Solving for Minimum Acceleration in an Elevator with a Hanging Monkey

[needhelp83]

I am not even sure where to start with this question. I believe I need to incorporate this equation into this problem, but not too sure.

w=mg+ma

A 16.0-kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a tension of 200 N and breaks as the elevator acclerates. What is the elevator's minimum acceleration(magnitude and direction)?


Thanks for any help

 

[PhanthomJay]

I am not even sure where to start with this question. I believe I need to incorporate this equation into this problem, but not too sure.

w=mg+ma

A 16.0-kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a tension of 200 N and breaks as the elevator acclerates. What is the elevator's minimum acceleration(magnitude and direction)?


Thanks for any help

The monkey and rope are attached to the elevator, so they accelerate at the same rate of the elevartor until the rope breaks. Examine the forces acting on the monkey. A tension of 200 N up. A weight of mg acting down. Use Newtons 2nd law to solve for a. Your equation is correct except use T instead of w on the left side. Is that what you meant?

 

[edavey8205]

It always helps to draw out a schematic of the system. Draw the monkey and then the forces in each directions and that will help a lot in an problem dealing with summing the forces (Newton's 2nd law).

 

[needhelp83]

How do I solve for the magnitude and direction?

 

[needhelp83]

a= 12.5 m/s^2


The monkey would be moving up since he is not "weightless"

Is this correct?

 

[PhanthomJay]

a= 12.5 m/s^2


The monkey would be moving up since he is not "weightless"

Is this correct?

NO.
You must draw the free body diagram of the monkey. The rope is pulling up on it with a force of 200N. Gravity is pulling down on it with a force of
mg= 16(10) = 160N. The NET force is therefore (200-160) = 40N upward.
From Newton's 2nd law, the NET unbalanced force acting on the monkey produces an acceleration in the direction of that force
F_net = ma
40=16a
a=2.5m/s/s
This is the minimum acceleration of the monkey.
It is also the minimum acceleration of the elevator.
magnitude is 2.5m/s/s
direction is UP (which you answered correctly).