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Conservation of energy of an elevator cable

1. Homework Statement
he cable of the 1,800 kg elevator cab in Fig. 8-51 snaps when the cab is at rest at the first floor, where the cab bottom is a distance d = 3.7 m above a cushioning spring whose spring constant is k = 0.15 MN/m. A safety device clamps the cab against guide rails so that a constant frictional force of 4.4 kN opposes the cab's motion. (a) Find the speed of the cab just before it hits the spring. (b) Find the maximum distance x that the spring is compressed (the frictional force still acts during this compression). (c) Find the distance (above the point of maximum compression) that the cab will bounce back up the shaft.


2. Homework Equations



3. The Attempt at a Solution

I got (a)...b and c are where I hit problems.

for b i used the equation....

mg(h+x) + F(h+x)(cos(pi))=.5kx^2

Now when I tried to solve for x i got an equation with huge numbers:

49012+13247x-0.075x^2
It just doesn't seem reasonable to solve for x here, and I know i must be doing something wrong I just cant see it.

for (c):

I used 1/2k(x)^2 - F(x+h) = (1800)(9.81)(x+h)

I would plug in x from c, its just x that I can't find.

Thanks!
 

Answers and Replies

Doc Al
Mentor
44,827
1,083
for b i used the equation....

mg(h+x) + F(h+x)(cos(pi))=.5kx^2
Looks good.

Now when I tried to solve for x i got an equation with huge numbers:

49012+13247x-0.075x^2
Check those numbers. (Realize that MN = 1 000 000 N.)
 
im still getting an answer of about 6 million...when the answer is 90...
 
Doc Al
Mentor
44,827
1,083
Check again.
Now when I tried to solve for x i got an equation with huge numbers:

49012+13247x-0.075x^2
The numbers in your first two terms are reasonable, but that 0.075 is WAY off. (It's a million times too small.)
 
i adjusted that term by a million. and now that term is too large. I plugged in the correct answer of 90, hoping that the entire term will go to zero. But it goes to a number greater than -6 million.

I am supposed to be solving for the x^2 and the x in the equation correct?
 
Doc Al
Mentor
44,827
1,083
Not sure where that answer of 90 m comes from. I solved it, getting a much smaller number. (I might certainly have made an arithmetic error.)

Correct that last term and solve the quadratic equation.
 

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