfused123, with a little nudging, solved that in post #26.