Monkey want bananas!

Relevant wiki: Newton's Third Law

Let the masses of the bananas and monkey be $M$ and $m$ respectively. Then the weights of the bananas and monkey are $W = Mg$ and $w = mg$ respectively.

Let the acceleration the monkey climbing up the rope be $a$ . Then using equation of motion ,

$\begin{aligned} s & = ut + \frac 12 at^2 \\ 16 & = (0)(2) + \frac 12 a(2^2) \\ \implies a & = 8 \text{ ft/s}^2 \end{aligned}$

Since the bananas are stationary, by Newton's third law:

$\begin{aligned} m(g+a)&=Mg \\ (32+8)m & = 32M \\ m &= \frac 45 M \\ mg &= \frac 45 Mg \\ \implies w &= \boxed{\dfrac {4W}5} \end{aligned}$

Technically, the problem is missing the information that the monkey covers 16 ft in 2 sec from a stationary state. Changing from "climbs" to "starts climbing" should hopefully fix that.

Suppose the weight of the monkey is $W'$ and acceleration of the monkey is $f$

So,

$S = ut+\dfrac 12 ft^2$

$16 = 0 \times 2 + \dfrac 12 f(2)^2 \implies f = 8 ~ \text{ft/s}^2$

If the tension is $T,$ then the equation of motion of the monkey will be:

$T - W' = \dfrac{W'}{g} \times f \implies T - W'= \dfrac{W'}{32} \times 8$

$\Rightarrow T = W' + \dfrac{W'}{4} = \dfrac{5W'}{4}$

So $T = \dfrac{5W'}{4} .......... \color{#3D99F6}(1)$

Since bananas always remain at rest,

$T - W = 0 \implies T = W .......... \color{#20A900} (2)$

From $\color{#3D99F6} (1)$ and $\color{#20A900} (2),$ we get

$\dfrac{5W'}{4} = W \implies W' = \boxed{\dfrac{4W}{5}}$

So I assumed that the monkey starts from rest, someone please inform me if there is a solution that does not assume this. I don't think there is but anyways, I digress.

We start by denoting a few things.

$M_m$ is the mass of the monkey.

$F_c$ is the force exerted on the string by the monkey's pulling.

$a_m$ is the acceleration of the monkey.

$T$ is the tension in the string.

$a_b$ is the acceleration of the bananas.

First let's find $a_m$ :

$s = ut + \frac{1}{2}at^2$

$2s = a_mt^2$

$32 = 4a_m$

$a_m = 8 ft$ $s^{-2}$

Do Newton's Second Law with respect to the bananas:

$T - W = 0$ since the bananas remain at rest.

$T = W$

Now do Newton's Second Law with respect to the monkey:

$T - F_c - M_mg = M_ma_b$ where $F_c = M_ma_m$

$W - M_ma_m - M_mg = 0$ since $a_b = 0$ and $T = W$

$W = M_m(32 + 8)$

$M_m = \frac{W}{40}$

Since $Weight_{monkey} = M_mg$

$Weight_{monkey} = \frac{W}{40} * 32 = \boxed{\frac{4W}{5}}$

Since the bunch of bananas remain at rest, the weight of the climbing monkey has to be less. Therefore force at the monkey end will be its mass Mb (4/4)g plus the force due to its acceleration up Mm (1/4) g, which amounts 5/4 Mm g in total . At the banana side the force values Mb g.

Balancing both Mmg=4/5Mb g or in term of weights Wm= 4/5 Wb

I bet a $\frac{4 W}{3}$ solution option would trick some people :)

Oops... didn't see it's already been answered...sorry.

F(net)=0

So acc=0

M(G+A)=WG M=(G+A)/G M=4W/5