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!

Forces and Newton's Law

  1. Mar 19, 2016 #1
    • Member advised to use the formatting template for questions posted to the homework forums
    Three blocks, each of mass 13 kg are on a frictionless table. A hand pushes on the left most box (A) such that the three boxes accelerate in the positive horizontal direction as shown at a rate of a = 1.4 m/s2.

    1)What is the magnitude of the force on block A from the hand?

    2)What is the net horizontal force on block B?


    For number 1 I was able to get the answer by adding up all the blocks and multiplying that by the acceleration but I am confused by why this is right. Before I did this I thought that the answer would be the mass of A * 1.4. Why does the force by the hand on block A include the other blocks as well? Is it because the whole system is accelerating and for that to happen the force has to take into account the other blocks as well?

    For number 2 I found the force done on the system to be (13+13+13)*1.4 = 54.6N. I then found that the force that object C does on block B and the force that object A does on block B would 13*1.4= 18.2N. I then added the force done on the system by the hand with block A and subtracted that by the force by C on B to get 54.6, but this is not the right answer. What should I be looking for in this problem?

    Thanks
     
  2. jcsd
  3. Mar 19, 2016 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    For 1, the hand doesn't care whether it's three separate blocks that happen to be adjacent, or three blocks glued together, or a single block mass 39kg. Alternatively, consider the force pair (action and reaction) at each interface. Each block accelerates according to the net force on it.

    For 2), all block B "knows" is that it experiences two forces, one from block A and one from block C, of opposite sign. The net force on B is simply the sum of those two. Alternatively, consider that for each block separately ΣF=Fnet=ma, where the sum is over the forces acting directly on the block. You tried using the first part of that, ΣF=Fnet, to find Fnet (but summed the wrong forces); the alternative is to use the second part, Fnet=ma.
     
    Last edited: Mar 20, 2016
  4. Mar 20, 2016 #3
    For 2) wouldn't there also be an equal and opposite force on block B by block C?
     
  5. Mar 20, 2016 #4

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Do you care about that?

    When using Newtons law F=ma what does F stand for? Hint.... the answer is NOT just "Force".
     
  6. Mar 20, 2016 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I made a typo in my post. Now corrected.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Forces and Newton's Law
  1. Forces and Newton's Laws (Replies: 11)

Loading...