Sine Law

1. Jun 18, 2011

mom2maxncoop

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$^{-}$0292176186

would that be correct??

2. Jun 18, 2011

Staff: Mentor

3. Jun 20, 2011