Force, magnitude, and direction HELP

[bellawella]

Homework Statement

A 322-kg boat is sailing 14.5° north of east at a speed of 1.84 m/s. Thirty seconds later, it is sailing 34.2° north of east at a speed of 4.40 m/s. During this time, three forces act on the boat: a 33.6-N force directed 14.5° north of east (due to an auxiliary engine), a 24.3-N force directed 14.5° south of west (resistance due to the water), and F_W (due to the wind). Find the magnitude and direction of the force F_W. Express the direction as an angle with respect to due east.

Homework Equations

The Attempt at a Solution

Vo= 1.84cos(14.5) + 1.84sin(14.5) = 2.24
Vf= 4.4cos(34.2) + 4.4sin(34.2) = 6.11

(6.11-2.24)/30 = a = .13 m/s²

F_total = 322 x .13 = 41.86 N

F_engine= 33.6cos(14.5) + 33.6sin(14.5)= 40.9
F_resistance= -24.3cos(14.5) - 24.3cos(14.5)= -29.6

F_total=F_engine+F_resistance+F_w
41.86=40.9+(-29.6)+F_W
F_W= 30.6

But its wrong ! PLEASE help me by explaining with numbers. I've rechecked my calculations so many times. and I can't get to the second part without the first, so please help. Thanks!

 

[willem2]

split all forces and speeds in north/south and east/west components. Use F=ma
in the north/south and in the east/west direction.

Vo= 1.84cos(14.5) + 1.84sin(14.5) = 2.24

What you do here doesn't make sense. You compute the 2 components of the speed and
then add them to get another speed?

 

[bellawella]

heyy. no i didnt compute them to get a total speed. i used deltaV/time and got acceleration. then i multiplied acceleration times mass to get the total force. and thanks for the advice, I am not sure ill do it right but ill give it a shot. words usually confuse me!

 

[bellawella]

or maybe i did? ahhh i don't even know what i did now :\