Question

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 1

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 =

   = $\sqrt{(76.5^2 + 59^2)}$

   = 96.61 N

$tan \theta = \dfrac{f_y}{f_x}$

$tan \theta = \dfrac{76.5}{59}$

$\theta = tan^{-1}(1.30)$

$\theta = 52.35^0$

Force is acting of magnitude F= 96.61 N

angle (θ) = 52.35°