# Forces and magnitude physics

1. Sep 29, 2009

### triplel777

1. The problem statement, all variables and given/known data

A 331-kg boat is sailing 14.6° north of east at a speed of 1.98 m/s. Thirty seconds later, it is sailing 37.5° north of east at a speed of 4.24 m/s. During this time, three forces act on the boat: a 28.8-N force directed 14.6° north of east (due to an auxiliary engine), a 24.7-N force directed 14.6° 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.

2. Relevant equations

3. The attempt at a solution

V initial= 1.98cos14.6 + 1.98sin14.6=2.42
V final= 4.24cos37.5 + 4.24sin37.5=5.94
delta V= 3.52
a=v/t so a=3.52/30=0.12m/s2
Ftotal=ma
Ftotal= 331*0.12=39.4
Ftotal= Fw+Fp+Fd
Fp= 28.8cos14.6+28.8sin14.6=35.13
Fd=24.7cos14.6+24.7sin14.6= 30.13
39.4= Fw+35.13+30.13
Fw=-25.86

what am i doing wrong?

2. Sep 29, 2009

### kuruman

You are not adding vectors correctly. In the above example, you are adding the x and y components of the velocities together. That is absolutely wrong. You have to keep the components separate. The same applies to the addition of the force components. You need to set up two separate sets of equations, one for the x-direction and one for the y-direction. Using unit vector notation helps you do that.