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 530.0 N directly northward.
Part A
Use vector components to find the magnitude of each of these pulls. Assume that the smaller pull has a component directed to the north.
Enter your answers numerically separated by a comma?
B) Use vector components to find the direction of the smaller pull. Assume that the smaller pull has a component directed to the north.
To find the magnitude of each pull, we can use vector components. Let's call the magnitude of the smaller pull "F" and the magnitude of the larger pull "2F" (since the larger pull is twice as hard).
The resultant of the two pulls is directed northward. This means that the northward component of the larger pull must cancel out the northward component of the smaller pull. So the northward component of the larger pull is equal to the northward component of the smaller pull, making it "F".
Using this information, we can set up the following equation:
F + F = 530 N
Simplifying the equation, we get:
2F = 530 N
Solving for F, we divide both sides by 2:
F = 265 N
So the magnitude of the smaller pull is 265 N, and the magnitude of the larger pull is 2F = 2 * 265 N = 530 N.
For part B, we need to find the direction of the smaller pull. We are given that the larger pull is directed at 25.0° west of north. Since the two pulls are in opposite directions, the smaller pull must be directed in the opposite direction of the larger pull.
Therefore, the direction of the smaller pull is 180° + 25.0° = 205.0° east of north.