F1 and F2 are two horizontal forces of 70 N each, acting on a wagon. They are at 45° to the x-axis and at right angles to each other. A third force, F3, also 70 N, acts along the positive x-axis. What if, instead, F2 acted in the opposite direction from what is shown? Find the magnitude and direction of the force that is now F1 + F3 - F2?

To find the magnitude and direction of the force when F2 acts in the opposite direction, we can follow these steps:

Step 1: Resolve the forces F1, F2, and F3 into their x-components and y-components.

Given:
F1 = F2 = 70 N
F3 = 70 N

For F1 and F2 at 45° to the x-axis and at right angles to each other, their x-components and y-components can be found using trigonometry:

F1x = F1 * cos(45°)
F1y = F1 * sin(45°)

F2x = F2 * cos(45°)
F2y = F2 * sin(45°)

Since F2 is acting in the opposite direction, the x-component will have negative sign.

F2x(opposite) = -F2 * cos(45°)

F3 is along the positive x-axis, so its x-component will be equal to the magnitude of F3, and y-component will be 0.

F3x = F3
F3y = 0

Step 2: Calculate the resultant x-component and y-component of the force (F1+F3-F2).

Resultant x-component = F1x + F3x - F2x(opposite)
Resultant y-component = F1y + F3y + F2y

Step 3: Calculate the magnitude and direction of the resulting force using the resultant x-component and y-component.

Magnitude of the force = sqrt[(Resultant x-component)^2 + (Resultant y-component)^2]

Direction of the force = atan(Resultant y-component / Resultant x-component)

Let's now substitute the values and calculate the result.

F1x = 70 N * cos(45°) = 49.497 N
F1y = 70 N * sin(45°) = 49.497 N

F2x(opposite) = -70 N * cos(45°) = -49.497 N
F2y = 70 N * sin(45°) = 49.497 N

F3x = 70 N
F3y = 0 N

Resultant x-component = 49.497 N + 70 N - (-49.497 N) = 168.491 N
Resultant y-component = 49.497 N + 0 N + 49.497 N = 98.994 N

Magnitude of the force = sqrt[(168.491 N)^2 + (98.994 N)^2] ≈ 193.727 N
Direction of the force = atan(98.994 N / 168.491 N) ≈ 29.377°