HELP PO!

Submitted by Kenneth Bryan … on

A MAN OF MASS 80 KG ENTERS AN ELEVATOR ABOUT TO DESCEND. WHAT PRESSURE DOES HE EXERT ON THE FLOOR:

a.) what it is at rest?
b.)when it descends with an acceleration of 1m/s
c.) when it is brought to a stop with a deceleration of 1m/s

Jhun Vert Mon, 04/20/2015 - 08:21

Let us call the required pressure Normal Force

letter (a)

Normal Force = Weight
Normal Force = mg = 80(9.81) = 784.8 N.
This is also true when the elevator is running at constant velocity (zero acceleration). The pressure he exerted on the floor will change when the acceleration of the elevator is not zero.

letter (b)

Reverse Effective Force, REF = ma = 80(1) = 80 N upward (Note: REF is always opposite the acceleration)
Weight = 784.8 N downward
Normal Force = 784.8 - 80 = 704.8 N

letter (c)

Reverse Effective Force, REF = ma = 80(1) = 80 N downward
Weight = 784.8 N downward
Normal Force = 784.8 + 80 = 864.8 N

Conclusion: Exercise may reduce the weight by 10 kg in less than a month. :)