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Tug a War Contest

  1. Nov 29, 2011 #1
    1. The problem statement, all variables and given/known data

    There is a tug a war match and everyone is pulling on the object?

    Person A is pulling with 45N at 45degrees, person B is pulling with 48 N at 110degrees, and person c is pulling at 40N at 200degrees.
    1)What is the net force acting on the object based on the pulling of the peoples?
    2) If the object has a weight of 36N, what acc do the persons give to the object?

    2. Relevant equations
    1)sqrt of (Rx^2+Ry^2)
    2)a=f/m
    f=force
    w=mg
    w=weight
    g=gravity

    3. The attempt at a solution
    1)45cos45=31.82
    48cos110=-16.42
    40cos200=-42.29
    Rx=57.69

    45sin45=31.82
    48sin110=45.11
    40sin200=-13.68
    Ry=63.25

    sqrt of (63.25^2+57.69^2)=85.61N
    85.61+40+45=170.61N


    2)a=170.61/m

    w=mg
    36N=m9.8
    m=3.67

    a=170.61/3.67
    a=46.49 m/s^2
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 30, 2011 #2

    PhanthomJay

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    make corrections....what about angle of force and acceleration?
     
  4. Nov 30, 2011 #3
    I assume all the forces are in a plane parallel to the horizontal?? And there is no friction involved? (the reason I asked was upon reading the problem I visualized people pulling diagonally upward for some reason) And yes you can't describe the force or the acceleration without including magnitude and direction.
     
  5. Nov 30, 2011 #4

    PhanthomJay

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    yes
    no friction on the object being pulled, but plenty on the pullers, which is not necessary to know in solving this problem.
     
  6. Nov 30, 2011 #5
    1)For the first one I got144.23 N. For angle, tan-1(63.25/-22.19)=-70.67+360=289.33deg.
    Not really sure if it's correct though
    2)I didn't lnow there had to be angle for acceleration. Is it 45, 110, and 200deg?not sure.
     
  7. Nov 30, 2011 #6

    PhanthomJay

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    You need to draw a graphical sketch of the problem roughly to scale, using the 'tail to head' graphical method when adding up the 3 vectors. This will give you a good idea of the approximate magnitude and angle of the resultant force vector.

    Checking your maths, the Rx component of the resultant vector is

    Rx = 45cos45 + 48cos110 + 40cos200 = 31.82 - 16.42 - 37.59 = -15.02

    The Ry component of the resultant vector is

    Ry = 45sin45 + 48sin110 + 40sin200 = 31.82 + 45.11 - 13.68 = + 63.25

    Since Rx is negative and Ry is positive, the resultant vector lies in the 2nd quadrant
    R = sqrt of (-15.02^2 + 63.25^2) = 65 N

    Now to get the angle of the resultant force vector, that's theta = tan^-1(Ry/Rx).. Use your calculator to find that angle, but use your sketch to get the angle it makes with the x axis (the vector lies in the 2nd quadrant). The acceleration is always in the direction of the net resultant force , so the object will accelerate in the same direction as the resultant force.
     
  8. Nov 30, 2011 #7
    for Rx i got -22.19. so, the acc angle is the same as the force angle
     
  9. Nov 30, 2011 #8

    PhanthomJay

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    yes, indeed...
    yes. Since f_net =ma, and m is a scalar, then 'f_net' and 'a' must always have the same direction.
     
  10. Nov 30, 2011 #9
    i got : tan ^-1(63.25/-22.19)= -70.67
    this angle lies in the 4th quandrant. did i do something wrong?
     
  11. Dec 1, 2011 #10

    PhanthomJay

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    Yes, you forgot to draw your sketch. The calculator doesn't tell you which axis the angle is measured from. You know it must be in the 2nd quadrant. Use sohcahtoa. The 70 degree angle is measured from which axis??
     
  12. Dec 1, 2011 #11
    don't ever do a vector problem without drawing a diagram so you have an intuitive idea of what and where the answer will be. And don't try to get your direction strictly from whatever sign you get in your answer. Too easy to make a mistake. You should have at least two ways of at least approximating your answer.
     
  13. Dec 1, 2011 #12
    a) 67.022N
    I found the components of each vector and summed to find the net force in each direction. Then I made a right triangle out of the two resultants and the hypotenuse measured 67.022N


    b) 18.25m/s^2
    I used F=MA:
    67.022 = (36/9.8)*A
     
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