Consider two vectors where F1 = 31 N, F2 = 67 N,and theta 1 = 240 degrees and theta 2 = 25 degrees, measured from the positive x-axis with counter-clockwise being positive. What is the magnitude of the equilibriant?
Help will be provided if work is shown
31^2+67^2=5186
sqrt(5186)=72.01388
ANSWER =72.01388
To find the magnitude of the equilibrant, we first need to find the resultant vector. The resultant vector is the vector sum of the given vectors F1 and F2.
To find the resultant vector (R), we can use the following equation:
R = F1 + F2
Now, let's break down the given vectors into their x and y components.
For F1:
Magnitude (F1) = 31 N
Angle (θ1) = 240 degrees
Calculating the x-component of F1:
F1x = F1 * cos(θ1)
= 31 N * cos(240 degrees)
Calculating the y-component of F1:
F1y = F1 * sin(θ1)
= 31 N * sin(240 degrees)
Similarly, for F2:
Magnitude (F2) = 67 N
Angle (θ2) = 25 degrees
Calculating the x-component of F2:
F2x = F2 * cos(θ2)
= 67 N * cos(25 degrees)
Calculating the y-component of F2:
F2y = F2 * sin(θ2)
= 67 N * sin(25 degrees)
Now, we can find the resultant vector R by adding the x and y components of F1 and F2:
Rx = F1x + F2x
= (31 N * cos(240 degrees)) + (67 N * cos(25 degrees))
Ry = F1y + F2y
= (31 N * sin(240 degrees)) + (67 N * sin(25 degrees))
To find the magnitude of the resultant vector, we can use the Pythagorean theorem:
|R| = sqrt(Rx^2 + Ry^2)
Substituting the values of Rx and Ry, we can calculate the magnitude of the resultant vector R.
Once we have the magnitude of the resultant vector R, we can find the magnitude of the equilibrant, which is equal in magnitude but opposite in direction.
Therefore, the magnitude of the equilibrant is the same as the magnitude of the resultant vector R.