Two workers pull horizontally on a heavy box, but one pulls twice as hard as the other. The larger pull is directed at 25 degree west of north and the resultant of these two pulls is 350 netwon directly northward. Use vector components to find the magnitude of each of these pulls and the direction of the small pull.
See previous post: Wed, 11-2-16, 9:11 PM.
To solve this problem using vector components, we need to break down the forces into their vertical and horizontal components.
Let's assume that the magnitude of the smaller pull is F and the magnitude of the larger pull is 2F.
Given that the resultant of these two pulls is 350 Newton directly northward, we can conclude that the vertical components of both pulls add up to 350 Newton.
The vertical component of the smaller pull is given by F*sinθ, where θ is the angle it makes with the north direction. We don't have the value of θ, but we have the information that the larger pull is directed 25 degrees west of north.
Therefore, the angle of the smaller pull with the north direction is (25 - 180) degrees or -155 degrees. When measuring angles clockwise from the positive y-axis, negative angles indicate a downward direction. So, we will use -155 degrees for θ.
Using the information above, we can write the equation:
F*sin(-155) + 2F*sin(25) = 350
We will now solve this equation to find the value of F.
sin(-155) = -sin(155), so the equation becomes:
-F*sin(155) + 2F*sin(25) = 350
Now, solving for F, we get:
F * (-sin(155) + 2*sin(25)) = 350
F = 350 / (-sin(155) + 2*sin(25))
Substituting the values and calculating, we find F ≈ 486.3 Newton.
The magnitude of the smaller pull is approximately 486.3 Newton.
Finally, to find the magnitude of the larger pull, we multiply the magnitude of the smaller pull by 2:
Magnitude of larger pull = 2 * 486.3 ≈ 972.6 Newton.
Therefore, the magnitude of the larger pull is approximately 972.6 Newton.
The direction of the smaller pull is -155 degrees, indicating it is directed 155 degrees downward from the north direction.
So the magnitude of the smaller pull is approximately 486.3 Newton, and the direction is 155 degrees downward from the north. And the magnitude of the larger pull is approximately 972.6 Newton.