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Force, magnitude, and direction HELP!

  1. Oct 5, 2009 #1
    1. The problem statement, all variables and given/known data

    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 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

    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

    Ftotal = 322 x .13 = 41.86 N

    Fengine= 33.6cos(14.5) + 33.6sin(14.5)= 40.9
    Fresistance= -24.3cos(14.5) - 24.3cos(14.5)= -29.6

    Ftotal=Fengine+Fresistance+Fw
    41.86=40.9+(-29.6)+FW
    FW= 30.6

    But its wrong !! PLEASE help me by explaining with numbers. Ive rechecked my calculations so many times. and I cant get to the second part without the first, so please help. Thanks!
     
  2. jcsd
  3. Oct 6, 2009 #2
    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?
     
  4. Oct 6, 2009 #3
    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, im not sure ill do it right but ill give it a shot. words usually confuse me!
     
  5. Oct 6, 2009 #4
    or maybe i did? ahhh i dont even know what i did now :\
     
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