Net Force on Log: Solving Using Cosine & Sine Laws

[mom2maxncoop]

Homework Statement

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°.

Homework Equations

Cosine law:
c2=a2+b2-2abCosC

Sine Law:
sinA/a= SinB/b

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??

 

[Doc Al]

Looks good. So what's your final answer for the angle?

 

[mom2maxncoop]

The answer I got was

Fnet= 132 N [17°counter clock-wise] from the original vector of 80.0N.

 

[Doc Al]

Looks good.