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Another force and acceleration problem

  1. Apr 24, 2006 #1
    In the picture above, a person is standing on a forcemeter (normally called a "scale") like those found in many bathrooms. This metric forcemeter reads the amount of upward force applied to a person's feet in Newtons. In other words, the reading on the scale is exactly equal to the normal force. Note that the upward normal force will equal the downward force of gravity only when their is no vertical acceleration. The situation is on Earth.

    If the mass of the person is 67 kg, what would be the reading on the scale as the upward moving elevator slows from its cruising speed of 8.2 m/sec, coming to a stop 2.3 seconds after beginning to slow down?

    I initially thought that I would find the acceleration which is v final- v initial / time. That got me -8.2/2.3 =-3.565 approx. Then I did F= ma so F=67 kg x -3.565 m/s^2=-238.855. This appears to be incorrect. What did I do wrong? What do I need to do?
     
  2. jcsd
  3. Apr 24, 2006 #2
    OK, think about it this way...
    There's more than one force acting on the person-- and acceleration causes a force, correct?
    Now, by convention, the sum of all forces (the net force-- the NET effect) = ma
    According to your question, there are forces acting on a person.... the normal force N and mg... N acting up and mg acting down. Also, it says that the elevator the person is in slows down with that acceleration you found-- the NET acceleration.
    Thus, you can write:
    N - m|g| = -m|a|
    And solve for N.
    Also, there's a - in front of the mg and ma because by convention, downward forces are - and upward ones are +.... but it's up to you really.
     
  4. Apr 24, 2006 #3
    thanks, I got it now!
     
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