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= 50.0 N and F2 = 90.0 N. F1 acts under an angle of 60 degrees with respect to the x-axis and F2 is directed along the x-axix. Find the magnitude and direction of F3 such that the resultant force acting on the ball is zero.

To find the magnitude and direction of F3, we need to calculate the x and y components of F3 and their resultant magnitude. The resultant force will be zero when the sum of all three forces is equal to zero.

Let's break down F1 and F2 into their x and y components:

F1x = F1 * cos(60°)
F1y = F1 * sin(60°)

F1x = 50.0 N * cos(60°) = 25.0 N
F1y = 50.0 N * sin(60°) = 43.3 N

Since F2 is directed along the x-axis, F2y = 0 and F2x = F2:

F2x = F2 = 90.0 N

For the resultant force to be zero, the x and y components of F3 should cancel out the x and y components of F1 and F2. In other words:

F1x + F2x + F3x = 0
F1y + F2y + F3y = 0

Substituting the known values:

25.0 N + 90.0 N + F3x = 0
43.3 N + 0 + F3y = 0

Simplifying the first equation:

115.0 N + F3x = 0
F3x = -115.0 N

Simplifying the second equation:

F3y = -43.3 N

Now we can find the magnitude and direction of F3:

Magnitude of F3 = √(F3x² + F3y²)
Magnitude of F3 = √((-115.0 N)² + (-43.3 N)²)
Magnitude of F3 = √(13225 N² + 1875.89 N²)
Magnitude of F3 = √(15100.89 N²)
Magnitude of F3 ≈ 122.91 N

To find the direction of F3, we can use the following formula:

θ = tan⁻¹(F3y / F3x)

θ = tan⁻¹((-43.3 N) / (-115.0 N))
θ = tan⁻¹(0.3765)

θ ≈ 21.1°

Therefore, the magnitude of F3 is approximately 122.91 N and it is directed at an angle of approximately 21.1°.

To find the magnitude and direction of F3, we need to use vector addition. Since the resultant force acting on the ball is zero, the vector sum of F1, F2, and F3 should equal zero.

Let's break down F1 and F2 into their x and y components:
F1x = F1 * cosθ1
F1y = F1 * sinθ1

Where θ1 is the angle F1 makes with the x-axis.

In this case, F1 makes an angle of 60 degrees with the x-axis, so:
F1x = F1 * cos(60°)
F1y = F1 * sin(60°)

Now, F2 is directed along the x-axis, so there is no y-component:
F2x = F2
F2y = 0

Next, we need to sum up the x and y components of the forces (including F3) separately:

For the x-components:
ΣFx = F1x + F2x + F3x = 0
=> (F1 * cos(60°)) + F2 + F3x = 0 (Substituting the values)

For the y-components:
ΣFy = F1y + F2y + F3y = 0
=> (F1 * sin(60°)) + 0 + F3y = 0 (Substituting the values)

From the x-component equation, we can solve for F3x:
F3x = -(F1 * cos(60°)) - F2

And from the y-component equation, we can solve for F3y:
F3y = -(F1 * sin(60°))

Finally, we can find the magnitude and direction of F3 using these components:
Magnitude of F3: |F3| = sqrt(F3x^2 + F3y^2)
Direction of F3: θ3 = arctan(F3y / F3x)

You can substitute the values of F1, F2, θ1, and calculate F3x, F3y, |F3|, and θ3 to find the answers.

78