Can anyone help me with a basic physics question.
Suppose a 75Kg person is standing in an elevator and the elevator is accelerating upwards at 20m/s^2 before stopping.
How do I work out the acceleration required to lift the person off the elevator floor?
Sorry, I realise that is not info info.
Forget that.
What if a 75Kg person standing was standing inside the elevator, and the elevator moved upwards of just 10mm in one quick motion.
How do I work out the acceleration necessary to lift the person off the elevator floor?
Forget that.
What if a 75Kg person standing was standing inside the elevator, and the elevator moved upwards of just 10mm in one quick motion.
How do I work out the acceleration necessary to lift the person off the elevator floor?
one quick motion = 0.5 seconds
Sorry again, physics is not my thing
Obviously the time in which it moves upwards of 10mm dictates the acceleration, so time is an unknown.
So we have an elevator at rest, with someone inside at 75Kg. The elevator suddenly moves upwards of just 10mm.
I'm assuming there needs to be a large acceleration followed by a large deceleration.
How do I work that out relative to the persons weight?
Hopefully this is the last time I reply to myself.
Obviously the time in which it moves upwards of 10mm dictates the acceleration, so time is an unknown.
So we have an elevator at rest, with someone inside at 75Kg. The elevator suddenly moves upwards of just 10mm.
I'm assuming there needs to be a large acceleration followed by a large deceleration.
How do I work that out relative to the persons weight?
Hopefully this is the last time I reply to myself.
QUOTE (hardlines+Jun 27 2009, 09:15 AM)
What if a 75Kg person standing was standing inside the elevator, and the elevator moved upwards of just 10mm in one quick motion.
How do I work out the acceleration necessary to lift the person off the elevator floor?
It depends on how far off the floor you want to lift him.Any such acceleration would lift him, it's just that you might not notice it if you were watching.
How do I work out the acceleration necessary to lift the person off the elevator floor?
It depends on how far off the floor you want to lift him.Any such acceleration would lift him, it's just that you might not notice it if you were watching.
The upwards acceleration just presses the man down against the floor - it's the downwards acceleration that lifts him off the floor. You can imagine cutting the rope holding the lift up - the man still doesn't actually leave the floor - so you need a downwards acceleration of >g (~9.81m/s/s) to lift him off the floor.
I'm guessing this was a 'made up' question - not from a physics or engineering textbook - am I right?
-C2.
Edit .. the weight of the person makes no difference to the answer.
I'm guessing this was a 'made up' question - not from a physics or engineering textbook - am I right?
-C2.
Edit .. the weight of the person makes no difference to the answer.
QUOTE (Confused2+Jun 27 2009, 03:48 PM)
The upwards acceleration just presses the man down against the floor - it's the downwards acceleration that lifts him off the floor. You can imagine cutting the rope holding the lift up - the man still doesn't actually leave the floor - so you need a downwards acceleration of >g (~9.81m/s/s) to lift him off the floor.
I'm guessing this was a 'made up' question - not from a physics or engineering textbook - am I right?
-C2.
Edit .. the weight of the person makes no difference to the answer.
Good one!
I'm guessing this was a 'made up' question - not from a physics or engineering textbook - am I right?
-C2.
Edit .. the weight of the person makes no difference to the answer.
QUOTE (hardlines+)
What if a 75Kg person standing was standing inside the elevator, and the elevator moved upwards of just 10mm in one quick motion.
He's referring to the momentum imparted by a sudden upward surge.
As I already said, the amount of acceleration depends on how high you want to lift the person.
He's referring to the momentum imparted by a sudden upward surge.
As I already said, the amount of acceleration depends on how high you want to lift the person.
QUOTE (Confused2+Jun 27 2009, 08:48 PM)
The upwards acceleration just presses the man down against the floor - it's the downwards acceleration that lifts him off the floor. You can imagine cutting the rope holding the lift up - the man still doesn't actually leave the floor - so you need a downwards acceleration of >g (~9.81m/s/s) to lift him off the floor.
I'm guessing this was a 'made up' question - not from a physics or engineering textbook - am I right?
-C2.
Edit .. the weight of the person makes no difference to the answer.
Imagine you have a tennis ball resting on the strings of a tennis racquet, if you apply an upward force, then suddenly stop, won't the ball lift off the strings due to inertia?
Wouldn't that same ball be require more acceleration if it was heavier?
I guess another example could be applying the brakes in the car and our body continues forward against the deceleration of the car, only in my elevator example the same is happening vertically, where in addition, gravity is applying a downward force to keep our feet planted on the floor.
Is there not an equation to work out how much acceleration followed by deceleration is required to lift a person in such a situation?
I'm guessing this was a 'made up' question - not from a physics or engineering textbook - am I right?
-C2.
Edit .. the weight of the person makes no difference to the answer.
Imagine you have a tennis ball resting on the strings of a tennis racquet, if you apply an upward force, then suddenly stop, won't the ball lift off the strings due to inertia?
