Question

A 15 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 23 kg package on the ground (see the 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, what are (b) the magnitude and the direction of the monkey's acceleration (choosing the positive direction up), and (c) what is the tension in the rope? Units () Number Units (a) Number (c) Number Units

Answer 1

To lift the package the monkey has to apply atleast a force that can be greater or equal to the force of gravity applied on the package

that is using free body diagram

$\frac{(M - m)g}{m} = a$

$\frac{(23 - 15)9.8}{15} = a$

So the acceleration of the monkey should be 5.23 m/s2

-------------------- Part b --------------------------

free body diagram

$\frac{(M- m)g}{ (m+M)} =a$

$\frac{(23- 15)9.8}{ (23+15)} =a$

Acceleration of monkey is +2.06 m/s2

----------------- Solution of Part c --------------------------

tension in the string is

Tension in the string is 177.95 N