Consider two vectors where F1 = 45 N, F2 = 63 N, and \theta _{1} = 240o and \theta _{2} = 25o, measured from the positive x-axis with counter-clockwise being positive. What is the magnitude of the equilibriant? [to 3 decimal places]
I explained how to do this already, a couple of hours ago. I assumed you know how to add vectors. If you don't, review your course work, or search the internet for a tutorial on vector addition by adding components. You need to make an effort.
To find the magnitude of the equilibriant vector, we first need to convert the given vectors into their x and y components. Using the given magnitudes and angles, we can use trigonometry to find the x and y components for each vector.
For F1:
F1x = F1 * cos(theta1)
F1y = F1 * sin(theta1)
Substituting the given values:
F1x = 45 * cos(240)
F1x ≈ -22.227 N (rounded to three decimal places)
F1y = 45 * sin(240)
F1y ≈ -38.784 N (rounded to three decimal places)
Similarly, for F2:
F2x = F2 * cos(theta2)
F2y = F2 * sin(theta2)
Substituting the given values:
F2x = 63 * cos(25)
F2x ≈ 57.341 N (rounded to three decimal places)
F2y = 63 * sin(25)
F2y ≈ 26.366 N (rounded to three decimal places)
Now, to find the equilibriant, we need to find the negative sum of the x and y components of F1 and F2.
Equilibriantx = -F1x - F2x
Equilibrianty = -F1y - F2y
Substituting the values we calculated earlier:
Equilibriantx = -(-22.227) - 57.341
Equilibriantx ≈ 79.568 N (rounded to three decimal places)
Equilibrianty = -(-38.784) - 26.366
Equilibrianty ≈ 12.418 N (rounded to three decimal places)
Finally, we can calculate the magnitude of the equilibriant using Pythagorean theorem:
Magnitude of the equilibriant = sqrt(Equilibriantx^2 + Equilibrianty^2)
Magnitude of the equilibriant = sqrt((79.568)^2 + (12.418)^2)
Magnitude of the equilibriant ≈ 80.070 N (rounded to three decimal places)
Therefore, the magnitude of the equilibriant vector is approximately 80.070 N.