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 degrees west of north, and the resultant of these two pulls is 350.0 N directly northward.

Use vector components to find the magnitude of each of these pulls and the direction of the smaller pull.

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To solve this problem using vector components, we can break down the forces into their x and y components.

Let's assume the magnitude of the smaller pull is represented by F1, and the magnitude of the larger pull is represented by F2.

Given:
F2 = 2 * F1

Let's break down the forces into their x and y components:

For F1:
The smaller pull is directed at an angle of 25.0 degrees west of north.
The x-component of F1 = F1 * cos(25.0°)
The y-component of F1 = F1 * sin(25.0°)

For F2:
Since F2 is twice as strong as F1, we can say that:
The magnitude of the x-component of F2 is twice as much as the x-component of F1: F2x = 2 * (F1 * cos(25.0°))
The magnitude of the y-component of F2 is twice as much as the y-component of F1: F2y = 2 * (F1 * sin(25.0°))

To find the resultant force, we need to sum the x-components and the y-components separately:

Fx = F1 * cos(25.0°) + F2x
Fy = F1 * sin(25.0°) + F2y = 350N (since the resultant force is directly northward)

Knowing the values of Fx and Fy, we can solve for F1 and F2:

Fx = F1 * cos(25.0°) + 2 * (F1 * cos(25.0°))
Fy = F1 * sin(25.0°) + 2 * (F1 * sin(25.0°)) = 350N

Simplifying the equations:

Fx = F1 * (cos(25.0°) + 2 * cos(25.0°))
Fy = F1 * (sin(25.0°) + 2 * sin(25.0°)) = 350N

Fx = F1 * (3 * cos(25.0°))
Fy = F1 * (3 * sin(25.0°)) = 350N

Now, we can solve for F1:

F1 = Fy / (3 * sin(25.0°))
F1 = 350N / (3 * sin(25.0°))

Calculating this value will give us the magnitude of the smaller pull, F1.

To solve this problem using vector components, we need to break down the forces into their horizontal and vertical components.

Let's denote the larger pull by F1 and the smaller pull by F2.

Given:
Resultant force (F) = 350.0 N directed northward
Angle of F1 (θ1) = 25.0 degrees west of north

To find the magnitude of each pull:
1. Find the vertical component of the resultant force by using the sine function:
Fv = F * sin(θ1)

2. Since the larger pull is twice as strong as the smaller pull, let's denote the magnitude of the smaller pull as F2.
F2 = F1/2

3. We can now find the magnitude of the larger pull by using the Pythagorean theorem:
F1 = sqrt(F^2 - F2^2)

Now, let's calculate the magnitudes:

Step 1:
Fv = 350.0 N * sin(25.0 degrees)
Fv ≈ 148.15 N

Step 2:
F2 = F1/2

Step 3:
F1 = sqrt(F^2 - F2^2)
= sqrt((350.0 N)^2 - (F1/2)^2)
= sqrt(122500 N^2 - (F1^2)/4)

To simplify this equation, let's introduce a new variable x = F1^2:
x = F1^2

We can now rewrite the equation as:
x = 4 * 122500 N^2 - x/4

Multiplying through by 4 gives us:
4x = 488000 N^2 - x

Rearranging terms:
5x = 488000 N^2

Solving for x:
x = 488000 N^2 / 5
x = 97600 N^2

Taking the square root:
F1 = sqrt(x)
F1 = sqrt(97600 N^2)
F1 ≈ 312.45 N

Now that we have found the magnitude of F1, we can substitute it back into the equation for F2:
F2 = F1/2
F2 ≈ 312.45 N / 2
F2 ≈ 156.23 N

Therefore, the magnitudes of the larger pull (F1) and the smaller pull (F2) are approximately 312.45 N and 156.23 N, respectively.

To find the direction of the smaller pull:
The direction of the smaller pull can be found by subtracting the angle of the larger pull from 180 degrees since it pulls in the opposite direction. In this case, the angle of the smaller pull can be calculated as:
Angle of F2 = 180 degrees - θ1
Angle of F2 = 180 degrees - 25.0 degrees
Angle of F2 ≈ 155.0 degrees

Therefore, the direction of the smaller pull is approximately 155.0 degrees.