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.
answer plz
First, let's label the larger pull as F₁ and the smaller pull as F₂.
From the problem statement, we know that F₁ is twice as large as F₂.
Let's break down the larger pull (F₁) into its vertical and horizontal components using trigonometry.
The vertical component (F₁y) can be found using the sine function:
F₁y = F₁ * sin(25°)
The horizontal component (F₁x) can be found using the cosine function:
F₁x = F₁ * cos(25°)
Now, let's write down the equation for the resultant of the two pulls in the vertical direction:
F₁y + F₂ = 350 N
Since F₁ is twice as large as F₂, we can rewrite this equation as:
2F₂ * sin(25°) + F₂ = 350 N
Simplifying this equation, we get:
2sin(25°) * F₂ + F₂ = 350 N
Combining like terms, we have:
(2sin(25°) + 1) * F₂ = 350 N
Now we can solve for F₂:
F₂ = 350 N / (2sin(25°) + 1)
Calculating this value, we find:
F₂ ≈ 149.5 N
Since F₁ is twice as large as F₂, we can determine that:
F₁ = 2 * F₂ = 2 * 149.5 N = 299 N
Now, to find the direction of the smaller pull (F₂), we can use trigonometry again. We know that the larger pull (F₁) is directed at 25 degrees west of north. Since the two pulls are in the same direction, the smaller pull (F₂) will have the same angle:
Direction of F₂ = 25° west of north
Therefore, the magnitude of the larger pull (F₁) is 299 N, the magnitude of the smaller pull (F₂) is 149.5 N, and the direction of the smaller pull is 25° west of north.
To find the magnitude of each pull and the direction of the smaller pull, we can break down the forces into their respective vector components.
Let's denote the magnitude of the smaller pull as F1 and the magnitude of the larger pull as F2.
The larger pull is directed at 25.0 degrees west of north, so its northward component would be F2 * cos(25°), and its westward component would be F2 * sin(25°).
Since the resultant of these two pulls is 350.0 N directly northward, this means the northward components of the two pulls must add up to 350.0 N. Therefore, we have:
F1 * sin(θ) + F2 * cos(25°) = 350.0 N ---(1)
where θ is the angle of the smaller pull with the north direction.
The smaller pull is directed towards the north, so its northward component would be F1 * cos(θ), and its westward component would be F1 * sin(θ).
Since the net westward pull is zero, the westward components of the two pulls must cancel each other out. Therefore, we have:
F1 * sin(θ) - F2 * sin(25°) = 0 ---(2)
Now we can solve equations (1) and (2) simultaneously to find the magnitude of each pull (F1 and F2) and the direction of the smaller pull (θ).
Let's solve these equations:
From equation (2), we can isolate F1 * sin(θ):
F1 * sin(θ) = F2 * sin(25°)
Substituting this into equation (1):
F2 * sin(25°) + F2 * cos(25°) = 350.0 N
Combining like terms:
F2 * (sin(25°) + cos(25°)) = 350.0 N
Dividing both sides by sin(25°) + cos(25°):
F2 = 350.0 N / (sin(25°) + cos(25°))
Calculating this value:
F2 ≈ 206.6 N
Now we can substitute this value back into equation (2) to find F1:
F1 * sin(θ) = F2 * sin(25°)
F1 = F2 * sin(25°) / sin(θ)
Substituting the known values:
F1 = 206.6 N * sin(25°) / sin(θ)
Since we know that F1 is half the magnitude of F2, we have:
F2 / 2 = 206.6 N * sin(25°) / sin(θ)
Simplifying:
206.6 N * sin(25°) / sin(θ) = 2 * 206.6 N
sin(25°) / sin(θ) = 2
Using trigonometric identities, we can rewrite this as:
sin(θ) = sin(25°) / 2
Taking the inverse sine of both sides:
θ = arcsin(sin(25°) / 2)
Calculating this angle:
θ ≈ 14.5°
Finally, we have found the magnitudes and direction of each pull:
Magnitude of smaller pull (F1) ≈ 103.3 N
Magnitude of larger pull (F2) ≈ 206.6 N
Direction of the smaller pull (θ) ≈ 14.5° north of the north direction.