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Calculating useful force or force at an angle

  1. Sep 16, 2013 #1
    1. The problem statement, all variables and given/known data

    I don't understand my textbook.

    2. Relevant equations

    Fh = Fa Cos θ

    Where Fa is some pulling force applied, in-between horizontal and vertical.

    Where Fh is force in the horizontal.

    3. The attempt at a solution

    Intuitively, if the angle of the force applied is at 45°, then the force in the horizontal and vertical is the same, and half of Fa.

    Cos θ for 45° is 0.707. I was expecting 0.5.

    What am I getting wrong? Thanks.
     
  2. jcsd
  3. Sep 16, 2013 #2

    Doc Al

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    Staff: Mentor

    You are correct that the horizontal and vertical components must be the same, but that doesn't mean that they are each half the total force. Realize that they are perpendicular. Imagine a 45°-45° right triangle with sides of 1 unit length. What's the hypotenuse equal?
     
  4. Sep 16, 2013 #3

    gneill

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    Staff: Mentor

    Force components sum the way the legs of a right triangle "sum" to find the hypotenuse.

    As you noted, if the angle is 45° then cos(θ) is 1/√2. But sin(θ) is also 1/√2, so the vertically directed force component is indeed equal to the horizontally directed component. But neither one is half of the net force!

    For a right triangle with side lengths A and B, the hypotenuse C is given by the relationship:

    C2 = A2 + B2

    Force components add in the same fashion.

    EDIT: Doc Al beat me to it!
     
  5. Sep 16, 2013 #4
    Yes, at that angle, the two components of the original force have the same magnitude.

    BUT since these two components are not in the same direction the magnitude of each of them is not half the magnitude of the original force.

    Two forces, each of magnitude F/2, can only give a resultant of magnitude F if these two forces are in the same direction.
     
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