When will the bolt pass the elevator floor?

In summary: Ok.Are you familiar with using different inertial frames? It is a little simpler if you work in the reference frame of the elevator. The cancellation of the initial velocity of the bolt and the uniform motion of the elevator results in the equation of motion beingx[bolt] = 8t - (v_elevator + av_wall)This is in the elevator's reference frame.
  • #1
Wrangler_77
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A 5m tall elevator travels uniformly up an elevator shaft at 8m/s. At some point in time, a bolt falls from the top of the elevator and falls freely down the shaft. You can ignore the effect of air resistance.

How many seconds after the fall will the bolt pass the bottom of the elevator?
y=yi + vit + (1/2)at^2 I tried to use the above equation for both the elevator and the bolt separately. The bolt takes 1.01 seconds to fall 5m and the elevator moves up 5m in 0.625 seconds.
 
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  • #2
Wrangler_77 said:
I tried to use the above equation for both the elevator and the bolt separately. The bolt takes 1.01 seconds to fall 5m and the elevator moves up 5m in 0.625 seconds.
Do you think that will work? How about the fact that the bolt and the bottom of the elevator are not going to meet up at the time when the bolt has dropped 5m because by then the elevator floor will have moved upwards.
 
  • #3
phinds said:
Do you think that will work? How about the fact that the bolt and the bottom of the elevator are not going to meet up at the time when the bolt has dropped 5m because by then the elevator floor will have moved upwards.
Not really. I am not able to grasp this I guess.
 
  • #4
Wrangler_77 said:
Not really. I am not able to grasp this I guess.
Well, you had best think about it some, else you will never solve the problem.
 
  • #5
phinds said:
Well, you had best think about it some, else you will never solve the problem.
I have been thinking about it all weekend. I know it must take less than 1 second because the floor is moving toward the bolt as it falls. The reason I signed up for this forum is to be given some insight about the problem. Can you explain to me how to solve this?
 
  • #6
Wrangler_77 said:
I have been thinking about it all weekend. I know it must take less than 1 second because the floor is moving toward the bolt as it falls. The reason I signed up for this forum is to be given some insight about the problem. Can you explain to me how to solve this?
What you need is the equation of motion for the bottom of the elevator and the equation of motion for the motion of the bolt and then you solve them for the place where they intersect. The equation of motion for the bottom of the elevator is trivial since it is uniform motion. What is the equation of motion for the bolt?
 
  • #7
phinds said:
What you need is the equation of motion for the bottom of the elevator and the equation of motion for the motion of the bolt and then you solve them for the place where they intersect. The equation of motion for the bottom of the elevator is trivial since it is uniform motion. What is the equation of motion for the bolt?
0=5 -4.9t^2 ?
 
  • #8
Wrangler_77 said:
0=5 -4.9t^2 ?
OK, now you need both equations using the same coordinate system.
 
  • #9
phinds said:
OK, now you need both equations using the same coordinate system.
This where I am getting confused. So, I need 5=8t and 0=5-4.9t^2 combined as 4.9t^2 + 8t - 10=0? That gives me 0.83 seconds.
 
  • #10
Wrangler_77 said:
This where I am getting confused. So, I need 5=8t and 0=5-4.9t^2 combined as 4.9t^2 + 8t - 10=0? That gives me 0.83 seconds.
I think you need to draw a diagram and label your variables
 
  • #11
Wrangler_77 said:
I know it must take less than 1 second because the floor is moving toward the bolt as it falls.
When the bolt falls, does it fall from the elevator's top or from the adjacent wall? What is its initial velocity?
 
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  • #12
phinds said:
I think you need to draw a diagram and label your variables

As it turns out, the bolt has the same initial velocity of the elevator. Rather than noticing this earlier I assumed the initial velocity of the bolt was zero.

Using the roof of the elevator as my origin my two equations are

x[bolt] = 8t - 4.9t^2
and
x[floor] = -5 + 8t

Setting them equal to each other eliminates the 8t from both sides and I am left with

4.9t^2 = 5

t = 1.01s
 
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  • #13
Wrangler_77 said:
As it turns out, the bolt has the same initial velocity of the elevator. Rather than noticing this earlier I assumed the initial velocity of the bolt was zero.

Using the roof of the elevator as my origin my two equations are

x[bolt] = 8t - 4.9t^2
and
x[floor] = -5 + 8t

Setting them equal to each other eliminates the 8t from both sides and I am left with

4.9t^2 = 5

t = 1.01s
Ok.
Are you familiar with using different inertial frames? It is a little simpler if you work in the reference frame of the elevator. The cancellation of the initial velocities is immediate.
 
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  • #14
haruspex said:
Ok.
Are you familiar with using different inertial frames? It is a little simpler if you work in the reference frame of the elevator. The cancellation of the initial velocities is immediate.

For future reference to these types of questions, terms like "falls" and "is dropped" can be very misleading, since they imply the object immediately starts moving down. In fact the "breaks free from" or "is released" is a much better descriptor - and you can use that even if the original question doesn't.
The "frame of reference" mentioned above is good. If you have access to a lift in a tall building, you should try the following.
Drop (release) an object as you stand in a stationary lift (hold the door open and it won't move) and note how long it takes to reach the floor.
Now start the lift - either up or down - and wait until it is traveling at constant speed (perhaps going past the 5th floor of 20) and again drop (release) the object and compare the time taken to reach the floor.
 
  • #15
Hello Wrangler_77, did you solve the second question?
Now assume that the bolt falls from the *wall* of the elevator shaft, and then falls freely without air resistance.

How many seconds after the fall will the bolt pass the bottom of the elevator?
 