Wouldn't that same ball be require more acceleration if it was heavier?
I guess another example could be applying the brakes in the car and our body continues forward against the deceleration of the car, only in my elevator example the same is happening vertically, where in addition, gravity is applying a downward force to keep our feet planted on the floor.
Is there not an equation to work out how much acceleration followed by deceleration is required to lift a person in such a situation?
QUOTE (hardlines+Jun 27 2009, 04:28 PM)
Is there not an equation to work out how much acceleration followed by deceleration is required to lift a person in such a situation?
There is not. There is an equation to determine how far an object of a given weight will continue to travel once the elevator stops in such a situation.
Let me repeat myself: Any acceleration on the part of the elevator will impart momentum to the person. When the elevator stops, the person will continue for a short while afterwords. It doesn't matter how quickly the elevator accelerated, how heavy the person is, how long the elevator accelerated for. The elevator doesn't even need to accelerate the whole time, it can accelerate then travel at a steady speed for any length of time before it stops.
There is not. There is an equation to determine how far an object of a given weight will continue to travel once the elevator stops in such a situation.
Let me repeat myself: Any acceleration on the part of the elevator will impart momentum to the person. When the elevator stops, the person will continue for a short while afterwords. It doesn't matter how quickly the elevator accelerated, how heavy the person is, how long the elevator accelerated for. The elevator doesn't even need to accelerate the whole time, it can accelerate then travel at a steady speed for any length of time before it stops.
If you are proposing 'stopping' as an instantaneous change of velocity then the acceleration would be infinite .
Clearly Inf > g and (as already explained) the man would leave he floor.
Clearly Inf > g and (as already explained) the man would leave he floor.
For obvious reasons - elevator designers (and users) try to avoid large jerks.
QUOTE (Confused2+Jun 28 2009, 02:25 AM)
If you are proposing 'stopping' as an instantaneous change of velocity then the acceleration would be infinite .
It would be better to phrase it as "the deceleration would be infinite." The acceleration that imparted the initial upwards impetus would still be finite.
That's pretty much what I've been saying this whole time.
It would be better to phrase it as "the deceleration would be infinite." The acceleration that imparted the initial upwards impetus would still be finite.
QUOTE
Clearly Inf > g and (as already explained) the man would leave he floor.
That's pretty much what I've been saying this whole time.
To assist you in your first question, i would first suggest drawing some free body diagrams of all forces this person is subject to. Secondly, in order to be lifted up, you want your normal force = 0. So i think while neglecting air friction (m)(g) - (m)(a) = 0 , you will find the acceleration required to lift one's body off of the floor.
That is how i would go about solving your question anyways. Depending on your different variables you wish to implement into the problem, you might have to split this up into a couple steps, or if you plan on adding air friction you would have to calculate a couple other things i think.
Also, accelerating upwards, as mentioned will apply further force on the person, which you could calculate in G's. Too lift off the ground, one must essentially be in a state of freefall (Fn = 0).
For your 2nd question (neglecting air friction), moving upward you would use (in simplest terms):
Fn =(m)(g)+(m)(a)
To convert from your distance of 10mm (.01m) you would require enough information to get the acceleration. I would opt to use
d = (1/2)(a)(t^2) , as timing one's self is simple enough.
From there , use that as your acceleration in the formula.
Please correct me if i've made an error, i just quickly thought these up
!
That is how i would go about solving your question anyways. Depending on your different variables you wish to implement into the problem, you might have to split this up into a couple steps, or if you plan on adding air friction you would have to calculate a couple other things i think.
Also, accelerating upwards, as mentioned will apply further force on the person, which you could calculate in G's. Too lift off the ground, one must essentially be in a state of freefall (Fn = 0).
For your 2nd question (neglecting air friction), moving upward you would use (in simplest terms):
Fn =(m)(g)+(m)(a)
To convert from your distance of 10mm (.01m) you would require enough information to get the acceleration. I would opt to use
d = (1/2)(a)(t^2) , as timing one's self is simple enough.
From there , use that as your acceleration in the formula.
Please correct me if i've made an error, i just quickly thought these up
!
QUOTE (Michael J+Jun 28 2009, 11:44 PM)
To assist you in your first question, i would first suggest drawing some free body diagrams of all forces this person is subject to. Secondly, in order to be lifted up, you want your normal force = 0. So i think while neglecting air friction (m)(g) - (m)(a) = 0 , you will find the acceleration required to lift one's body off of the floor.
That is how i would go about solving your question anyways. Depending on your different variables you wish to implement into the problem, you might have to split this up into a couple steps, or if you plan on adding air friction you would have to calculate a couple other things i think.
Also, accelerating upwards, as mentioned will apply further force on the person, which you could calculate in G's. Too lift off the ground, one must essentially be in a state of freefall (Fn = 0).
