- #1

suffian

edit::

nm, i figured it out. i forgot to account for the weight of elevator in the "If static friction.. " part. thx anyway. i'm sorry i can't delete the post.

::edit

I hit a brick wall over yet another problem concerning energy methods, basically a mismatch between my answer and the book's. I would appreciate whoever has the time to look this over and guess where I might be going wrong.

Anyway, here's the problem verbatim:

[tex] \begin{align*}

\Delta K =& \, W_\text{gravity} + W_\text{friction} + W_\text{spring} \\

0 - \frac{1}{2}mv^2 =&\, mgX - fX - \frac{1}{2}kX^2

\end{align*} [/tex]

Quadratic formula to find compression distance X (neglect neg ans):

[tex]

X = \frac{1}{k} (mg - f + \sqrt{ (mg-f)^2 +kmv^2 })

[/tex]

If static friction will hold elevator at X, then it follows:

[tex] F = kX \leq f \text{ or } X \leq \frac{f}{k} [/tex]

[tex]

X = \frac{1}{k} (mg - f + \sqrt{ (mg-f)^2 +kmv^2 }) \leq \frac{f}{k} [/tex]

Manipulate to find assertion about spring constant:

[tex] k \leq \frac{f}{mv^2} (3f - 2mg) [/tex]

[tex]

k \leq \frac{[17000 \text{ N}]}{[2000 \text{ kg}][25 \text{ m/s}]^2} (3[17000 \text{ N}] - 2[2000 \text{ kg}][9.80 \text{ m/ss}])

[/tex]

[tex] k \leq 160 \text{ N/m} [/tex]

So, answer is not the same.

edit::

nm, i figured it out. i forgot to account for the weight of elevator in the "If static friction.. " part. thx anyway. i'm sorry i can't delete the post.

::edit

nm, i figured it out. i forgot to account for the weight of elevator in the "If static friction.. " part. thx anyway. i'm sorry i can't delete the post.

::edit

I hit a brick wall over yet another problem concerning energy methods, basically a mismatch between my answer and the book's. I would appreciate whoever has the time to look this over and guess where I might be going wrong.

Anyway, here's the problem verbatim:

Begin with work-energy formula:Redesign the elevator safety system of Example 7-11 [A falling elevator stopped by a powerful spring] so that the elevator does not bounce but stays at rest the first time its speed becomes zero. The mass of the elevator is 2000 kg and its speed when it first touches the spring is 25 m/s. There is a kinetic friction force of 17,000 N and the maximum static friction on the elevator is also 17,000 N. The mass of the spring can be neglected. a) What spring constant is required, and what distance is the spring compressed when the elevator is stopped? Do you think the design is practical? Explain. b) What is the maximum magnitude of the acceleration of the elevator?

Book Ans: a) 919 N/m, 39.8 m b) 17.0 m/s^2

[tex] \begin{align*}

\Delta K =& \, W_\text{gravity} + W_\text{friction} + W_\text{spring} \\

0 - \frac{1}{2}mv^2 =&\, mgX - fX - \frac{1}{2}kX^2

\end{align*} [/tex]

Quadratic formula to find compression distance X (neglect neg ans):

[tex]

X = \frac{1}{k} (mg - f + \sqrt{ (mg-f)^2 +kmv^2 })

[/tex]

If static friction will hold elevator at X, then it follows:

[tex] F = kX \leq f \text{ or } X \leq \frac{f}{k} [/tex]

[tex]

X = \frac{1}{k} (mg - f + \sqrt{ (mg-f)^2 +kmv^2 }) \leq \frac{f}{k} [/tex]

Manipulate to find assertion about spring constant:

[tex] k \leq \frac{f}{mv^2} (3f - 2mg) [/tex]

[tex]

k \leq \frac{[17000 \text{ N}]}{[2000 \text{ kg}][25 \text{ m/s}]^2} (3[17000 \text{ N}] - 2[2000 \text{ kg}][9.80 \text{ m/ss}])

[/tex]

[tex] k \leq 160 \text{ N/m} [/tex]

So, answer is not the same.

edit::

nm, i figured it out. i forgot to account for the weight of elevator in the "If static friction.. " part. thx anyway. i'm sorry i can't delete the post.

::edit

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