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Sine Law

  1. Jun 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Two ropes are attached to a log that is floating in the water. A force of 80.0 N is applied to one rope and a force of 60.0 N is applied to the other rope, which is lying at an angle of 40° from the first rope. What is the net force on the log?

    We know that the angle between the two known vectors is 140°.

    2. Relevant equations
    Cosine law:
    c2=a2+b2-2abCosC

    Sine Law:
    sinA/a= SinB/b

    3. The attempt at a solution
    Cosine Law:
    c²=a²+b²-2abCosC
    c²=[(80.0)²+[60.0]²-2(80.0)x(60.0)cos140]
    c=[80)²+(60)²-2(80)x(60)cos14]^½
    c=131.346828
    c=132 N

    Now to find the angle between the resultnt force vector and the 80.0 N vector, use sine law

    sinA/a = sinB/b = sinC/c

    sinA/60.0 N = sin140°/132 N
    sinA/60.0 N x 60.0 = sin140/132 x60.0
    sinA =sin140/132 x 60.0
    sinA = 0.292176186
    A= sin[itex]^{-}[/itex]0292176186


    would that be correct??
     
  2. jcsd
  3. Jun 18, 2011 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good. So what's your final answer for the angle?
     
  4. Jun 20, 2011 #3
    The answer I got was

    Fnet= 132 N [17°counter clock-wise] from the original vector of 80.0N.
     
  5. Jun 20, 2011 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good.
     
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