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Homework Help: Applied Forces

  1. Feb 6, 2006 #1
    Can anyone help with what formulas to apply/use here ?

    An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 6.5 N; a second force has a magnitude of 3.5 N and points in the negative y direction. Find the direction ° (counterclockwise from theC +x axis) and magnitude (N) of the third force acting on the object.
  2. jcsd
  3. Feb 6, 2006 #2


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    The fact that the object moves with 'constant' velocity means that it is not accelerating, which would imply a change in velocity. The fact that it is not accelerating indicates that the net force or sum of forces is nil (0).

    Therefore the third force must have a component equal to but opposite of the force in the +ve x-direction, and a y component which is equal to but oppositve the force in the -ve y-direction.

    Then the angle is simply determined by an inverse trig function based on ratio of either the x or y component for the third force and the total or resultant force.
  4. Feb 6, 2006 #3

    i got the answer of 7.38 for the first part (which is correct) but i'm still unsure about the second part. can anyone help ?
  5. Feb 6, 2006 #4


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    The vector is in the upper left quadrant (of the Cartesian coordinate system) since its x-component must be - to offset the + x-force, and its y-component must be + to offset the - y-force.

    7.38 is correct! This should simply be [itex]\sqrt{3.5^2\,+\,6.2^2}[/itex].

    Think of a vector F which has two components, Fx and Fy, and both components are orthogonal.

    The magnitude of F = [itex]\sqrt{(F_x^2\,+\,F_y^2)}[/itex], and Fx = F cos[itex]\theta[/itex] and Fy = F sin[itex]\theta[/itex], where [itex]\theta[/itex] is the angle between F and the x-axis.

    Think of the definition of the cos of an angle and how it relates to the legs and hypotenuse of a right triangle.
    Last edited: Feb 6, 2006
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