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A mass m slides in a moving reference frame

  1. Mar 28, 2015 #1
    1. The problem statement, all variables and given/known data
    An inclined plane, fixed to the inside of an elevator, makes a ## 32 ^\circ ## degree angle with the floor. A mass m slides on the plane without friction. What is its acceleration relative to the plane if the elevator accelerates upward at 0.50 g?

    2. Relevant equations
    The effective value of acceleration due to gravity ##g_e = g + a_e ## The e on the left stands for effective while the one on the right represents the elevator.
    Relative acceleration = ## g_e sin \theta ##


    3. The attempt at a solution
    I understand that ## a_r ## is the force of gravity along the incline plane and I understand how to derive it. However
    my textbook never taught me the "effective value of acceleration" but it uses it to solve the problem. How can acceleration be "effective"? When is this equation used? How do I derive it?

    Going back to the original question, my solution guide states that I should plug in
    ##g_e = g + 0.5g = 1.5g ##
    Then put the answer here ##a_r = g_e sin \theta = 1.5gsin(32 ^\circ ) ## Why is it addition if a and g are in opposite directions?

    Thank you (The main question I want answered is highlighted in RED)
     
  2. jcsd
  3. Mar 28, 2015 #2
    Effective value of g in general is g-a not g+a and the sign is important.
    Example, if a ball is hanging on a thread in an elevator and the elevator falls down with acceleration a, then w=-g-(-a)=0, since both g and a are in -j direction.
     
  4. Mar 28, 2015 #3
    G effective is just the acceleration of a body with respect to a non inertial frame.
     
  5. Mar 28, 2015 #4

    haruspex

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    The two accelerations are not both happening. g is the acceleration that would happen were there no upward force to oppose it. a is the acceleration that is happening because the upward force exceeds what would be necessary to prevent the downward acceleration. So the upward force is related to the sum of the two.
    The sign of g can be confusing, though. My preference is to say the acceleration due to gravity is g, regardless of the sign convention adopted for the vertical axis. If up is positive then the value of g is negative. But it may be more common to keep the value of g as positive and write -g in the equation.
     
  6. Mar 28, 2015 #5
    I am trying to understand this problem as well.

    So the block acceleration and elevator add?

    The relatI've acceleration equation is a vector equation. Is it not? So the -j component for the block would be the W*sin (theta) as the OP stated and then of course would have the acceleration of the elevator acting in the -j direction as well. My bad... I reread OP's problem statement...
     
  7. Mar 28, 2015 #6

    TSny

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    Calpalned,
    A common way to arrive at an “effective g” in the elevator is to start with the idea that the acceleration of the block relative to the earth (##\vec{a}##) equals the acceleration of the block relative to the elevator (##\vec{a}\:'##) plus the acceleration of the elevator relative to the earth (##\vec{a}_e##): $$\vec{a} = \vec{a}\:’+ \vec{a}_e$$
    In the reference frame of the earth, Newton’s second law applied to the block on the incline is $$\vec{N} + m\vec{g} = m\vec{a}$$ where ##\vec{N}## is the normal force.

    Substitute for ##\vec{a}## using the relative acceleration formula. Show that you can rearrange terms to get $$\vec{N} + m( \vec{g} - \vec{a}_e) = m\vec{a}\:’$$ If you define an effective g as $$\vec{g}_{eff} = \vec{g}-\vec{a}_e$$ then you can see that the acceleration of the block relative to the elevator satisfies the second law in the form$$\vec{N} + m\vec{g}_{eff} = m\vec{a}\:’$$
    Hope this helps.
     
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