Solving Elevator Problems w/ Fnet=ma: 10m/s & 2.4 m/s2

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In the discussion about the forces acting on a girl standing on a scale in a descending elevator, key points include the understanding of constant speed versus acceleration. When the elevator descends at a constant speed of 10 m/s, the net force is zero, and the scale reads her weight. However, when the elevator accelerates downward at 2.4 m/s², the scale reads less than her weight due to the additional downward acceleration. The conversation emphasizes the importance of recognizing that even when moving downward, a decreasing speed indicates an upward acceleration, affecting the forces at play. Overall, correctly interpreting acceleration and direction is crucial for solving these problems.
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Homework Statement


A 56.0 kg girl weighs herself by standing on a scale in an elevator. What is the force exerted by the scale when the elevator is descending at a constant speed of 10 m/s?

What is the force exerted by the scale if the elevator is accelerating downward with an acceleration of 2.4 m/s2?

If the elevator's descending speed is measured at 10 m/s at a given point, but its speed is decreasing by 2.4 m/s2, what is the force exerted by the scale?



Homework Equations


Fnet=ma

How do we find acceleration without having a time?
 
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In all cases you are given the acceleration (or all the information needed to figure it out). You don't need the time.
 
BuBbLeS01 said:

Homework Statement


A 56.0 kg girl weighs herself by standing on a scale in an elevator. What is the force exerted by the scale when the elevator is descending at a constant speed of 10 m/s?

What is the force exerted by the scale if the elevator is accelerating downward with an acceleration of 2.4 m/s2?

If the elevator's descending speed is measured at 10 m/s at a given point, but its speed is decreasing by 2.4 m/s2, what is the force exerted by the scale?



Homework Equations


Fnet=ma

How do we find acceleration without having a time?

Draw a free body diagram and then apply \sum F_y = m a_y. That is all there is to it, really. Just be careful to include the correct sign of a_y.
 
Okay so I have F=(m*-a)+w for the second part
 
Now for the first part I am not sure what to do with the speed??
 
BuBbLeS01 said:
Now for the first part I am not sure what to do with the speed??
All you care about is acceleration.
 
I don't understand
 
Reread the statement of the first problem. How is the velocity changing?
 
it's descending at 10 m/s
 
  • #10
BuBbLeS01 said:
it's descending at 10 m/s
Acceleration is the rate of change of velocity. How is the velocity changing? (Hint: Read it carefully. :wink:)
 
  • #11
it's not its constant
 
  • #12
BuBbLeS01 said:
it's not its constant
Exactly! So what is the acceleration?
 
  • #13
0 m/s^2
 
  • #14
BuBbLeS01 said:
0 m/s^2
You got it.
 
  • #15
ohhh so the only force acting is the weight.
 
  • #16
BuBbLeS01 said:
ohhh so the only force acting is the weight.
No. If the only force acting on her was her weight, the girl would be in free fall.
 
  • #17
so how is the third part any different from the second if the acceleration is the same?
 
  • #18
BuBbLeS01 said:
so how is the third part any different from the second if the acceleration is the same?
The accelerations are not the same. Direction counts!
 
  • #19
but they are both going down?
 
  • #20
BuBbLeS01 said:
but they are both going down?
No. The second one has acceleration going down (it tells you that), but the third one you have to figure out the direction of acceleration by reading carefully.
 
  • #21
ok so since the speed is decreasing then the force is acting to slow it down and bring the elevator eventually to a stop?
 
  • #22
BuBbLeS01 said:
ok so since the speed is decreasing then the force is acting to slow it down and bring the elevator eventually to a stop?
Sure. And which way must it be accelerating, since it happens to be moving downward as it slows?
 
  • #23
it is accelerating up
 
  • #24
BuBbLeS01 said:
it is accelerating up
Right!
 

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