Consider four vectors ~F1, ~F2, ~F3, and ~F4, where their magnitudes are F1= 57 N, F2= 38 N, F3= 25 N, and F4= 67 N. Let θ1= 130◦, θ2=−130◦, θ3= 29◦, and θ4=−64◦, 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=~F1+~F2+~F3+~F4?
Answer in units of N.
x components of each
57 cos 130 = -36.6
38 cos -130= -24.4
25 cos 29=
67 cos -64=
sum of x components = sumx =
y components of each
57 sin 130=
38 sin -130=
25 sin 29=
67 sin -64=
sum of y components = sumy =
|F|^2 = sumx^2 + sumy^2
To find the magnitude of the resultant vector ~F, we'll use the concept of vector addition.
First, let's break down each force vector into its x and y components. Given the magnitudes and angles provided, we can use trigonometry to calculate these components.
For ~F1:
Magnitude: F1 = 57 N
Angle: θ1 = 130°
Using trigonometric functions, we can find the x and y components of ~F1:
Fx1 = F1 * cos(θ1)
Fy1 = F1 * sin(θ1)
Similarly, for ~F2:
Magnitude: F2 = 38 N
Angle: θ2 = -130°
Components of ~F2:
Fx2 = F2 * cos(θ2)
Fy2 = F2 * sin(θ2)
For ~F3:
Magnitude: F3 = 25 N
Angle: θ3 = 29°
Components of ~F3:
Fx3 = F3 * cos(θ3)
Fy3 = F3 * sin(θ3)
Finally, for ~F4:
Magnitude: F4 = 67 N
Angle: θ4 = -64°
Components of ~F4:
Fx4 = F4 * cos(θ4)
Fy4 = F4 * sin(θ4)
Next, we'll sum up the individual x and y components of the force vectors:
ΣFx = Fx1 + Fx2 + Fx3 + Fx4
ΣFy = Fy1 + Fy2 + Fy3 + Fy4
Now, we can combine the x and y components to find the resultant magnitude:
~F = sqrt(ΣFx^2 + ΣFy^2)
Substituting the calculated values, we can find the magnitude of ~F.