Three forces act on a moving object. One force has a magnitude of 76.5 N and is directed due north. Another has a magnitude of 59.0 N and is directed due west. What must be the magnitude and direction of the third force, such that the object continues to move with a constant velocity?

Answer :

Answer:

F= 96.61 N

angle (θ) = 52.35°

Explanation:

given,

One Force F₁ = 76.5 N

second force F₂ = 59 N

That means F₃ must equal –(F₁+F₂) so the net sum of the three is zero.

F₁+F₂,  since they are orthogonal, is easy.

F₁+F₂ =

   = [tex]\sqrt{(76.5^2 + 59^2)}[/tex]

   = 96.61 N

[tex]tan \theta = \dfrac{f_y}{f_x}[/tex]

[tex]tan \theta = \dfrac{76.5}{59}[/tex]

[tex]\theta = tan^{-1}(1.30)[/tex]

[tex]\theta = 52.35^0[/tex]

Force is acting of magnitude F= 96.61 N

angle (θ) = 52.35°

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