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## 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[itex]^{-}[/itex]0292176186

would that be correct??