- #1

- 23

- 0

**Problem Text:**

A 10kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15kg 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 climbing and hold onto the rope, what are

B)The magnitude and

C) Direction of the monkey's acceleration, and

D) What is the tension on the rope?

The figure given looks something like this.

_____

|LIMB|

|......|

|......|

|......|

|......|

[M] [P]

M

_{M}=10kg

M

_{P}=15kg

**What I've done**

A) The monkey, at rest, pulls on the rope with a downward force of 10kg[itex]\bullet[/itex]9.8m/s

^{2}=98N

The package pulls on the rope 15kg[itex]\bullet[/itex]9.8m/s

^{2}=147N

To put the package in motion, a net force has to be applied (i.e. the monkey cancels out the normal force). I take this to mean that the monkey must apply a downward force of at least 147N-98N=49N.

To create a force of downward force of 49N, the 10kg monkey must climb upwards at [itex]\frac{49N}{10kg}[/itex]=4.9m/s

^{2}.

B) When the monkey stops climbing, the only forces are 147N and 98N by the package and monkey respectively.

The net force will be 49 N downwards on the package side.

The acceleration will therefore be [itex]\frac{49N}{10kg+15kg}[/itex]=1.96m/s

^{2}

C) Up

D) F

_{net}=1.96m/s[itex]\bullet[/itex]15kg=29.4N

F

_{tension}=F

_{gravity}-F

_{net}

=147N-29.5N

=117.6N