posted by marissa .
A 316-kg boat is sailing 12.0° north of east at a speed of 1.60 m/s. 39.0 s later, it is sailing 36.0° north of east at a speed of 4.00 m/s. During this time, three forces act on the boat: a 31.4-N force directed 12.0° north of east (due to an auxiliary engine), a 20.7-N force directed 12.0° south of west (resistance due to the water), and (due to the wind). Find the (a) the magnitude and (b) direction of the force . Express the direction as an angle with respect to due east.
Fn = 31.4N @ 12o,CCW + 12.7N @ 192o,CCW.
a. Mag.=sqrt((18.3)^2+(3.89)^2=18.7 N
b. tanA = Y/X = 3.89/18.3 = 0.21257
A = 12o, CCW = 12o North of East.