Question

An \(11-\mathrm{kg}\) monkey is climbing a massless rope attached to a \(15-\mathrm{kg}\) log over a frictionless tree limb. (a) With what minimum acceleration must the monkey climb up the rope so that it can raise the \(15-\mathrm{kg}\) log off the ground? If, after the \(\log\) has been raised off the ground, the monkey stops climbing and hangs on to the rope, what will now be \((b)\) the monkey's acceleration and \((c)\) the tension in the rope?

Short answer

Minimum monkey acceleration = \((m_l \cdot g - m_m \cdot g) / m_m\), monkey's acceleration after stopping = \((m_m + m_l) \cdot g / (m_m + m_l)\), tension in rope = \(m_l \cdot (g + a_m')\)

Step-by-step solution

Determine the minimum acceleration of the monkey

Let's denote \(a_m\) as the minimum acceleration of the monkey. From Newton's second law, we know that the net force equals the mass times this acceleration. From the monkey's perspective, the weight of the monkey and the upward force due to climbing must be equal or greater to the weight of the log, which results in the following inequality: \(m_m \cdot g + m_m \cdot a_m \geq m_l \cdot g\). Here, \(m_m\) and \(m_l\) refer to the monkey and log's masses respectively, and \(g\) is the acceleration due to gravity. Solving this inequality for \(a_m\) gives \(a_m \geq (m_l \cdot g - m_m \cdot g) / m_m\)

Find the monkey's acceleration when it stops climbing

When the monkey stops climbing, its acceleration, denoted as \(a_m'\), can be obtained from Newton's second law by knowing that the net force is the mass times the acceleration. In this scenario, the weight of the monkey and the log both act downward, while the tension in the rope acts upward. Thus, \(m_l \cdot g + m_m \cdot g = (m_m + m_l) \cdot a_m'\). Simplifying this equation gives \(a_m' = (m_m + m_l) \cdot g / (m_m + m_l)\)

Determine the tension in the rope

The tension in the rope, denoted as \(T\), can be calculated from Newton's second law by knowing that the net force equals mass times acceleration. For the log in this case, the tension upwards and weight downwards results in \(m_l \cdot g = T - m_l \cdot a_m'\). Solving this equation for \(T\) gives \(T = m_l \cdot (g + a_m')\)

Key concepts

Frictionless Surfaces

When dealing with physics problems, frictionless surfaces often appear to simplify calculations. Imagine a smooth pathway without any resistance, which means that friction, the force that usually opposes motion, is absent. This creates a perfect scenario for tracking motion:
- Without friction, objects move more freely, acting only under the influence of other forces, like tension or gravity.
- In our example with the monkey, the tree limb is a frictionless surface. This means the forces at play do not include the usual frictional force that would act against the monkey's movement or the log’s motion.
- It simplifies calculations of the monkey climbing the rope, as the only forces to consider are tension in the rope and gravitational pull.
Understanding frictionless situations allows us to focus on basic principles of motion without the added complexity of counteracting forces caused by friction. This helps us deeply learn the relationship between motion, forces, and mass, without interference.

Tension in Ropes

The concept of tension is essential when we discuss problems involving ropes, strings, or chains. Tension is the pulling force transmitted along the length of a string or rope when it is pulled tight by forces acting at both ends. Follow these pointers to grasp how tension functions:
- It occurs only when the rope is taut, meaning there is no slack.
- The tension must be equal throughout if the rope is considered massless, as in our monkey problem.

As the monkey climbs, the tension in the rope changes depending on the forces applied. While climbing, the monkey exerts a force that must not only hold its own weight but also lift the log from the ground. Here, tension is crucial since it counteracts gravitational forces pulling the monkey and the log downwards.

When the monkey stops climbing, the tension balances out the gravitational forces acting both on the monkey and the log. This showcases how tension adjusts to maintain equilibrium in changing scenarios.

Acceleration Due to Gravity

Gravity is the force that attracts two bodies towards each other, with Earth providing a consistent pull towards its center. The acceleration due to gravity, denoted as "g," is approximately 9.81 m/s² on Earth's surface. This is a constant factor in motion-related calculations:
- It's the reason objects fall or move downwards when unsupported.
- In physics problems, it's crucial to consider how this acceleration influences other forces in play.

In our problem involving the monkey and the log, gravity acts on both, pulling them downwards and influencing the calculations for the minimum acceleration needed by the monkey to lift the log.

This gravitational pull must be overcome by the tension force in the rope to lift the log. Additionally, understanding the constant pull of gravity helps determine the forces at equilibrium, such as when the monkey stops climbing and the system reaches a steady state.
Knowing how gravity interplays with other forces gives us a comprehensive understanding of motion and dynamics, particularly in environments without friction."