Calculating Time of Fall: Elevator Problem

In summary, the problem involves an elevator ascending with a uniform velocity of 2 m/s and a loose bolt falling from the ceiling to the floor. The goal is to find the time required for the bolt to fall from the ceiling to the floor, with a given value of acceleration due to gravity (g=10 m/s2). The attempt at a solution involves using the relative velocity of the bolt and the equation s=ut+1/2 gt2, where s is the displacement, u is the initial velocity, and t is the time. The solution involves putting u=-2 m/s, as the bolt had an initial upward velocity, while the displacement is in the downward direction. However, it is also valid to use the
  • #1
Prathamesh
20
1

Homework Statement


An elevator is ascending with a uniform velocity 2 m/s when a loose bolt gets detached from the ceiling and falls on the floor. If the ceiling is 3 m above the floor , find the time required by the bolt to fall from the ceiling to the floor? (g=10 m/s2)

Homework Equations

The Attempt at a Solution


Relative velocity of bolt with respect to floor before detaching=0 m/s Hence u = 0 m/s
So ,
s=ut + 1/2 at2
=1/2 gt2
Hence , t=0.77 s
but solution says that
put u= -2 m/s
Why this is so??
 
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  • #2
Perhaps the textbook is solving this from the viewpoint of motion relative to the building, i.e., relative to the building's basement?
 
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  • #3
Prathamesh said:

Homework Statement


An elevator is ascending with a uniform velocity 2 m/s when a loose bolt gets detached from the ceiling and falls on the floor. If the ceiling is 3 m above the floor , find the time required by the bolt to fall from the ceiling to the floor? (g=10 m/s2)

Homework Equations

The Attempt at a Solution


Relative velocity of bolt with respect to floor before detaching=0 m/s Hence u = 0 m/s
So ,
s=ut + 1/2 at2
=1/2 gt2
Hence , t=0.77 s
but solution says that
put u= -2 m/s
Why this is so??
The bolt had an upward initial velocity, while the displacement is in the downward direction.
 
  • #4
Prathamesh said:

Homework Statement


An elevator is ascending with a uniform velocity 2 m/s when a loose bolt gets detached from the ceiling and falls on the floor. If the ceiling is 3 m above the floor , find the time required by the bolt to fall from the ceiling to the floor? (g=10 m/s2)

Homework Equations

The Attempt at a Solution


Relative velocity of bolt with respect to floor before detaching=0 m/s Hence u = 0 m/s
So ,
s=ut + 1/2 at2
=1/2 gt2
Hence , t=0.77 s
but solution says that
put u= -2 m/s
Why this is so??
It's perfectly valid and preferable to use the frame of reference of the lift, as you did. Using the ground frame makes u=-2ms/s, but then the rest of the problem gets messier because you have to take into account the moving floor. If the textbook gets a different answer it is wrong.
 
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  • #5
Prathamesh said:

Homework Statement


An elevator is ascending with a uniform velocity 2 m/s when a loose bolt gets detached from the ceiling and falls on the floor. If the ceiling is 3 m above the floor , find the time required by the bolt to fall from the ceiling to the floor? (g=10 m/s2)

Homework Equations

The Attempt at a Solution


Relative velocity of bolt with respect to floor before detaching=0 m/s Hence u = 0 m/s
So ,
s=ut + 1/2 at2
=1/2 gt2
Hence , t=0.77 s
but solution says that
put u= -2 m/s
Why this is so??

Since the velocity of the elevator is constant, all laws of Physics (and of mechanics, in particular) are the same as if the elevator were not moving at all! In other words, from the point-of-view of measurements made inside the elevator, you cannot tell if it is moving or stationary. Therefore, you can neglect the elevator's motion in this question, and treat it as a simple problem of a bolt falling from ceiling to floor in your bedroom.

For more on that "not moving" analogy, see, eg.,
http://psi.phys.wits.ac.za/teaching/Connell/phys284/2005/lecture-01/lecture_01/node5.html
which concerns the so-called Galilean invariance of mechanical laws.
 
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  • #6
Ray Vickson said:
Since the velocity of the elevator is constant, all laws of Physics (and of mechanics, in particular) are the same as if the elevator were not moving at all! In other words, from the point-of-view of measurements made inside the elevator, you cannot tell if it is moving or stationary. Therefore, you can neglect the elevator's motion in this question, and treat it as a simple problem of a bolt falling from ceiling to floor in your bedroom.
As I read the OP, that's what Prathamesh did. He asks why the book didn't. What is not clear is whether the book arrived at a different answer.
 
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  • #7
Thank you very much haruspex and ray vickson...
I just wanted to make sure that i m going in r8 direction
 

1. How do you calculate the time of fall for an elevator?

The time of fall for an elevator can be calculated by using the formula t = √(2h/g), where t is the time in seconds, h is the distance the elevator falls in meters, and g is the acceleration due to gravity (9.8 m/s²).

2. What is the acceleration of an elevator?

The acceleration of an elevator is typically equal to the acceleration due to gravity, which is 9.8 m/s². However, this can vary depending on the design and operation of the elevator.

3. Does the weight of the elevator affect the time of fall?

Yes, the weight of the elevator does affect the time of fall. According to the formula t = √(2h/g), the time of fall is inversely proportional to the acceleration due to gravity. This means that a heavier elevator will have a longer time of fall compared to a lighter elevator, assuming all other factors are constant.

4. How can you determine the distance the elevator falls?

The distance the elevator falls can be determined by using the formula h = ½gt², where h is the distance in meters, g is the acceleration due to gravity, and t is the time in seconds. Simply plug in the time of fall and solve for h to find the distance the elevator falls.

5. Can you use the same equation to calculate the time of fall for an elevator going up?

No, the equation t = √(2h/g) is specifically for calculating the time of fall for an elevator going down. To calculate the time of fall for an elevator going up, a different equation must be used, taking into account the initial velocity and acceleration due to gravity.

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