Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum 
Apr 17th 2018, 04:06 PM

#1  Junior Member
Join Date: Apr 2018
Posts: 6
 The Elevator and Bolt Problem!! (Again =[)
Hello Guys... I'm sorry if this is a very simple question but, I recently started to review Physics basics, I have a very poor basis, and I'm struggling with the following exercise. ANY help or advice are welcome...
A 5m tall elevator travels uniformly up an elevator shaft at 8 m/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.
a) How many seconds after the fall will the bolt pass the bottom of the elevator?
b) 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?
For the first question (a), I was able to solve the problem... See:
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?

 
Apr 17th 2018, 08:13 PM

#2  Senior Member
Join Date: Apr 2017
Posts: 453

(a).....You can forget about the speed of the elevator ... it's a closed system experiencing gravity. Inside the elevator the bolt falls with initial velocity Zero
..... s=5 .... g=9.81 ....s= 1/2 g (t squared) .... t=1.01 secs
(b) ... now the bolt falls from the wall of the shaft (presumably at the moment the roof of the elevator passes it )
Now our frame of reference is the shaft ...we don't know the distance of fall , the floor of the elevator is rising to meet the falling bolt at a rate of 8t m/sec ....
the fall distance is 5 meters minus the distance the elevator travels .. s = 5  8t
We use this equation again ..... s= 1/2 g (t squared) .... s= 4.905 (t squared)
so then ... 5  8t = 4.905 (t squared)
so 5 = 8t + 4.905 (t squared ) ...t comes out to a little less than half a second .

 
Apr 18th 2018, 03:59 AM

#3  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 1,009

Why do you think the bolt and the elevator floor travel the same distance?
Hint : Afterall they start from different locations and arrive at a common location when?
Incidentally why do you not use conventional symbols for your kinematic variables?

 
Apr 18th 2018, 04:50 AM

#4  Junior Member
Join Date: Apr 2018
Posts: 6

Originally Posted by oz93666 (a).....You can forget about the speed of the elevator ... it's a closed system experiencing gravity. Inside the elevator the bolt falls with initial velocity Zero
..... s=5 .... g=9.81 ....s= 1/2 g (t squared) .... t=1.01 secs
(b) ... now the bolt falls from the wall of the shaft (presumably at the moment the roof of the elevator passes it )

That was the whole problem!!! OMG!!!!For some reason, I honestly dont know why... I didn't get that...
I was thinking that the elevator started at some distance from the wall of the shaft and, I was imaginating that I should figure out when they would cross each other.... Anyway.. in one of my drafts (looking at my notebook after your explanation) I already had calculated that.. The roots of the 2nd order equation were 2.11 and 0.4824... But again.. for some reason.. I just ignore that and started to think that the problem was more complicated than it really is!!!
After reading your explanation... I just used the same equation and set the initial velocity of the bolt as 0...
Ahhh.. and thanks for the tip related to closed system experiencing gravity it is really simpler than the other way...
Thank you so much for your patiente and detailed explanation! I will change the title to [Solved]!!!! 
 
Apr 18th 2018, 04:59 AM

#5  Junior Member
Join Date: Apr 2018
Posts: 6

Originally Posted by studiot Why do you think the bolt and the elevator floor travel the same distance?
Hint : Afterall they start from different locations and arrive at a common location when?
Incidentally why do you not use conventional symbols for your kinematic variables? 
Honestly... I dont know why... I get involved with such distance... and started "to divagate" so far from the real problem...
As I wrote at the previous post, I was able to solve after realizing that I had to consider the time "presumably at the moment the roof of the elevator passes it "... Anyway... thanks for your Hints!!
Ahh.. and regarding the notation of variables... I'm sorry for that... I'm taking an online course: Mechanics Part 1, for Rice University and, at least until now, they are using such names for 1D Kinematics... I imagine with time they will introduce the standard names... SOrry for using those ones....

 
May 8th 2018, 12:44 PM

#6  Junior Member
Join Date: Apr 2018
Posts: 6

(a).....You can forget about the speed of the elevator ... it's a closed system experiencing gravity. Inside the elevator the bolt falls with initial velocity Zero
..... s=5 .... g=9.81 ....s= 1/2 g (t squared) .... t=1.01 secs
(b) ... now the bolt falls from the wall of the shaft (presumably at the moment the roof of the elevator passes it )...
Guys... first of all.. I really sorry for this long period to writing this reply (AGAIN).... I'd already answered in the same day you replied me... But for some reason.. Until now... it wasn't approved by the moderator!!!!
Anyway... sorry again....
Returning to the question... (presumably at the moment the roof of the elevator passes it ) <<< That was the role problem!!! For some reason, I didn't realize that!! I was thinking about to compute or estimate the High between the elevator and the wall of the elevator shaft.... I was imagining a lot of things... and was diverging from the true problem / question...
In fact, I'd already computed the equation according to your tips... but again... for some reason.. I set them aside...
So... I just would like to thank you for your patience and detailed explanation about the problem... And once more... sorry for taking too long... I don't know why me previous answers weren't approved....

 
May 8th 2018, 12:47 PM

#7  Junior Member
Join Date: Apr 2018
Posts: 6

Originally Posted by studiot Why do you think the bolt and the elevator floor travel the same distance?
Hint : Afterall they start from different locations and arrive at a common location when?
Incidentally why do you not use conventional symbols for your kinematic variables? 
First of all.. as I said in my previous comment... Sorry for taking too long to answer... I'd already answered both of you on the same day you replied the post, but, for some reason, until now that replies weren't approved by the moderator....
Honestly, I don't know why... For some reason, I was thinking about the high between the wall of the elevator shaft and the elevator... Trying to find some formula.... I don't know... I just diverged from the problem too much...
Regarding the conventional symbols... sorry for that... I'm taking a course and at this moment, they are using such symbols for kinematics... I think they will probably introduce the conventional notation at the right moment.
Anyway... thanks for replying me... for your hints and specially, patience!!!

 
May 8th 2018, 12:48 PM

#8  Junior Member
Join Date: Apr 2018
Posts: 6

Originally Posted by studiot Why do you think the bolt and the elevator floor travel the same distance?
Hint : Afterall they start from different locations and arrive at a common location when?
Incidentally why do you not use conventional symbols for your kinematic variables? 
Honestly, I don't know why... For some reason, I was thinking about the high between the wall of the elevator shaft and the elevator... Trying to find some formula.... I don't know... I just diverged from the problem too much...
Regarding the conventional symbols... sorry for that... I'm taking a course and at this moment, they are using such symbols for 1D kinematics... I think they will probably introduce the conventional notation at the right moment.
Anyway... thanks for replying me... for your hints and specially, patience!!!

 
May 8th 2018, 12:59 PM

#9  Forum Admin
Join Date: Apr 2008 Location: On the dance floor, baby!
Posts: 2,554

I think I got all the extra copies in the thread.
@physicsnoobmaster (and anyone else reading that may have this problem)
In the future if you can't see your posts either for Moderation or other issues, please either report the post (a button in the lower left corner of the screen) or PM me. Don't just keep posting over and over again. It does nothing but make a mess for me.
Thanks!
Dan
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May 8th 2018, 06:44 PM

#10  Senior Member
Join Date: Apr 2017
Posts: 453

Originally Posted by physicsnoobmaster Honestly, I don't know why... For some reason, I was thinking about... 
I really don't think you were at fault here .
It seems so many questions are written badly and open to misinterpretation .
Not only is it frustrating to the student but gives them the example that it's OK to be sloppy and imprecise , the exact opposite of what science is about .

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