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Two dogs pull horizontally on ropes attached to a post; the angle problem 1 between the ropes is 60.0 degrees. If dog A exerts a force of 270N and dog B exerts a force of 300N find the magnitude of the resultant force and the angle it makes with dog A's rope.

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4 answers
  1. You have a SAS triangle problem. You need the third side, and an angle.

    Draw this out in a head to tail vector on vector.

    one vector 270 at zero degrees (dogA). Next, starting at the end of that vector, a 300N at 60 deg upward (the angle then in the triangle is 120 deg).

    First, the length of the other side:
    Law of cosines..
    R^2=a^2+b^2 -2abcos120
    you know a, b find R

    Then the angle between the dogA and the resultant.
    SinAngle/opposite side=Sin120/R
    where the opposite side in this triangle is 300
    solve for the angle.

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    bobpursley
  2. F = 270[0o] + 300[60o]
    F = 270+300*Cos60 + 300*sin60=420+259.8i
    = 494N.[31.74o].

    F = 31.74o above dog A.

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  3. F=
    270
    2
    +300
    2
    +2(270)(300)cos60

    =494 N
    \theta=\arctan{\frac{(300)\sin{60}}{270+(300)\cos{60}}}=31.7\degreeθ=arctan
    270+(300)cos60
    (300)sin60

    =31.7°

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  4. F=
    270
    2
    +300
    2
    +2(270)(300)cos60

    =494 N
    \theta=\arctan{\frac{(300)\sin{60}}{270+(300)\cos{60}}}=31.7\degreeθ=arctan
    270+(300)cos60
    (300)sin60

    =31.7°

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