Hooke's Law Elevator Spring Question

In summary: In general, when a force is applied to an object, the object will accelerate in the direction of the force. However, if you're standing on a spring, the upward force will cause the spring to lengthen, not decrease in length.
  • #1
MMVS
5
0

Homework Statement


Elevator initially at rest.
Equilibrium length L0=40.0cm
60-kg person stands on spring
L1= 32.0cm

The elevator than speeds upwards at 2.50 m/s2
What is the new length (L2)

Homework Equations


Fnet=ma
Fsp=-kdeltaX
FG=mg

The Attempt at a Solution


Taking down as positive y hat direction
Before:
Fnet=Fsp+FG
0=-k(0.32-0.40)+(60.0)(9.81)
K=-7358
After:
Fnet=Fsp+FG
ma=-kdeltaX+mg
m(-2.5)=-(-7358)(L2-0.40)+60(9.81)
L2=[{60(-2.5)-60(9.81)}/-(-7358)]+0.40
=0.30m from equilibrium

I am confused since i don't know how this makes sense according to the coordinate system. (I know the answer is right as my prof posted the answer, not the work, just answer)
At rest change in x = -0.08
When accelerating change in x= -0.10
But since my coordinate system is set up so Y-hat is positive is the downward direction doesn't this mean my spring is being displaces upward?
This doesn't make sense to me , can someone please explain very clearly?
 
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  • #2
MMVS said:
Fnet=Fsp+FG
0=-k(0.32-0.40)+(60.0)(9.81)
K=-7358

Take a look at your signs again. Are they consistent with the defined direction?
 
  • #3
JeremyG said:
Take a look at your signs again. Are they consistent with the defined direction?
I kinda see what you are getting at
 
  • #4
MMVS said:

Homework Statement


Elevator initially at rest.
Equilibrium length L0=40.0cm
60-kg person stands on spring
L1= 32.0cm

The elevator than speeds upwards at 2.50 m/s2
What is the new length (L2)

Homework Equations


Fnet=ma
Fsp=-kdeltaX
FG=mg

The Attempt at a Solution


Taking down as positive y hat direction
Before:
Fnet=Fsp+FG
0=-k(0.32-0.40)+(60.0)(9.81)
K=-7358
After:
Fnet=Fsp+FG
ma=-kdeltaX+mg
m(-2.5)=-(-7358)(L2-0.40)+60(9.81)
L2=[{60(-2.5)-60(9.81)}/-(-7358)]+0.40
=0.30m from equilibrium

I am confused since i don't know how this makes sense according to the coordinate system. (I know the answer is right as my prof posted the answer, not the work, just answer)
At rest change in x = -0.08
When accelerating change in x= -0.10
But since my coordinate system is set up so Y-hat is positive is the downward direction doesn't this mean my spring is being displaces upward?
This doesn't make sense to me , can someone please explain very clearly?
If you're standing in an elevator when it starts to go up, do you feel heavier or lighter when the car starts to move? Is the spring therefore going to get longer or shorter as a result?

You seem to have sprinkled a generous helping of negative signs throughout your calculations without systematically considering coordinate systems.
 

Related to Hooke's Law Elevator Spring Question

1. What is Hooke's Law?

Hooke's Law is a principle in physics that states that the force required to stretch or compress a spring is directly proportional to the distance the spring is stretched or compressed from its original length. This relationship is known as the spring constant and is denoted by the letter k.

2. How does Hooke's Law apply to elevator springs?

In an elevator, Hooke's Law applies to the spring that is used to counterbalance the weight of the elevator car and its occupants. As the elevator goes up or down, the spring stretches or compresses, respectively, in accordance with the amount of force applied to it. This allows for a smooth and controlled movement of the elevator.

3. What factors affect the spring constant in an elevator spring?

The spring constant of an elevator spring is affected by several factors, including the material and thickness of the spring, the number and arrangement of coils, and the amount of weight that the spring needs to support. A higher spring constant means that the spring is stiffer and requires more force to stretch or compress.

4. How does Hooke's Law affect the safety of an elevator?

Hooke's Law plays a crucial role in ensuring the safety of an elevator. If the spring constant is not properly calibrated, the elevator may not be able to support its maximum weight capacity or may move too quickly, causing discomfort or danger to passengers. Regular maintenance and monitoring of the spring are necessary to maintain its proper function and ensure the safety of elevator passengers.

5. Can Hooke's Law be applied to other types of springs besides elevator springs?

Yes, Hooke's Law can be applied to any elastic material, including other types of springs such as those used in mattresses, trampolines, and shock absorbers. It can also be applied to other materials such as rubber bands and bungee cords. Hooke's Law is a fundamental principle in physics and has many practical applications in various industries.

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