At a picnic, there is a contest in which hoses are used to shoot water at a beach ball from three different directions. As a result, three forces act on the ball, F1, F2, and F3. The magnitudes of F1 and F2 are F1=50.0N and F2=90.0N. F1 acts under the angle of 60degrees with respect to the x axis and F2 is directed along the x-axis. Find the magnitude and direction of F3 such that the resultant force acting on the ball is zero.

F₃ = 122.88 N

θ₃ = 20.63°

Explanation:

First we find the components of F₁:

For x-component:

F₁ₓ = F₁ Cos θ₁

F₁ₓ = (50 N) Cos 60°

F₁ₓ = 25 N

For y-component:

F₁y = F₁ Sin θ₁

F₁y = (50 N) Sin 60°

F₁y = 43.3 N

Now, for F₂. As, F₂ acts along x-axis. Therefore, its y-component will be zero and its x-xomponent will be equal to the magnitude of force itself:

F₂ₓ = F₂ = 90 N

F₂y = 0 N

Now, for the resultant force on ball to be zero, the sum of x-components of the forces and the sum of the y-component of the forces must also be equal to zero:

F₁ₓ + F₂ₓ + F₃ₓ = 0 N

25 N + 90 N + F₃ₓ = 0 N

F₃ₓ = - 115 N

for y-components:

F₁y + F₂y + F₃y = 0 N

43.3 N + 0 N + F₃y = 0 N

F₃y = - 43.3 N

Now, the magnitude of F₃ can be found as:

F₃ = √F₃ₓ² + F₃y²

F₃ = √[(- 115 N)² + (- 43.3 N)²]

F₃ = 122.88 N

and the direction is given as:

θ₃ = tan⁻¹(F₃y/F₃ₓ) = tan⁻¹(-43.3 N/-115 N)

θ₃ = 20.63°

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