Tension on a Cable

  • Thread starter PhysicsCCR
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  • #1
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Homework Statement


An elevator in a tall building is allowed to reach a maximum speed of 3.89 m/s going down. What must the tension be in the cable to stop this elevator over a distance of 2.70 m if the elevator has a mass of 1320 kg including occupants?



Homework Equations


Vfy^2 = Viy^2 + 2ay(Yf-Yi)

F = ma (Newton's Second Law in the y-direction)


The Attempt at a Solution


Vi = 3.89
T = ?
Yf = 2.70 m
Yi = 0
m = 1320 kg
a = ?

Using the first equation: 0 = (3.89)^2 + 2(ay) (2.70)
ay = -2.80 m/s^2

Applying Newton's Second Law, I got: T - mg = ma

T = mg + ma = (1320)(9.81) + (1320)(-2.80) = 9250.24 N

However, the system says it's incorrect. Any help would be appreciated!
 

Answers and Replies

  • #2
ehild
Homework Helper
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1,912

Homework Equations


Vfy^2 = Viy^2 + 2ay(Yf-Yi)

F = ma (Newton's Second Law in the y-direction)


The Attempt at a Solution


Vi = 3.89

Yf = 2.70 m
Yi = 0
m = 1320 kg
a = ?

Using the first equation: 0 = (3.89)^2 + 2(ay) (2.70)
ay = -2.80 m/s^2

Applying Newton's Second Law, I got: T - mg = ma

T = mg + ma = (1320)(9.81) + (1320)(-2.80) = 9250.24 N

Think over the signs. Considering the upward direction positive, the displacement of the elevator is negative, -2.7 m. The acceleration points upward, it is positive.

ehild
 
Last edited:
  • #3
24
0
Oh, okay. Thank you, Ehild!
 

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