Question

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?

Short answer

1. Magnitude of least acceleration monkey should have to lift the box is${\mathrm{a}}_{\mathrm{m}}=4.9\mathrm{m}/{\mathrm{s}}^2$
2. Magnitude of monkey’s acceleration,$\mathrm{a}=2.0\mathrm{m}/{\mathrm{s}}^2$
3. Direction of Monkey’s acceleration is upward.
4. Tension in the rope,$T=120\mathrm{N}$

Step-by-step solution

Given information

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

Understanding the concept of Newton’s law

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.

Write the force equations

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)

(a) Calculate the magnitude of the least acceleration the monkey must have if it is to lift the package off the ground

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 $4.9\mathrm{m}/{\mathrm{s}}^2$.

(b) Calculate the magnitude of the monkey’s acceleration

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$2.0\mathrm{m}/{\mathrm{s}}^2$ .

(c) Calculate the direction of the monkey’s acceleration

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

(d) Calculate the tension in the rope

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.