Elevator Problem: Find Scale Reading in Falling Elevator

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AI Thread Summary
In a scenario where an 80.0 kg person stands on a scale in a falling elevator at a constant speed of 3.5 m/s, the acceleration is zero. Consequently, the net force acting on the person is also zero, leading to the scale reading equal to the person's weight, which is 882 Newtons. However, the discussion reveals confusion about the correct application of formulas when acceleration is involved. It emphasizes the importance of defining a positive direction for acceleration and gravitational force when calculating net force. The conversation highlights the need for clarity in understanding the relationship between speed, acceleration, and scale readings in a falling elevator.
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Homework Statement



A 80.0 kg person stands on a scale in an elevator.
(c) What does it read when the elevator is falling at 3.5 m/s?

Homework Equations



EF = MA

The Attempt at a Solution



I can't even attempt the solution because I don't know acceleration...
 
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Toxage said:
A 80.0 kg person stands on a scale in an elevator.
(c) What does it read when the elevator is falling at 3.5 m/s?
…
I can't even attempt the solution because I don't know acceleration...

Hi Toxage! :smile:

If the speed is constant, then the acceleration is … ? :wink:
 
tiny-tim said:
Hi Toxage! :smile:

If the speed is constant, then the acceleration is … ? :wink:

Then the acceleration is equal to 0.

EF = MA

EF = M(0)

EF = 0

Fn - Fg = 0

Fn = Fg

80(9.8) = 882 Newtons

882/9.8 = 80kg

Doesn't work... Webassign says its the wrong answer...


Did I do something wrong in my math?
 
No your maths looks fine. What is the quoted answer? If your standing on scales and not accelerating the scales will tell you your weight. When you are accelerating use this formula:

F_{net} = m(a + g)

But MAKE SURE YOU DEFINE A POSITIVE DIRECTION AND STICK TO IT. So if you define up as positive, g = -9.81. And then when accelerating up, a is positive and vice versa.
 

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