Can someone help me with my physics hw, plz?

Filed under: portal.xn--b8qr95do2b.com — jack @ March 9, 2010 edit
  • The force acting on a sailboat are 390N north and 180 N east. If the boat has a mass of 270 kg, what are the magnitude and direction of the boat's aceleration?


  • The forces are perpendicular to each other.
    Let net force = F
    Magnitude of F = sqrt(390^2 + 180^2) = 429.5 N

    Let angle made by F be theta north of east.
    theta = tan-1(north component of F/east component of F)
    = tan-1(390/180) = 65.2 deg

    Acceleration = force/mass
    = 429.5/270 = 1.6 m/s^2

    Direction of acceleration is the same as the direction of the net force.
    Ans: 1.6 m/s^2 at 65.2 deg North of East


  • F=(390^2 +180^2)^1/2=429.53NE
    a=429.53/270=1.59m/sec^2NE


  • The net force acting on the boat is sqrt(390^2 + 180^2) = 429.5 N
    The magnitude of the acceleration = F/m = 429.5/270 = 1.59 m/s^2
    The direction of the boat's acceleration = arctan(390/180) =
    65.22 degrees CCW







    • #If you have any other info about this subject , Please add it free.#
    Your name:
    E-mail:
    Telphone:

    Your comments:


    If you have any other info about Can someone help me with my physics hw, plz? , Please add it free.