A 325-kg boat is sailing 15.0° north of east at a speed of 2.00 m/s.
Thirty seconds later, it is sailing 35.0° north of east at a speed of 4.00 m/s.
During this time, three forces act on the boat: a 31.0-N force directed 15.0°
north of east (due to an auxiliary engine), a 23.0-N force directed 15.0° south
of west (resistance due to the water), and FW
→ (due to the wind). Find the
magnitude and direction of the force FW
→ . Express the direction as an angle
with respect to due east.
it's easy to use complex numbers to handle all these pairs of x-y values.
starting with the vector
a = 2 cis15°
and remembering that F = ma and v = at, we have acceleration vectors
b = 31/325 cis15°
c = 23/325 cis215°
Now, at time 30 seconds, we should have a resultant velocity of
r = a + (b+c)t = 2cis 15° + 30/325 (31cis15° + 23cis215°) = 2.957+0.405i
But we are told that the boat is moving with a velocity of
v = 4cis35° = 2.277+2.294i
That means that FW = v-r = -0.68+1.889i = 2.008 cis109.8°