What is the force on a 1000 kg elevator that is falling freely at 9.8 m/sec2? And How do I draw a Free Body Diagram of this?

Answer 1

The only force on it is its weight, #w = 9800 N#.

The only force acting on it is its weight, w, as it is falling freely.

#w = m*g = 1000 kg*9.8 m/s^2 = 9800 N#

Draw an elevator cage with a downward force of 9800 N to create a free body diagram (I'm sure you'd get a lot of points for drawing it in such detail).

Hope this is helpful, Steve.

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Answer 2

The force on the elevator is its weight, given by the formula F = m * g, where m is the mass (1000 kg) and g is the acceleration due to gravity (9.8 m/s^2).

To draw a Free Body Diagram:

  1. Represent the elevator as a box.
  2. Draw an arrow pointing downward from the center of the box to indicate the force of gravity (mg).
  3. If there are other forces involved, like tension in a cable, represent them with additional arrows. In this case, only gravity is acting.
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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