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!

Normal Force

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data
    A 3.72 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 11.5 N at an angle θ = 20.5° above the horizontal. The coefficient of kinetic friction between the block and the floor is 0.070. What is the speed of the block 6.10 s after it starts moving?


    2. Relevant equations
    F=ma
    vf = at


    3. The attempt at a solution
    What is the normal force in this case, because it's not mg, as I tried that and got it wrong.
     
  2. jcsd
  3. Sep 22, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Zhalfirin88! :smile:
    That's right … in this case, there are three forces with a (non-zero) component in the vertical direction: F N and W, instead of the usual 2.

    Obviously, you need the three components to add to zero. :wink:
     
  4. Sep 22, 2009 #3

    Doc Al

    User Avatar

    Staff: Mentor

    To figure out the normal force, add up all the vertical force components. What must they add up to?
     
  5. Sep 22, 2009 #4
    I have no idea what you said after there are 3 non-zero components in the vertical direction.

    Since it's not moving in the vertical direction it'd be zero right?
     
  6. Sep 22, 2009 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    There's the force from the cord (F), the weight of the block, and the normal force.

    Their components (in any direction) have to add to zero.
     
  7. Sep 22, 2009 #6
    Okay, I have no patience for this.

    0 = FN + Fg + FT

    FN = -mg - FT

    FN = -(3.72)(9.8) - 11.5sin(20.5)

    FN = -36.45 - 3.8523

    FN = -40.30 N

    But, you plug that in to static friction equation and you get fs = -2.821

    So, 11.5sin(20.5) + 2.821 = 6.8484N/3.72 kg

    a = 1.841 m/s2

    vf = at

    vf = 1.841 * 6.1

    vf = 11.23 and that is wrong.

    edit: I also did:

    FN = +36.45 - 3.8523 Because down is the negative direction.

    FN = 32.5977 N

    But, you plug that in to static friction equation and you get fs = 2.821

    So, 11.5sin(20.5) - 2.821 = 1.7455N/3.72 kg

    a = .4692 m/s2

    vf = at

    vf =.4692 * 6.1

    vf = 2.86 and that is wrong. But that was my final try at the question so it's wrong for good now.
     
    Last edited: Sep 22, 2009
  8. Sep 22, 2009 #7

    Doc Al

    User Avatar

    Staff: Mentor

    You made your mistake in your 2nd line.
    ΣF = FN + Fg + FT
    ΣF = FN - mg + 11.5sin(20.5)

    Since gravity acts down it gets a negative sign, while FT gets a positive sign.

    Since ΣF = 0:
    FN = mg - 11.5sin(20.5)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Normal Force
  1. Normal Force! (Replies: 3)

  2. Normal force (Replies: 5)

Loading...