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Simple Physics Mass pulled by a Force

  1. Feb 27, 2008 #1
    [SOLVED] Simple Physics Mass pulled by a Force

    1. The problem statement, all variables and given/known data
    Two part problem:
    Part 1:

    Show by calculations that the mass can not be lifted up from the ground

    F=27.0
    m=3.00kg
    Angle= 30degrees

    Part 2:
    show by calculations whether the mass moves level along the ground, if the coefficient of static friction between the surface and the mass is .750 and the coefficient of kinetic friction is .400

    attached is drawing...

    2. Relevant equations




    3. The attempt at a solution

    i draw the FBD and get everything in terms of x an y

    Fx=F cos 30 = m*a; 27.0 cos 30 = m*a; a = 7.79 m/s^2
    Fy= Fn sin 30 - m*g = 0 ;Fn = (m*g)/(sin 30) , which equaled 58.8N

    uhmm... or do i

    w = m*g; 3.00*9.80 = 29.4
    f sin 30 = 13.5

    13.5> 29.4 so, obj doesn't move up or down?

    im confused....

    Part 2:

    F=27.0
    m= 3.00kg
    Fs=.750
    Fk=.400

    I draw FBD and

    F cos 30 = 23.4

    Fx= F cos 30 - Fk = m*a; a = (27.0 cos 30 -.400)/3 = 7.66m/s^2

    so the object moves right?
     

    Attached Files:

  2. jcsd
  3. Feb 27, 2008 #2

    Doc Al

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    Staff: Mentor

    Just compare the upward component of the force with the weight. (Which is what you did in your second attempt.)

    For the second part, realize that

    Fs & Fk are the coefficients of friction, not the force of friction. Hint: What's the normal force?
     
  4. Feb 27, 2008 #3
    so since the weight is more than the upward force(f sin 30) there is no upward movement?

    does a normal force come into play here? is it equal to m*g?


    i know they are the coefficients of friction but, i didn't know how to type uhmm... Mu? (spelling)...do you know how to type that im noob so i don't...

    thnx doc!
     
    Last edited: Feb 27, 2008
  5. Feb 27, 2008 #4

    Doc Al

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    Staff: Mentor

    Right. In order to lift the mass, the force must support its weight. But it weighs too much for that.

    The normal force does not equal mg. (It would, if there were no applied force.) To find the normal force, add up all the vertical force components. What must the sum equal?
    I'm not concerned with how you write it, but that you don't leave out the normal force when calculating friction.
     
  6. Feb 27, 2008 #5
    well the question ask if the box moves right along the level of the ground...with the given coefficents of friction. so, i need to figure out if the force pulling the box is greater than the force restricting the box(normal force?) correct?
     
  7. Feb 27, 2008 #6

    Doc Al

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    Staff: Mentor

    You need to compare the horizontal component of the applied force with the force of friction. Start by finding the normal force, which you'll need in order to calculate the friction force.
     
  8. Feb 27, 2008 #7
    okay how about,

    Fn = F sin (theta) - m*g?
     
  9. Feb 27, 2008 #8

    Doc Al

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    Almost. (Check your signs.)
     
  10. Feb 27, 2008 #9
    lol............F sin (theta) + m*g?

    i would assume that would be correct...then once i have the normal force where would i go from there?...
     
  11. Feb 27, 2008 #10

    Doc Al

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    That's not correct. The way to do it is to add up the vertical force components (each with the correct sign) and set the sum equal to zero. Then you can solve for Fn.

    How does friction relate to the normal force and mu?
     
  12. Feb 27, 2008 #11
    im sorry this problem has me stumped...

    the only vertical force would be...f sin theta and the normal force... opposite force would be m*g

    horizontal forces would be friction to the left...and f cos theta?

    would it be Fn = m*g - F sin theta
     
  13. Feb 27, 2008 #12

    Doc Al

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    Good.

    Now compare F cos theta to the friction force. First figure out the maximum value of static friction and check if that horizontal force is enough to overcome it.
     
  14. Feb 27, 2008 #13
    is Fn = 15.9?


    so f cos theta = 27.0 cos 30 = 23.4

    so... Fx = F1 cos 30 - f(mu) = ma?

    then solve for accel?
     
  15. Feb 27, 2008 #14

    Doc Al

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    Good.

    Good.

    First figure out the friction forces.
     
  16. Feb 27, 2008 #15
    sorry it has been along time since physics...uhmm...by that you mean... like


    Fs= Fn*(mu)s

    and

    Fk= Fn*(mu)k

    ??????...lol
     
  17. Feb 27, 2008 #16

    Doc Al

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    That's what I mean. :wink:
     
  18. Feb 27, 2008 #17
    which could only mean...

    Fs = 11.9 and Fk = 6.36
    are these newtons?


    ...then do i compare to f cos theta?
     
  19. Feb 27, 2008 #18

    Doc Al

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    Yes.
     
  20. Feb 27, 2008 #19
    would the sum of all x forces be? Fx = F cos theta - (mu)k = m*a

    since f cos theta > Fs and f cos theta > Fk

    since those are magnitudes of the friction forces that means the box will move...

    and its displacement is given by? (not part of the problem....just curious)
    y = Vot + (.5)(a)(t^2)....

    i think...and we must solve for acceleration first?
     
    Last edited: Feb 27, 2008
  21. Feb 28, 2008 #20

    Doc Al

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    Staff: Mentor

    Right. Since the applied force is greater than static friction, the box will move.

    To find the acceleration of the box, just find the net force and apply Newton's 2nd law.

    Good. In this case, we presume the box is initially at rest.
     
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