Two workers pull horizontally on a heavy box, but one pulls twice as hard as the other. The larger pull is directed at 25.0 west of north, and the resultant of these two pulls is 430.0 directly northward.

365.8

To solve this problem, you can break down the forces into their x and y components. The x-component represents the horizontal force, and the y-component represents the vertical force.

Let's assume the smaller force is F1, and the larger force is F2.

Given:
F1 = x-component of smaller force
F2 = x-component of larger force = 2F1 (since one pull is twice as hard)
Angle of F2 with respect to north = 25.0° west of north
Resultant force = 430.0 N directly northward

Step 1: Calculate the x and y-components of F2.
Since the angle of F2 with respect to north is west of north, we need to consider the horizontal component as negative.

F2x = F2 * cos(theta) = 2F1 * cos(25.0°) [negative, since west of north]
F2y = F2 * sin(theta) = 2F1 * sin(25.0°)

Step 2: Calculate the x-component of F1.
Since the horizontal components of F1 and F2 are in the same direction, the x-component of F1 would be positive.

F1x = F1

Step 3: Calculate the y-component of F1.
Since the vertical components of F1 and F2 are in opposite directions, subtract F2y from F1y.

F1y = F2y - Resultant force = 2F1 * sin(25.0°) - 430.0

Step 4: Substitute F1x and F1y back into the equation F1x + F2x = 0.
F1 + 2F1 * cos(25.0°) = 0

Step 5: Solve for F1.
F1 + 2F1 * cos(25.0°) = 0
F1(1 + 2cos(25.0°)) = 0
F1 = 0 [ignored] or F1 = 0 / (1 + 2cos(25.0°))

Now, you have the value of F1. To find F2, multiply F1 by 2.

F2 = 2F1 =
2 * [0 / (1 + 2cos(25.0°))]

Simplifying the equation will give you the final values of F1 and F2.