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Newton's 2nd Law Problem, Angle exerted on block by surface

  1. Feb 11, 2017 #1
    1. The problem statement, all variables and given/known data
    Problem:
    A 12.0-kg block is pushed to the left across a rough horizontal surface by a force that is angled 30.0◦ below the horizontal. The magnitude of the force is 75.0 N and the acceleration of the block as it is pushed is 3.20 m/s^2 . What angle does the force exerted on the block by the surface make with the horizontal?

    For this I have :
    m= 12kg
    theta= 30 degrees
    applied force: 75 N
    acceleration= 3.20 m/s^2

    2. Relevant equations
    For this problem I didn't use any equations, I only drew a picture and free body diagram (attached).

    3. The attempt at a solution

    fbd and picture.JPG Sorry if the image is big

    I guess the main reason I didn't use any equations was because I am a little confused on what the question is asking for. Would the angle in question simply be 90 degrees? Isn't the force exerted on the block by the surface the normal force, which would be perpendicular to the horizontal?
    Thank you for all the help.
     
  2. jcsd
  3. Feb 11, 2017 #2

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

    Hello @Icycub,

    Welcome to PF! :smile: :welcome:


    I think the problem statement is asking you to combine the normal force and the frictional force into a single "surface" force [Edit: it's a vector, don't forget]. Then, calculate the angle of this surface force with respect to the horizontal.
     
    Last edited: Feb 11, 2017
  4. Feb 11, 2017 #3
    Yes, you are supposed to consider the reaction force to be the vectorial sum of the normal and frictional force (which some books call "tangent reaction")
     
    Last edited: Feb 11, 2017
  5. Feb 11, 2017 #4

    gneill

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    Staff: Mentor

    Hi Naso,

    wtpf.png

    Sorry, that's not how things work here. If you'd like help checking your work you should start your own thread and post what you've done. Helpers will be happy to give you a hand once you've shown your work.
     
  6. Feb 11, 2017 #5
    Oh alright, Edited!
     
  7. Feb 11, 2017 #6
    Alright thanks for the help, it makes sense that I have to combine the friction and normal force. I've worked out the problem, but i'm not sure if my answer is correct. In order to find the normal and kinetic force and then add them together, I set up two equations:

    (note: I reversed the positive and negative signs for the x axis) --> left=positive right=negative

    Σfx=ma
    0-fk + facos(theta)=ma
    fk=-ma +facos(theta)
    fk= -(12)(3.2)+75cos(30)= 26.6 N

    Σfy=0
    N-fg-fasin(theta)=0
    N=fg+fasin(theta)
    N= (12)(9.8)+75sin(30)= 155 N

    I'm confused because kinetic friction is positive. With my chosen coordinate system, I have it pointing in the negative x direction. Is this the correct value for kinetic friction?
     
  8. Feb 11, 2017 #7
    That is the correct value; it is positive because you already considered it as being subtractive in the ∑Fx equation
     
  9. Feb 11, 2017 #8
    Alright thank you!
     
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