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= 40sin90 40cos90
f2= 25sin30 +25cos30
f3= +12sin290 12cos30
find each of these values but you can only add cosines with cosines and sines with sines, but then with those values you will need to find the magnitude and the angle of the values
in case that doesnt make sense.
[(40cos90+25cos30+12COS290)^2 +
(40sin90+25sin30+12sin290)^2]^(1/2)
thats what you need to do...remember this for physics
To find the magnitude and direction of the resultant force using vector components, we need to break each force down into its x and y components.
First, let's find the x and y components of each force:
For F1:
Magnitude: 40 N
Bearing: 90°
The x-component (F1x) can be found using the formula:
F1x = F1 * cos(θ)
where θ is the bearing. In this case, θ = 90°.
F1x = 40 N * cos(90°)
F1x = 0 N
The y-component (F1y) can be found using the formula:
F1y = F1 * sin(θ)
F1y = 40 N * sin(90°)
F1y = 40 N
So, the x-component of F1 (F1x) is 0 N, and the y-component (F1y) is 40 N.
For F2:
Magnitude: 25 N
Bearing: 30°
The x-component (F2x) can be found using the formula:
F2x = F2 * cos(θ)
where θ is the bearing. In this case, θ = 30°.
F2x = 25 N * cos(30°)
F2x = 21.65 N
The y-component (F2y) can be found using the formula:
F2y = F2 * sin(θ)
F2y = 25 N * sin(30°)
F2y = 12.5 N
So, the x-component of F2 (F2x) is 21.65 N, and the y-component (F2y) is 12.5 N.
For F3:
Magnitude: 12 N
Bearing: 290°
The x-component (F3x) can be found using the formula:
F3x = F3 * cos(θ)
where θ is the bearing. In this case, θ = 290°.
F3x = 12 N * cos(290°)
F3x = -9.73 N
The y-component (F3y) can be found using the formula:
F3y = F3 * sin(θ)
F3y = 12 N * sin(290°)
F3y = -5.88 N
So, the x-component of F3 (F3x) is -9.73 N, and the y-component (F3y) is -5.88 N.
Now, we can find the resultant force components by summing up the x-components and y-components of all the forces.
Resultant x-component (Rx) = F1x + F2x + F3x
Rx = 0 N + 21.65 N + (-9.73 N)
Rx = 11.92 N
Resultant y-component (Ry) = F1y + F2y + F3y
Ry = 40 N + 12.5 N + (-5.88 N)
Ry = 46.62 N
The magnitude (R) of the resultant force is given by the Pythagorean theorem:
R = sqrt(Rx^2 + Ry^2)
R = sqrt((11.92 N)^2 + (46.62 N)^2)
R = sqrt(142.01 N^2 + 2168.84 N^2)
R = sqrt(2310.85 N^2)
R ≈ 48.08 N
To find the direction (θ) of the resultant force, we use the inverse tangent function:
θ = tan^(-1)(Ry/Rx)
θ = tan^(-1)(46.62 N / 11.92 N)
θ ≈ 74.9°
Therefore, the magnitude of the resultant force is approximately 48.08 N, and the direction is approximately 74.9°.