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 600.0 N directly northward.
Use vector components to find the magnitude of each of these pulls. Assume that the smaller pull has a component directed to the north.??
Sorry. A force directed N and a force directed west of N cannot have a resultant directed N.
To find the magnitude of each pull, we can break down the given information into vector components and solve for the unknowns.
Let's start by assigning variables to the magnitudes of the two pulls. Suppose the smaller pull (P1) has a magnitude of x, while the larger pull (P2) has a magnitude of 2x.
Now, let's express the given directions as angles relative to the positive x-axis. Since the larger pull is directed at 25.0° west of north, its angle relative to the positive x-axis would be 90° - 25.0° = 65.0°. We can refer to this angle as θ2.
Since the resultant of the two pulls is directly northward, the angle of the resultant would be 0°. We can refer to this angle as θR.
Using these angles, we can now find the components of each pull. The y-component (northward component) of P2 is P2 * sin(θ2), while the x-component (eastward component) is P2 * cos(θ2). Since we know the angle, θR, of the resultant, the y-component of P2 should equal the magnitude of the resultant: P2 * sin(θ2) = 600 N.
Using the relationship between P1 and P2 (P1 = 2P2), we can express the components of P1 as an equation: 2P2 * sin(θ1) = x.
Now, let's solve for the unknowns:
1. From the equation P2 * sin(θ2) = 600 N, we can calculate the value of P2:
P2 = 600 N / sin(θ2) = 600 N / sin(65.0°).
2. Using the relationship P1 = 2P2, we find:
P1 = 2 * P2 = 2 * (600 N / sin(65.0°)).
Now, you can substitute the value of P2 back into the equation to find P1.