For your 2nd question (neglecting air friction), moving upward you would use (in simplest terms):
Fn =(m)(g)+(m)(a)
To convert from your distance of 10mm (.01m) you would require enough information to get the acceleration. I would opt to use
d = (1/2)(a)(t^2) , as timing one's self is simple enough.
From there , use that as your acceleration in the formula.
Please correct me if i've made an error, i just quickly thought these up !
Crap... mjolnirpants you are right, i didn't read all the thread, just skimmed through it.
Momentum yes, that is what you would use to calculate it out. duh, i'm dumb, sorry for not fully understanding the OP!
Or could you still use force to figure this out, and end it off with some kinematics?
Umm, im trying to picture this...
F of elevator transferred with 100% efficiency to person. Then gravitational acceleration minus the acceleration of the force transferred to the person inserted into maybe 0=Vo^2 + 2ad ? However you would also require the initial velocity before the sudden stop i think. V final = 0 as that is the point which the person stops mid air due to gravitational deceleration. So that would give you distance he/she flies up i'm guessing?
If you already know that, you could reverse the steps to find the required acceleration to reach X distance?
I need to think this one over, i remember doing a question like this on a worksheet, i cannot remember the method i used to figure it out...
edit again:
Nope, I think this makes more sense with momentum, unless using energy to figure this out?
kinetic right after the "jolt" = potential, as transfer will be 100% upon 0 velocity being reached mid air. This could be used again to calculate height reached, or if height is given, then the velocity required for the person to be lifted.
That is how i would go about solving your question anyways. Depending on your different variables you wish to implement into the problem, you might have to split this up into a couple steps, or if you plan on adding air friction you would have to calculate a couple other things i think.
Also, accelerating upwards, as mentioned will apply further force on the person, which you could calculate in G's. Too lift off the ground, one must essentially be in a state of freefall (Fn = 0).
For your 2nd question (neglecting air friction), moving upward you would use (in simplest terms):
Fn =(m)(g)+(m)(a)
To convert from your distance of 10mm (.01m) you would require enough information to get the acceleration. I would opt to use
d = (1/2)(a)(t^2) , as timing one's self is simple enough.
From there , use that as your acceleration in the formula.
Please correct me if i've made an error, i just quickly thought these up
! Crap... mjolnirpants you are right, i didn't read all the thread, just skimmed through it.
Momentum yes, that is what you would use to calculate it out. duh, i'm dumb, sorry for not fully understanding the OP!
Or could you still use force to figure this out, and end it off with some kinematics?
Umm, im trying to picture this...
F of elevator transferred with 100% efficiency to person. Then gravitational acceleration minus the acceleration of the force transferred to the person inserted into maybe 0=Vo^2 + 2ad ? However you would also require the initial velocity before the sudden stop i think. V final = 0 as that is the point which the person stops mid air due to gravitational deceleration. So that would give you distance he/she flies up i'm guessing?
If you already know that, you could reverse the steps to find the required acceleration to reach X distance?
I need to think this one over, i remember doing a question like this on a worksheet, i cannot remember the method i used to figure it out...
edit again:
Nope, I think this makes more sense with momentum, unless using energy to figure this out?
kinetic right after the "jolt" = potential, as transfer will be 100% upon 0 velocity being reached mid air. This could be used again to calculate height reached, or if height is given, then the velocity required for the person to be lifted.
Thanks for all the help guys, this was actually helping me with some homework.
A little late, but this does cause me to reflect on my visit to the Sear's Tower years ago. It was the fastest elevator I personally had ever been on. On the way up, one feels more pressed to the floor. It almost seemed like a slight jerk where the sensation was most noticeable when the elevator just started. On the way down, I felt lighter, the sensation being most noticed as the elevator just started. When it stopped, it seemed I was heavier suddenly for just a second, then normal. The beginnings and endings of the ride have a startling feel. On the way down, my innards felt a little strange, like they moved up higher in my body.
Sears Tower was in the news today. If you like to feel your physics, it is a great place to visit in Chicago.
http://www.msnbc.msn.com/id/21134540/vp/31698179#31698179
Sears Tower was in the news today. If you like to feel your physics, it is a great place to visit in Chicago.
http://www.msnbc.msn.com/id/21134540/vp/31698179#31698179
How are you doin' guys? you can call me pauline
I am a new member of this forum and I hope I'll have the chance to interact with you all!
have a nice day!
I am a new member of this forum and I hope I'll have the chance to interact with you all!
have a nice day!
You are not in the right place for introductions (There is an intro thread), but nonetheless, welcome. I hope you enjoy it here.
QUOTE (Confused2+Jun 28 2009, 10:06 AM)
For obvious reasons - elevator designers (and users) try to avoid large jerks.
Exactly, sometimes if there is a large jerk waiting to use the elevator I will take the stairs.
Exactly, sometimes if there is a large jerk waiting to use the elevator I will take the stairs.
Good one!
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