Consider four vectors F~ 1, F~ 2, F~ 3, and F~ 4, where their magnitudes are F1
ID: 1529536 • Letter: C
Question
Consider four vectors F~ 1, F~ 2, F~ 3, and F~ 4, where their magnitudes are F1 = 41 N, F2 = 39 N, F3 = 22 N, and F4 = 58 N. Let 1 = 140 , 2 = 140 , 3 = 20 , and 4 = 50 , measured from the positive x axis with the counter-clockwise angular direction as positive. What is the magnitude of the resultant vector F~ , where F~ = F~ 1 + F~ 2 + F~ 3 + F~ 4? Answer in units of N.
What is the direction of this resultant vector F~ ? Note: Give the angle in degrees, use counterclockwise as the positive angular direction, between the limits of 180 and +180 from the positive x axis. Answer in units of
Given two vectors F~ 1, and F~ 2. Where the magnitude of these vectors are F1 = 54 N , and F2 = 87 N . And where 1 = 169 , and 2 = 84 . The angles are measure from the positive x axis with the counter-clockwise angular direction as positive. What is the magnitude of the resultant vector kF~ k, where F~ = F~ 1 + F~ 2 ? Answer in units of N
. Note: Give the angle in degrees, use counterclockwise as the positive angular direction, between the limits of 180 and +180 from the positive x axis. What is the direction of this resultant vector F~ ? Answer in units of .
Explanation / Answer
1)
x-component of resultant force,
Fx = F1*cos(theta) + F2*cos(theta2) + F3*cos(theta3) + F4*cos(theta4)
= 41*cos(140) + 39*cos(-140) + 22*cos(20) + 58*cos(-50)
= -3.33 N
y-component of resultant force,
Fy = F1*sin(theta) + F2*sin(theta2) + F3*sin(theta3) + F4*sin(theta4)
= 41*sin(140) + 39*sin(-140) + 22*sin(20) + 58*sin(-50)
= -36.6 N
so, magnitude of resulstant force,
|F| = sqrt(Fx^2 + Fy^2)
= sqrt(3.33^2 + 36.6^2)
= 36.75 N <<<<<<<------------------Answer
direction : theta = tan^-1(Fy/Fx)
= tan^-1(36.6/3.33)
= 84.8 degrees below -x axis
= -95.2 degrees with +x axis(clockwise) <<<<<<<------------------Answer
2)
x-component of resultant force,
Fx = F1*cos(theta) + F2*cos(theta2)
= 54*cos(169) + 87*cos(84)
= -43.9 N
y-component of resultant force,
Fy = F1*sin(theta) + F2*sin(theta2)
= 54*sin(169) + 87*sin(84)
= 96.8 N
so, magnitude of resulstant force,
|F| = sqrt(Fx^2 + Fy^2)
= sqrt(43.9^2 + 96.8^2)
= 106.3 N <<<<<<<------------------Answer
direction : theta = tan^-1(Fy/Fx)
= tan^-1(-43.9/96.8)
= 24.4 degrees above +x axis
= 155.6 degrees with +x axis (counter clockwise) <<<<<<<------------------Answer
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