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Newton Applications. Help solving equation with sin and cos.

  1. Nov 28, 2009 #1
    1. The problem statement, all variables and given/known data

    At an airport, you pull a 18-kg suitcase across the floor with a strap that is at 45 degrees above the horizontal. Find the normal force and the tension on the strap, given that the suitcase moves with a constant speed and the coefficient of kinetic friction is 0.38.

    [tex]\mu[/tex]k = 0.38
    m = 18kg
    [tex]\sum[/tex]F = 0 (because speed is constant)
    angle = 45
    N = ?
    T = ?

    2. Relevant equations

    Fcos([tex]\theta[/tex]) - Ff = 0
    Ff = mg - Fsin([tex]\theta[/tex])([tex]\mu[/tex]k)

    3. The attempt at a solution

    Fcos([tex]\theta[/tex]) - [mg - Fsin([tex]\theta[/tex])([tex]\mu[/tex]k)]
    Fcos(42) - [(18)(9.81) - Fsin(42)](0.38) = 0

    I'm trying to solve for F but I don't know how to get rid of, solve equations with cos and sin.
    Help is appreciated!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Nov 28, 2009 #2
    you just need to evaulate cos(42) and sin(42) using your calculator. Be very careful with roundoff error (don't round too much too soon), and make sure your calculator is in degrees mode (if the problem is stated in degrees.. most intro physics problems are though)
     
  4. Nov 28, 2009 #3
    so Fcos(45) turns to .707107F?
     
  5. Nov 28, 2009 #4
    Yeah, but [tex] \cos(45^\circ) = \frac{\sqrt{2}}{2} [/tex], so maybe you can leave it like that until the very end, then you can approximate with decimals. It'll save you the hassle of round-off error. To be honest, it won't make much difference in this case, especially since it looks like you're rounding out to 6 places anyway, but it's not a bad idea in general.

    but the short answer is yes.
     
  6. Nov 28, 2009 #5
    I solved the equation,

    F = 68.763N
    then i found normal force
    N = 127.957N

    I have to find the tension on the strap, but i don't completely understand how to do that. Could someone explain the concept or how to find it?
     
  7. Nov 28, 2009 #6
    The question I have is should I add Force and Force of Friction since you have to overcome Force of Friction?
    But Force is on an angle, so does that mean i have to find Fx and then add them, and then combine Fx and Fy?
     
  8. Nov 29, 2009 #7

    HallsofIvy

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    Staff Emeritus
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    The horizontal component of force must be equal to the friction force so that there is no acceleration or deceleration.
     
  9. Nov 29, 2009 #8
    Okay, so how do I find tension on the strap?
     
  10. Nov 29, 2009 #9

    Doc Al

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    Staff: Mentor

    Set up two equations, one for vertical forces and one for horizontal force. You'll have two equations and two unknowns (Normal force and Tension)--which works out just right.
     
  11. Nov 29, 2009 #10
    I already know the normal force.
    I was trying to figure it out by myself and thought if i did this:
    Ty = 68.763sin(45)
    Tx = 68.763cos(45) + (127.957) (0.38)
    solve and combine x and y components to have the tension? not sure...at all.
     
  12. Nov 29, 2009 #11

    Doc Al

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    Staff: Mentor

    Looks to me like you already know the tension, if you're able to find it's y component. This equation says Ty = T sin(45) = 68.76 sin(45)
     
  13. Nov 29, 2009 #12
    The 68.783n is not the tension, its the force of the pull. Or is it the same thing?
    If it is the same thing, i have to add the force of friction to the x component of force since thats working against the movement?
     
  14. Nov 29, 2009 #13

    Doc Al

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    Staff: Mentor

    The force with which the strap is pulled is the tension.
    The sum of the forces in the x-direction will equal zero, if that's what you mean.
     
  15. Nov 29, 2009 #14
    Okay, thanks Doc Al!
    Sorry, my teacher is pretty confusing sometimes.
     
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