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Motion and Forces

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data
    A 3.54 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 10.6 N at an angle theta = 15.0° above the horizontal, as shown in the Figure. The coefficient of kinetic friction between the block and the floor is 0.08. What is the speed of the block 3.9 s after it starts moving?


    2. Relevant equations
    Xcomp= 10.6 cos 15
    Ycomp= 10.6 sin 15

    I don't know where to go from here :(
     
  2. jcsd
  3. Sep 16, 2007 #2
    How about Newton's second law?

    [tex] \sum_{i=1}^n F_i = ma[/tex]
     
  4. Sep 16, 2007 #3
    so do I have to find the resultant vector or no?
     
  5. Sep 16, 2007 #4

    learningphysics

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    Did you draw a diagram? Do [tex]\Sigma\vec{F} = ma[/tex] in the vertical and horizontal directions...
     
  6. Sep 16, 2007 #5
    I don't know? I am so lost!
     
  7. Sep 16, 2007 #6
    Okay so I have come up with the net force for the x and y directions...

    Fxnet = Fx + Wx + Nx = Fx + 0 + Nx = max
    Fynet = Fy + Wy + Ny = Fy - Wy + 0 = may
     
  8. Sep 16, 2007 #7
    Is that set up right?
     
  9. Sep 16, 2007 #8

    learningphysics

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    For the x-direction you wrote:

    [tex]F_x + N_x = ma_x[/tex]

    What is N_x ? Does the normal force act in the x - direction? what about the frictional force?

    For the y-direction you wrote:

    [tex]F_y - W_y + 0 = ma_y[/tex]

    Hmmm... why didn't you put a normal force here?
     
  10. Sep 16, 2007 #9
    Nx doesn't act in the x direction so I put it as 0...
    so I have...
    Fx/m = ax

    For the normal force in the y direction I put 0...
    (Fy-Wy)/m = ay
     
  11. Sep 16, 2007 #10

    learningphysics

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    Hmmm... you're missing forces.

    List all the forces acting in the x-direction.

    List all the forces acting in the y-direction.
     
  12. Sep 16, 2007 #11
    Forces in the x-direction:

    Tension
    Normal
    Fricton

    Forces in the y-direction:

    Weight
    Normal
    Friction
    Tension
     
  13. Sep 16, 2007 #12

    learningphysics

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    There's no normal force in the x - direction. There's no frictional force in the y-direction.

    So in the x-direction... you only have the x-component of tension, and the friction... write out the F=ma equation in the x-direction.

    In the y-direction... you have the y-component of tension, gravity and the normal force... write out the F=ma equation in the y-direction.
     
  14. Sep 16, 2007 #13
    Oh okay...that was a dumb mistake I made...

    Fxnet = Tx + Fx = ma
    Fynet = Ty - Wy + Ny = ma
     
  15. Sep 16, 2007 #14

    learningphysics

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    Cool.. try to substitute the values of tension, friction, weight etc... into your equations...
     
  16. Sep 16, 2007 #15
    I got the friction and weight but I m drawing a blank on how to calculate the tension?
     
  17. Sep 16, 2007 #16

    learningphysics

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    Actually you have the components of tension in your first post. :wink:
     
  18. Sep 16, 2007 #17
    okay so the fx = 10.6 cos 15 is actually the tension force not the friction force?
     
  19. Sep 16, 2007 #18

    learningphysics

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    yes, 10.6 is the force the cord exerts... ie tension... so 10.6cos15 is the horizontal component... 10.6sin15 is the vertical component.
     
  20. Sep 16, 2007 #19
    then how do I calculate the friction force. I know I have to use the Kinetic friction force of 0.08 but isn't there something else I have to do to it?
     
  21. Sep 16, 2007 #20

    learningphysics

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    Frictional force = normal force * uk

    uk is the coefficient of kinetic friction = 0.08
     
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