  • #16
physicsnoobmaster said:
Hello Wrangler_77, did you solve the second question?
Now assume that the bolt falls from the *wall* of the elevator shaft, and then falls freely without air resistance.

How many seconds after the fall will the bolt pass the bottom of the elevator?
He has not been here in 5 months so you're not likely to get an answer.
 
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  • #17
yeap... I see it now... =/ I'm struggling with the second part... I was able to solve the first part:

I made a draw to understand the "dynamics" of the environment and computed the Xf for the elevator and bolt:

Xfb = Xib + Vit + 1/2 at^2
= 0 - 8t + 9.8/2 t^2
Xfb = 4.9t^2 -8t

Xfe = 5 - 8t (constant speed)...

Then, by equaling them:

5 -8t = 4.9t^2 -8t
5 = 4.9t^2
5/4.9 = t^2
t = sqrt(1.02) = 1.01 ;Now... for the b question, I'm really struggling in relating the distance between the bolt and elevator... I tried to mount the equations based on the bolt frame:

Xfb = h + 0t + 9.8/2 t^2 (considering initial position at the elevator floor)...

Xfe = Xi - 8t;

I'm trying to relate the initial position of the elevator to h of the bolt... But, so far, no good... =/ ...

Could someone, please, bring me some light to this problem?
 
  • #18
physicsnoobmaster said:
yeap... I see it now... =/ I'm struggling with the second part... I was able to solve the first part:

I made a draw to understand the "dynamics" of the environment and computed the Xf for the elevator and bolt:

Xfb = Xib + Vit + 1/2 at^2
= 0 - 8t + 9.8/2 t^2
Xfb = 4.9t^2 -8t

Xfe = 5 - 8t (constant speed)...

Then, by equaling them:

5 -8t = 4.9t^2 -8t
5 = 4.9t^2
5/4.9 = t^2
t = sqrt(1.02) = 1.01 ;Now... for the b question, I'm really struggling in relating the distance between the bolt and elevator... I tried to mount the equations based on the bolt frame:

Xfb = h + 0t + 9.8/2 t^2 (considering initial position at the elevator floor)...

Xfe = Xi - 8t;

I'm trying to relate the initial position of the elevator to h of the bolt... But, so far, no good... =/ ...

Could someone, please, bring me some light to this problem?
When the bolt falls, it begins exactly 5m above the floor of the lift - it is level with the roof.
Draw a couple of position time graphs (at least in your mind). Let the bolt start at +5m, while the floor starts at 0m - then you won't get tied up with negative numbers.
The bolt follows the usual parabola of a falling object (reaching 0m at 1.01 seconds I think you had already calculated - but exactly 1.0s if you use a g value of 10, as is often done with simple introductory problems.
The floor proceeds up at a steady 8m/s
When the lines intersect you have your answer.
Now use your formulas to find the answer mathematically - complete with 2nd and 3rd decimal places..

Note: to save the time needed to say the 4 syllable word elevator, we call it a lift.
 
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  • #19
PeterO said:
When the bolt falls, it begins exactly 5m above the floor of the lift - it is level with the roof.
Draw a couple of position time graphs (at least in your mind). Let the bolt start at +5m, while the floor starts at 0m - then you won't get tied up with negative numbers.
The bolt follows the usual parabola of a falling object (reaching 0m at 1.01 seconds I think you had already calculated - but exactly 1.0s if you use a g value of 10, as is often done with simple introductory problems.
The floor proceeds up at a steady 8m/s
When the lines intersect you have your answer.
Now use your formulas to find the answer mathematically - complete with 2nd and 3rd decimal places..

Note: to save the time needed to say the 4 syllable word elevator, we call it a lift.

Thank you so much for your patience and detailed explanation. I was doing some mass and trying to figure out some distance between the lift and the bolt... Thinking that, even with the bolt starting with a initial velocity = 0 (the wall of shaft).. I was imagining that at some point, more specifically, at the moment that the bolt meet the roof of the elevator, what would be its velocity... anyway... I was doing such confusion and couldn't realize that what I had to do was just to consider the roof of the lift and the position of the bolt (at the wall of shaft), was my initial position!

Lastly... I could solve the problem with the same equations and draws I already had:
Xfe = 5 - 8t (considering down direction positive and bolt initial position as 0); then:

Xfb= 0 + 0t +1/2 * 9.8 * t^2;

so: 5-8t = 4.9t^2;
4.9t^2 + 8t -5 =0

t' = 0.4824
t'' = -2.1151

Since time can't be negative, t= 0.4824s...

Again.. Thank you for your patience and detailed explanation!
 

1. When will the bolt pass the elevator floor?

The time it takes for the bolt to pass the elevator floor depends on several factors, such as the speed of the elevator and the height of the floor. It can be calculated by dividing the height of the floor by the speed of the elevator.

2. How can we determine the speed of the elevator?

The speed of the elevator can be determined by measuring the time it takes for the elevator to travel a certain distance. This can be done by using a stopwatch and measuring the time it takes for the elevator to pass two different floors.

3. Does the weight of the bolt affect the time it takes to pass the elevator floor?

Assuming the bolt is small and light compared to the elevator, its weight will not significantly affect the time it takes to pass the elevator floor. The main factors that affect this time are the speed of the elevator and the height of the floor.

4. Is the time it takes for the bolt to pass the elevator floor constant?

No, the time it takes for the bolt to pass the elevator floor will vary depending on the height of the floor and the speed of the elevator. The higher the floor and the faster the elevator, the shorter the time will be.

5. Can we use this information to determine the height of the elevator floor?

Yes, if we know the speed of the elevator and the time it takes for the bolt to pass the floor, we can calculate the height of the floor. This can be useful for maintenance or repair purposes.

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