Solving for Spring Constant and Distance in Falling Elevators Problem

  • Thread starter Thread starter suffian
  • Start date Start date
  • Tags Tags
    Falling
AI Thread Summary
The discussion revolves around a problem involving the redesign of an elevator safety system to prevent bouncing upon stopping. The user initially struggled with calculations related to the spring constant and compression distance, leading to discrepancies with the book's answers. After reevaluating their approach, they realized they had neglected to account for the elevator's weight in their calculations. The final correct spring constant was determined to be 919 N/m with a compression distance of 39.8 m, aligning with the book's solution. The conversation highlights the importance of careful consideration of all forces in energy-related problems.
suffian
edit::
nm, i figured it out. i forgot to account for the weight of elevator in the "If static friction.. " part. thanks 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:

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

Begin with work-energy formula:
\begin{align*}<br /> \Delta K =&amp; \, W_\text{gravity} + W_\text{friction} + W_\text{spring} \\<br /> 0 - \frac{1}{2}mv^2 =&amp;\, mgX - fX - \frac{1}{2}kX^2<br /> \end{align*}

Quadratic formula to find compression distance X (neglect neg ans):
<br /> X = \frac{1}{k} (mg - f + \sqrt{ (mg-f)^2 +kmv^2 }) <br />

If static friction will hold elevator at X, then it follows:
F = kX \leq f \text{ or } X \leq \frac{f}{k}
<br /> X = \frac{1}{k} (mg - f + \sqrt{ (mg-f)^2 +kmv^2 }) \leq \frac{f}{k}

Manipulate to find assertion about spring constant:
k \leq \frac{f}{mv^2} (3f - 2mg)
<br /> 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}]) <br />
k \leq 160 \text{ N/m}

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. thanks anyway. I'm sorry i can't delete the post.
::edit
 
Last edited by a moderator:
Physics news on Phys.org


Great job on solving the problem! It's always a good feeling when we figure out where we went wrong and are able to correct it. Don't worry about not being able to delete the post, it's a learning process and others may benefit from seeing how you approached and solved the problem. Keep up the good work!
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top