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ichivictus
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My apologies for asking two questions in a short period of time, this is the last question I have. Hopefully one day I'll be skilled enough to answer everyone elses! :)
An empty elevator has a mass of 722 kg. It moves between floors at a maximum speed of 6.00 m/s. The elevator is stopped on the 20th floor of the building when someone pushes the call button in the lobby (the first floor).
Assuming that it takes 15.25 meters for the elevator to reach its maximum speed, and assuming constant acceleration, calculate the tension in the cable as the elevator car begins to descend. Take the acceleration due to gravity to be 9.81 m/s2. Remember to include units with your answer.
Kinematics:
y = Vot + (1/2)at2
Vf = Vo + at
Vf2 = Vo2 + 2ay
y = t * [ (Vo + Vf) / 2 ]
Newton's Second Law
F = ma
I read somewhere that tension would be Ft = m(a+g). Does this seem right?
Using kinematics I find the acceleration from which the elevator is at rest, to the point it reaches 6 m/s.
I need to solve for t to replace in another kinematic equation. I use the distance equation for it. (I am using y+ in from top to bottom).
15.25 = .5at^2
30.5 = at^2
sqrt(30.5/a)=t
I plug that into Vf = Vo + at
6 m/s = a * sqrt(30.5m/a)
36 m^2/s^2 = a^2 * 30.5m/a
36 m^2/s^2 = a * 30.5m
(36 m^2/s^2) / 30.5 m = a = 1.18 m/s^2
Then I plug that into
Ft = m(a+g)
Ft = 722kg(1.18 m/s^2 + 9.81 m/s^2) = 7934.78 N
I am really bad at tension problems and want to double check that I got this correct. Thanks :)
Homework Statement
An empty elevator has a mass of 722 kg. It moves between floors at a maximum speed of 6.00 m/s. The elevator is stopped on the 20th floor of the building when someone pushes the call button in the lobby (the first floor).
Assuming that it takes 15.25 meters for the elevator to reach its maximum speed, and assuming constant acceleration, calculate the tension in the cable as the elevator car begins to descend. Take the acceleration due to gravity to be 9.81 m/s2. Remember to include units with your answer.
Homework Equations
Kinematics:
y = Vot + (1/2)at2
Vf = Vo + at
Vf2 = Vo2 + 2ay
y = t * [ (Vo + Vf) / 2 ]
Newton's Second Law
F = ma
I read somewhere that tension would be Ft = m(a+g). Does this seem right?
The Attempt at a Solution
Using kinematics I find the acceleration from which the elevator is at rest, to the point it reaches 6 m/s.
I need to solve for t to replace in another kinematic equation. I use the distance equation for it. (I am using y+ in from top to bottom).
15.25 = .5at^2
30.5 = at^2
sqrt(30.5/a)=t
I plug that into Vf = Vo + at
6 m/s = a * sqrt(30.5m/a)
36 m^2/s^2 = a^2 * 30.5m/a
36 m^2/s^2 = a * 30.5m
(36 m^2/s^2) / 30.5 m = a = 1.18 m/s^2
Then I plug that into
Ft = m(a+g)
Ft = 722kg(1.18 m/s^2 + 9.81 m/s^2) = 7934.78 N
I am really bad at tension problems and want to double check that I got this correct. Thanks :)