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....
To find the magnitude and direction of the net force using vector components, we need to resolve each force into its horizontal and vertical components. Then, we can add up the horizontal and vertical components separately to get the components of the net force.
First, let's find the horizontal and vertical components of each force:
For F1:
- Horizontal component (F1x): F1x = F1 * cosθ1
= 40N * cos(90')
= 0N, since cos(90') = 0
- Vertical component (F1y): F1y = F1 * sinθ1
= 40N * sin(90')
= 40N
For F2:
- Horizontal component (F2x): F2x = F2 * cosθ2
= 25N * cos(30')
= 25N * (√3/2)
= 21.65N
- Vertical component (F2y): F2y = F2 * sinθ2
= 25N * sin(30')
= 25N * (1/2)
= 12.5N
For F3:
- Horizontal component (F3x): F3x = F3 * cosθ3
= 12N * cos(290')
= 12N * (-√3/2)
= -10.39N
- Vertical component (F3y): F3y = F3 * sinθ3
= 12N * sin(290')
= 12N * (-1/2)
= -6N
Now we have the horizontal (x) and vertical (y) components of each force. Let's add them up:
Sum of horizontal components (ΣFx) = F1x + F2x + F3x
= 0N + 21.65N + (-10.39N)
= 11.26N
Sum of vertical components (ΣFy) = F1y + F2y + F3y
= 40N + 12.5N + (-6N)
= 46.5N
Using these components, we can find the magnitude and direction of the net force:
Magnitude of the net force (|Fnet|) = √[(ΣFx)^2 + (ΣFy)^2]
= √[(11.26N)^2 + (46.5N)^2]
= √(126.6276N^2 + 2162.25N^2)
= √(2288.8776N^2)
= 47.86N (approximately)
Direction of the net force (θnet) = tan^(-1)(ΣFy / ΣFx)
= tan^(-1)(46.5N / 11.26N)
= tan^(-1)(4.13)
= 76.68° (approximately)
Therefore, the magnitude of the net force is approximately 47.86 N, and it is directed at an angle of approximately 76.68°.