Fnet= F1+F2+F3, F1=40N with bearing of 90' , F2=25N with bearing of 30' ,and F3= 12N with bearing of 290' what is the magnitude and direction ....using VECTOR COMPONENTS....
F1 = 0i+40j
F2 = 21.65i + 12.5j
F3 = 4.10i - 11.28j
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Fnet = 25.75i + 41.22j
= 48.60 at 50°
To find the magnitude and direction of the net force using vector components, we first need to break down each force into its x and y components.
Let's start with F1 = 40N with a bearing of 90°. Since the bearing is 90°, this force is purely in the y-direction. Therefore, the x-component of F1 is 0 and the y-component is 40N.
Next, let's consider F2 = 25N with a bearing of 30°. To find the x and y components, we can use trigonometry. The x-component of F2 is given by F2*cos(30°), and the y-component is given by F2*sin(30°):
F2x = 25N * cos(30°) = 21.65N
F2y = 25N * sin(30°) = 12.50N
Finally, let's consider F3 = 12N with a bearing of 290°. However, we need to adjust the bearing since the y-axis is positive in the upward direction. To do this, we subtract the given bearing from 360°. So, the adjusted bearing is 360° - 290° = 70°. Using the same trigonometric approach, we can find the components:
F3x = 12N * cos(70°) = 4.37N
F3y = 12N * sin(70°) = 11.24N
Now that we have the components of each force, we can add up the x-components and the y-components separately to find the net force:
Net Force in the x-direction:
Fnet,x = F2x + F3x = 21.65N + 4.37N = 26.02N
Net Force in the y-direction:
Fnet,y = F1 + F2 = 40N + 12.50N + 11.24N = 63.74N
To find the magnitude of the net force, we can use the Pythagorean theorem:
Magnitude of Fnet = sqrt(Fnet,x^2 + Fnet,y^2)
Magnitude of Fnet = sqrt((26.02N)^2 + (63.74N)^2) = 68.94N
To find the direction of the net force, we can use the inverse tangent function:
Direction of Fnet = arctan(Fnet,y / Fnet,x)
Direction of Fnet = arctan(63.74N / 26.02N) = 67.9°
Therefore, the magnitude of the net force is 68.94N, and the direction is 67.9°.