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 560.0 directly northward.
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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 directly northward.
To solve this problem, we can break down the given information into components and use vector addition to find the resultant magnitude and direction.
Let's denote the magnitude of the smaller pull as 'S' and the magnitude of the larger pull as '2S'. The direction of the larger pull is given as 25.0° west of north.
To find the resultant of these two pulls, we can break them down into their x and y components.
For the smaller pull:
The x-component is S * 0 (since it is directly north) = 0
The y-component is S
For the larger pull:
The x-component is 2S * cos(25°) (west is negative direction)
The y-component is 2S * sin(25°)
Now, we can find the sum of the x-components and y-components to get the resultant components:
x-component: 0 + 2S * cos(25°)
y-component: S + 2S * sin(25°)
Given that the resultant of the two pulls is 560.0 directly northward, we can set up the equation:
560.0 = sqrt((0 + 2S * cos(25°))^2 + (S + 2S * sin(25°))^2)
Simplifying this equation will allow us to solve for the magnitude S.
After finding the value of S, you can substitute it back into the component equations to determine the individual magnitudes and directions of the smaller and larger pulls.