An elevator (mass 4600 ) is to be designed so that the maximum acceleration is 0
ID: 2173509 • Letter: A
Question
An elevator (mass 4600 ) is to be designed so that the maximum acceleration is 0.0650 What is the maximum force the motor should exert on the supporting cable? What is the minimum force the motor should exert on the supporting cable?The correct formulas are:
m*a=T-m*g
if a=0.0650 g
4600*9.81*(1+0.0650)=T
T=~48000N
in the downward direction for the minimum T:
m*a=T-m*g
this is a=-0.0650*g
T=m*g*(1-0.0650)
T=4600*9.81*(1-0.0650)
T= 42100N
My question is, where did the 1 come from (4600*9.81*(1+0.0650)=T?
Explanation / Answer
(4600)(9.8 + .0650*9.8) = (9.8)(4600)(1 + .0650) when we common 9.8 from above equation then 1+.0650 remains in the equation