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Homework Help: Bathroom scale

  1. Mar 1, 2005 #1


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    Here's a question. I don't have any problem getting the answer, or identifying the forces, but I'm having trouble explaining why the scale reads what it does.

    A woman of mass 65 kg stands inside an elevator on a bathroom scale calibrated to read in Newtons. Calculate the scale reading in each of the following situations and explain in terms of forces acting on the scale why it reads as it does.

    As I work these, it's as if a scale reads only the force pushing up on it, rather than down.

    a) elevator stationary

    65 * 9.8 = 637
    force on the scale:
    mg down
    mg up

    the forces balance and the scale does not move. But with a net force of 0, why does it read 637?

    b) elevator accelerating upward at 2.0 m/s^2

    65 * 11.8 = 767 N
    forces on the scale:
    mg = 637 N down
    mg + m *2.0 = 767 N up

    total forces = 130 N up
    so to check, f=ma, 130 = 65 a, a = 2

    So why does the scale read 767 N? Does a scale just read the highest force pushing on either the bottom or the top?

    c) elevator accelerating downward at 2.0 m/s^2
    65 * 7.8 = 507N

    Forces on scale
    mg down, 637 N
    mg - m * 2.0 = 507 N up

    d) elevator decending with constant velocity
    same as part a
    e) elevator in freefall after the cable breaks
    0 N

    mg = 637 down
    mg - m*9.81 = 0 N up
    Last edited: Mar 2, 2005
  2. jcsd
  3. Mar 2, 2005 #2
    Because the woman is exerting a force of 637 newtons on the scale.

    You've answered your own question, yes the scale just reads the force, which is given my F=ma, which is just F = 65 * (9.8+2) = 767 N

    yes the scale again just reads the force, F=ma, which is F = 65 * (9.8-2) = 507 N

    Essentially all the scale does is read the force pushing on it.
  4. Mar 2, 2005 #3


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    But it's not just total force. There's always 2 forces acting on the scale, the force down, and the force up. Just like a book sitting on a table has mg down, and force normal (-mg) up. They balance, it does not move.

    I have to express my answer acknowledging both forces on the scale, not just the net force. The net force in example (a) is 0, mg down + (mg) up
  5. Mar 2, 2005 #4


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    You still seem confused about this. Erienion answered...twice:

    The scale reads the force pushing on it...i.e. how heavy the person/object standing on the scale's platform is. The total force on the scale is not really relevant.
  6. Mar 2, 2005 #5


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    The question asks explain in terms of forces acting on the scale
    where forces is plural.

    So I can't just say that the force acting on it is 637 N. Because there are 2 forces acting on it.

    In the case of the elevator sitting still, there is the 637 N of the woman standing on the scale, pushing down. But there is also 637 N on the scale of the floor pushing up. So there is a net force of 0. But the scale does not read 0.

    So for each question, (a-e) I have to talk about both the force pushing down on the scale, which is the woman, and the force pushing up on the scale, which is the elevator floor.

    This is what confuses me. Intuitively, I know what the scale will read, because intuitively I know that an elevator accelerating up at 2.0 m/s^2 will be just like standing on a planet where g = 11.8 m/s^2.

    But I dont know how to say:

    The force pushing down is ...
    The force pushing up is...
    Therefore, the scale reads...
  7. Mar 2, 2005 #6


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    Arrrgghhh! Not again! :rofl:

    Look...if you look at a free body diagram of the scale, then you will see that the net force on it is zero. FINE. The only thing that will help you to explain is why the scale is not going anywhere! BUT...a scale is not designed to show you the net force acting on it. If it were, it would always read zero (unless your bathroom is in the habit of accelerating up and down :rolleyes: )

    What does the reading on the scale show you? It shows you the force pushing down on that spring mounted platform or whatever the heck it is on the scale's surface that is designed to be depressed when somebody stands on it. Period. What is that equal to? It is equal to the force with which the person pushes down on the scale. Which is equal (in magnitude) to the force with which the scale pushes up on the person. So in all three cases, the scale reading is equal to the NORMAL force pushing up on the person's feet. Period. Full stop. So when they said to determine the reading of the scale in terms of the forces acting on it, they must have meant to consider only the apparent weight of the person pushing down on the scale, otherwise their instructions make no sense. That's perfectly justified if you think of the scale as part of the elevator floor (so you can't say the floor is pushing back up on it :wink: ), and like I said, just answer the question: with what force is the floor pushing up on the person's feet? You really need to look at all the forces acting on the person and find out what that is in each of the three cases. I think you did that.

    You would then say:

    The force (of the person) pushing down is....
    Therefore the scale reads....

  8. Mar 2, 2005 #7


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    That's a missinterpretation of Newton's third principle.The two forces acting in a two-body interaction are APPLIED ON DIFFERENT BODIES."mg" of the book sitting on the table is not becuase of the interaction book-table,but beause of interaction book-Earth.The book is pressing the table with a force.This is another force than gravity.In this case (in which the book is not acted by any other external force),it's simply equal to the force Earth exerts.The table acts on the book with the same force,but that force (called "reaction force is acting on the book).

    So that's the picture of the interaction book-table...

  9. Mar 2, 2005 #8


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    #1) Yes, there are two forces- one up, one down. They are the SAME, but oppositely directed so that the net force is 0, precisely because there is no acceleration. The scale reading is the upward force the scale is exerting on the person standing on the scale.
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