1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Force and acceleration (not actually homework)

  1. Mar 12, 2014 #1
    1. The problem statement, all variables and given/known data
    This is more of a conceptual problem than a mathematical one (Note that this is purely for conceptual purposes of my own and probably doesn't really hold any practical value). I was just curious to know that if we imagine a book of mass m1 resting on any flat surface on Earth and then put on top of it another book of mass m2, then the greater weight of the stack obviously seems a direct result of the additional mass (the book) added on top of the bottom book. But then a thought occurred to me: wouldn't the book on top pressing down on the book on bottom increase the downward acceleration of the bottom book? I turned to mathematics for the answer and I applied two slightly different approaches. Though, the two approaches give the same answer for the total weight of the stack, the values for acceleration that I got don't quite add up.

    The two approaches I applied are:

    1)Simply adding the two masses together and multiplying by g i.e

    2)Finding the extra acceleration a that is caused by the top book pushing down on the bottom book, then adding the resulting value to g and multiplying by the mass of the bottom book only i.e

    It seems the problem can be interpreted in two ways that appear, at least on the face of it, equivalent. The first approaches imagines a single system of two books. The second approach reduces the physical book to a downward force, so we're effectively dealing with a single book.

    But there must a single correct physical interpretation to this. Is the bottom book being accelerated by 9.8m/s^2 or something more? That's my question.

    P.S: I am aware that this is an absolutely pointless exercise, since the net accelerations, whatever the component accelerations, are always going to cancel out in this case no matter what approach we take.

    Thank you.
    Last edited: Mar 12, 2014
  2. jcsd
  3. Mar 12, 2014 #2

    Doc Al

    User Avatar

    Staff: Mentor

    If the books are resting on a surface their acceleration is zero.

    Think of 9.8 m/s^2 as a measure of the strength of Earth's gravitational field. It's the acceleration of a falling body when the only force acting is gravity, which is not the case here. But the weight of an object is mg, regardless of its acceleration.
  4. Mar 12, 2014 #3
    I meant that in the sense that we're only looking at the component acceleration only. The net acceleration would of course be zero (for example, 13m/s^2 of acceleration of a single object being cancelled by an equal and opposite force). But I wanted to know what these equal and opposite acceleration values are, if looked in isolation.

    I agree about the weight, but it's specifically the acceleration I'm concerned about.

    Thank you for your answer! :)
  5. Mar 12, 2014 #4


    User Avatar
    Gold Member

    Let's forget about the table.
    We will drop the two books from a building(air resistance is negligible)
    The acceleration of two books will be constant in this case.Isn't it?
    Even if we add the masses,the acceleration will still be constant.
  6. Mar 12, 2014 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Do not think in terms of 'component acceleration'. Think in terms of forces. Then apply Newton's 2nd law to find the resulting acceleration.

    To think that gravity creates a 9.8 m/s^2 acceleration downward and the normal force creates a 9.8 m/s^2 acceleration upward (for a net of zero) is a confusing an inaccurate way of viewing things.
  7. Mar 12, 2014 #6
    What the others said already.

    But if you REALLY want to go this way, there is no contradiction.
    That "a" in the formula is the acceleration of m1 under the weight of m2. So it will me (m2g)/m1
    Plug in and you get the same as in (1).
  8. Mar 13, 2014 #7
    I see. That's a very helpful instruction.

    Thanks a lot guys! :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted