A 10 kgmonkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kgpackage on the ground (Figure). (a) What is the magnitude of the least acceleration the monkey must have if it is to lift the package off the ground? If, after the package has been lifted, the monkey stops its climb and holds onto the rope, (b) What is the magnitude and (c) What is the direction of the monkey’s acceleration and (d) What is the tension

in the rope?

1. Mass of monkey is, $M_m=10\mathrm{kg}$
2. Mass of Package,$M_p=15\mathrm{kg}$

Newton’s second law states that the force acting on the object is directly proportional to the acceleration of the object. The net force is equal to the product of mass and acceleration. The resultant force on the object is equal to the vector sum of all the forces acting on the object.

Apply Newton’s second law to the monkey and box to solve for its acceleration.

Formula:

F = ma

Here, F is the net force, is the mass of the body, and is acceleration.

From the diagram, the force equation for the monkey can be written as,

$T-M_{m}g=M_m{a}_m$ (i)

In a similar way, the force equation for the package is,

$T+F_N-M_{P}g=M_p{a}_p$ (ii)

For the least acceleration, we $F_N=0$and $a_p=0$.

So, equation (ii) will become, $T=M_{p}g$

Substituting this, in eq. (1), and solving for the acceleration of the monkey, we get

$$\begin{gathered} a_m=\frac{\left(M_p-M_m\right)g}{M_m} \\ =\frac{\left(15\mathrm{kg}-10\mathrm{kg}\right)9.8\mathrm{m}/{\mathrm{s}}^2}{10\mathrm{kg}} \\ =4.9\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration of the monkey is $4.9\mathrm{m}/{\mathrm{s}}^2$.

This case will be similar to Atwood machine problem in which two boxes are hanging to the either side of pulley. So, the acceleration of one is equal and opposite to other.

We have, equation for acceleration for Atwood machine as,

$a=\frac{\left(m_2-m_1\right)g}{m_2+m_1}$

Here,$m_2=m_p=15kg$and$m_1=m_m=10kg$ . So, the acceleration of monkey is,

$$\begin{gathered} a=\frac{\left(m_p-m_m\right)g}{m_p+m_m} \\ =\frac{\left(15\mathrm{kg}-10\mathrm{kg}\right)9.8\mathrm{m}/{\mathrm{s}}^2}{\left(15\mathrm{kg}+10\mathrm{kg}\right)} \\ =2.0\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration of the monkey is$2.0\mathrm{m}/{\mathrm{s}}^2$ .

As the value of acceleration in part (b) is positive, the acceleration of monkey is in upward direction.

Applying Newton’s law of motion to the package, we get

$$\begin{gathered} \mathrm{T}={\mathrm{m}}_{\mathrm{p}}\left(\mathrm{g}-\mathrm{a}\right) \\ =15\mathrm{kg}\left(9.8\mathrm{m}/{\mathrm{s}}^2-2\mathrm{m}/{\mathrm{s}}^2\right) \\ =120\mathrm{N} \end{gathered}$$

Therefore, the tension in the rope is 120 N.