Answer to Question #165295 in Classical Mechanics for Pp

Question

A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground. (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?

Answer 1

Answer

For according to theory of question

Mass of particle

m=10kg

Mass of package M=15kg

Part (a)

Now Let T

be the tension in the string as the monkey climbs it up with the least acceleration

a(positive upwards). So to lift the package, the minimum tension in the string must equal the weight of the package:

According to newton law

$T=mg=15\times9.8=147N$

Now

$a=\frac{T-mg}{m}=\frac{147-10\times9.8}{10}=4.9m/s^2$

Part b)

After the package is lifted and the monkey is at rest, let T

T

 be the tension in the string and a

a

 be the acceleration of the monkey (as well as the package). Applying Newton's second law, we have:

$147-T=15\times a$ . ....... 1

And

$T-98=10\times a$ . . . . ... 2

Solving both equation

$a=1.96m/s^2$

Part c)

As the acceleration value computed in Part (b) is positive, the monkey is accelerating upwards.

Part d) putting a=1.96m/s²

In equation 2

Tension

$T=98+10\times1.96=117.6N$