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

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In summary, the force exerted by the scale on the 56.0 kg girl in the elevator is equal to her weight when the elevator is descending at a constant speed of 10 m/s. If the elevator is accelerating downward with an acceleration of 2.4 m/s2, the force exerted by the scale is given by F=(m*-a)+w. When 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, the force exerted by the scale is equal to her weight, but with an acceleration in the opposite direction. This can be determined by drawing a free body diagram and applying the equation \sum F_y
  • #1
BuBbLeS01
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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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  • #2
In all cases you are given the acceleration (or all the information needed to figure it out). You don't need the time.
 
  • #3
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 [tex] \sum F_y = m a_y [/tex]. That is all there is to it, really. Just be careful to include the correct sign of [tex] a_y [/tex].
 
  • #4
Okay so I have F=(m*-a)+w for the second part
 
  • #5
Now for the first part I am not sure what to do with the speed??
 
  • #6
BuBbLeS01 said:
Now for the first part I am not sure what to do with the speed??
All you care about is acceleration.
 
  • #7
I don't understand
 
  • #8
Reread the statement of the first problem. How is the velocity changing?
 
  • #9
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!
 

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

What is the formula for calculating net force?

The formula for calculating net force is Fnet = ma, where Fnet is the net force, m is the mass of the object, and a is the acceleration.

How do you apply the formula to solve elevator problems?

To solve elevator problems using the formula Fnet = ma, you first need to identify the mass of the elevator and the acceleration due to gravity, which is approximately 9.8 m/s2. Then, you can substitute these values into the formula to calculate the net force acting on the elevator.

What is the significance of the values 10m/s and 2.4 m/s2 in this equation?

The value of 10m/s represents the initial velocity of the elevator, while 2.4 m/s2 represents the acceleration of the elevator. These values are used to determine the net force acting on the elevator and ultimately, whether or not the elevator is moving at a constant velocity or accelerating/decelerating.

Can this formula be used to solve other types of problems besides elevator problems?

Yes, the formula Fnet = ma can be used to solve a wide range of problems involving motion and forces, such as calculating the net force acting on a moving car or a falling object.

How can the net force calculated from this equation be applied in real life?

The net force calculated from this equation can be used to understand and predict the motion of objects in real life. For example, using this formula, engineers can determine the necessary force and acceleration for an elevator to reach a certain floor within a specific time frame.

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