An 11-kg monkey is climbing a massless rope attached to a 15-kg log over a frictionless tree lim

An 11-kg monkey is climbing a massless rope attached to a...

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?

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?

Key Concepts

Dynamic versus Static Equilibrium

Understanding the difference between dynamic (moving, accelerating) and static (at rest or moving at constant velocity) conditions is key. In dynamic scenarios, acceleration is determined by net forces, while in static or equilibrium situations, the net forces balance to zero. This distinction is critical when analyzing transitions in system behavior.

System of Connected Bodies

For systems where multiple objects are connected by a rope, the motion of one body affects the motion of the other. Analyzing such systems requires accounting for the interactions and often leads to setting up simultaneous equations that relate the accelerations and forces acting on each object.

Newton's Second Law of Motion

This principle states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. It is essential for setting up equations of motion for each body in a system, allowing one to determine relationships between forces, masses, and resulting accelerations.

Tension in a Massless Rope

In problems involving ropes assumed to be massless, the tension is uniform along the rope. This simplifies the analysis because the same force is transmitted through the rope from one end to the other, which is crucial when connecting multiple bodies.

Frictionless Pulleys and Supports

When a rope passes over a frictionless pulley or tree limb, there is no energy loss due to friction, and the pulley does not alter the magnitude of the tension. This assumption allows the forces and accelerations of the connected objects to be directly related without additional frictional force considerations.

Transcript

00:01 Hello student, we are going to test this question here given mass of monkey that it is represented by small m mass of monkey it is given 11 kg, mass of log that it is represented by capital m which is 15 kg.

00:30 Let the required acceleration of the monkey be a, let the required acceleration of the monkey be a, let the required acceleration of the monkey required acceleration of monkey b a due to which log will lift from the ground now let's draw the black diagram here in this question we have supposed that monkeys moving in upward direction with acceleration a due to which due to which tension t will lift the log which which will lift the log which which is is having mass 15 kg means tension required to lift the log that is m into g t is equal to m into g due to this tension this log will lift in upward direction so required tension in the street m into g now substituting the value of m mass of log it is given 15 kg and acceleration of the monkey we have supposed a and here g is gravitational acceleration so value of g is 9 .8 meter per second square after solving we will get 147 newton so required tension is 147 newton now we can write we can write t minus m g is equal to m into a for the monkey.

02:29 Now here we have calculated tension in the string 147 minus mass of monkey that is 11 kgg and gravitational acceleration that is 9 .8 meter per second square is equal to 11 into 8 now further solving 147 minus 11 into 9 .8 we will get 108 .8 is equal to 118 after solving we will get a is equal to 39 .2 upon 11.

03:15 After solving we will get a is equal to 3 .56 meter per second square.

03:22 A is equal to 3 .56 meter per second square.

03:24 A is equal to 3 .56 meter per second square.

03:32 Now for bit be when monkey will stop to raise the height, then we have supposed the net acceleration, the net acceleration after raising the height or the resultant acceleration, the net acceleration for this case, for this case, b a a one now let's draw the block diagram for this question now with the help of this block diagram let's draw the equation write the equation for lower force 15g minus t is equal to 15 a 15 a1 let this is equation number now for monkey we can write t minus 11g is equal to 11 a1.

04:51 Then this is equation number 4...

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