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One other question i need help with please

  1. Apr 22, 2008 #1
    one other question i need urgent help with please

    1. The problem statement, all variables and given/known data
    A person standing on a bathroom scale in an elevator notices that when the elevator starts to move their weight decreases to 0.73 of its original reading. What is the acceleration of the elevator?


    2. Relevant equations



    3. The attempt at a solution
    I have no idea on where to get started with this one, i dont understand how the information provided can help work out the answer
     
  2. jcsd
  3. Apr 23, 2008 #2

    alphysicist

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    Hi inflames829,

    How did they know that their weight decreases to 0.73 of what is was before? What gave them that information? That will tell you which of the two forces that is acting on this person is actually changing here.

    Then draw force diagrams for the case of no elevator motion and for the case of the elevator moving. What relationships between forces and accelerations do you get?
     
  4. Apr 23, 2008 #3
    Thats all th information the question gave me. Using a force diagram i see that the sum of forces in the downward direction is 0 when the elevator is not moving and by my working out it says its 7.154 when it is moving down. I dont know if this is what you mean or if its right, I just tried to get the sum of the forces in the downward direction which is equal to the mass multiplied by the acceleration in that direction.
     
  5. Apr 23, 2008 #4

    alphysicist

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    In the first diagram, with the elevator is not moving, what two forces are acting on the man? Since the acceleration is zero, they must be cancelling so they must be equal in magnitude.

    Then in the second diagram, you have the same two foces, but one of them is smaller. What is Newton's law ([itex]F_{\rm net}=ma[\itex]) for this situation?
     
  6. Apr 23, 2008 #5
    equal and opposite forces? but i dont see how i can use the numbers given to get a numerical answer
     
  7. Apr 23, 2008 #6

    alphysicist

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    Right; they are equal and opposite. But what are they? They are the graviational force and the normal force from the scale. What you know for the case of no motion is that N=mg since the acceleration is zero.

    Now look at the moving case. You have the same forces: normal and gravitational. But which one is now smaller, and by how much? Once you put those forces into Newton's law we can get the acceleration.

    So what do you get for the second diagram when the elevator is in motion?
     
  8. Apr 23, 2008 #7
    [​IMG]

    If you are accelerating upward you feel heavier, and if you are accelerating downward you feel lighter. If the elevator cable broke, you would feel weightless since both you and the elevator would be accelerating downward at the same rate.
     
  9. Apr 24, 2008 #8
    doe sthat mean i put 0.73 as the mass and 9.8 as g? will the total of that be the acceleration
     
  10. Apr 24, 2008 #9

    alphysicist

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    No; the mass is not changed. What changes when the elevator accelerates is the normal force from the bathroom scale. It was equal to mg when the elevator was not moving. What is the normal force now that the elevator is moving?
     
  11. Apr 24, 2008 #10
    the normal force is equal to what it was originally multiplied by 0.73?
     
  12. Apr 24, 2008 #11

    alphysicist

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    Yes, that's right; so now with the elevator in motion the normal force is N=0.73 mg.

    Now Newton's law says the sum of the forces in some direction equals the mass time the acceleration in that direction. (Here the only direction required is vertical.) Write out Newton's law for these two forces (normal and gravitational) and set it equal to ma. Then solve for a. What do you get?

    Be sure to check for the right sign for the force components.
     
  13. Apr 24, 2008 #12
    but what do i put as the mass value to work out a
     
  14. Apr 24, 2008 #13
    or do i do 0.73 x mg=ma then work it out?
     
  15. Apr 24, 2008 #14

    alphysicist

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    You're missing the gravitational force on the left side of the equation. Since there are two forces acting you'll need two terms on the left side. The weight is equal to mg, and is downwards. So what does the left side become?
     
  16. Apr 24, 2008 #15
    a vector?
     
  17. Apr 24, 2008 #16

    alphysicist

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    I mean what is your new equation? If you have two forces, then newton's law in the y direction is:

    [tex] F_{1,y} + F_{2,y} = m a_y[/tex]

    where the force components on the left hand side can be positive or negative depending on their direction. But the equation you had written on post #13 only had the normal force on the left hand side.

    Once you include the graviational force in your force equation (with the proper sign), then you can solve for a.
     
  18. Apr 24, 2008 #17
    do i have to work it out using vectors? with that being the downward one
     
  19. Apr 24, 2008 #18
    dont pay attention to that last reply that was from before
     
  20. Apr 24, 2008 #19
    so now i have (0.73 x mg) which is the Normal force, then i have mg as the gravitational force so if i add those together do i get ma? Then if I rrearrange m to the other side does that cancel the other m's? and is that the final answer?
     
  21. Apr 24, 2008 #20

    alphysicist

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    That's right; but remember that the normal force is pushing upwards on the person and you had that as positive (so you are choosing upwards as the positive direction) and the gravitational force is downwards so it will be negative. Adding those positive and negative values together equals ma.

    What equation do you get?

    Then solve for a. If the number you get for a is positive, the elevator is accelerating upwards; if it is negative it is accelerating downwards.
     
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