A 10 kg monkey climbs up a massless rope that runs over a...

A 10 $\mathrm{kg}$ monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 $\mathrm{kg}$ package on the ground (Fig. $5-54 )$ . (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, what are the (b) magnitude and (c) direction of the monkey's acceleration and (d) the tension in the rope?

A 10 $\mathrm{kg}$ monkey climbs up a massless rope that runs over a
frictionless tree limb and back
down to a 15 $\mathrm{kg}$ package on the
ground (Fig. $5-54 )$ . (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, what are the (b)
magnitude and (c) direction of the
monkey's acceleration and (d) the
tension in the rope?

Key Concepts

System of Connected Bodies and Motion Constraints

When bodies are connected by an inextensible rope, the acceleration of one body is directly related to the acceleration of the other. This interconnection, or constraint of motion, is vital in solving problems since it links the movement of both masses, facilitating the determination of conditions such as the minimum acceleration required for one body to lift the other, as well as the behavior of the system when it reaches equilibrium.

Tension in a Massless Rope with a Frictionless Pulley

In an idealized system with a massless rope and a frictionless pulley, the tension that develops in the rope is constant throughout its length. This uniform tension simplifies the analysis because it allows the forces on the monkey and the package to be directly related, which is essential when writing the equations of motion and solving for the unknowns.

Free-Body Diagram Analysis

Drawing free-body diagrams for each object helps in visually organizing all the forces acting on them, such as gravitational forces, tension, and any applied forces. This step is crucial for setting up the correct equations of motion for both the monkey and the package, ensuring that no force is overlooked in the dynamic analysis.

Newton’s Second Law

Newton’s second law, summarized by F = ma, is fundamental in analyzing the forces and resulting accelerations in mechanical systems. For connected bodies, such as a monkey and a package linked by a rope, applying the law to each component allows one to determine how the applied forces, including gravity and tension, result in the acceleration of the system.

Transcript

00:01 A quick free bite diagram for the monkey and the package.

00:04 We know that applying the newton second law for the forces of the monkey, this would be force f minus the mass of the monkey times g, equaling the mass of the monkey times the acceleration of the monkey.

00:21 And so here, since the rope is massless, we can say that f is going to be equal to the tension force t.

00:28 And so we can then apply newton's second law for the package.

00:33 This would be the force f plus the normal force minus the mass of the package times the acceleration, rather mass the package times g, equalling the mass of the package, acceleration of the package.

00:50 And so we know that if now if f is the minimum force required to lift the package, we can say that the normal force is going to be zero.

01:02 It'd be better to it.

01:04 You can say normal force equaling zero newton's, the acceleration of the package equaling zero meters per second squared.

01:12 And that is giving us that the force is equaling the mass of the package times g.

01:20 And so we can say for part a, substituting the mass of the package times g for f in the, equation for the monkey we have m sub p g minus m sub m g equaling m sub m g equaling m sub m a sub m and so to find the acceleration of the monkey this would be equaling the mass of the package minus the mass of the monkey multiplied by g divided by the mass of the monkey and so the acceleration of the monkey is to is found to be 15 kilograms minus 10 kilograms multiplied by 9 .8 meters per second squared and then this will be divided by 10 kilograms and so this is giving us 4 .9 meters per second squared this would be our answer for part a for the acceleration of the monkey now we know that for part b f minus m sub p g equals m sub p a sub p prime.

02:39 We're saying a sub p prime because now there is some sort of acceleration.

02:44 And then this will be f minus m sub m g equaling m a sub m prime.

02:52 So now here the acceleration of the package is downwards.

02:59 The acceleration of the monkey is upwards.

03:02 So the acceleration of the monkey prime is equaling the negative acceleration of the package prime.

03:08 And so we can then see that the force f is going to be equaling to the mass of the package multiplied by g plus the acceleration of the package prime.

03:19 And this is equaling the mass of the package multiplied by g minus the acceleration of the monkey prime...

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