Compute algebraically the resultant Of following coplanar force : 100N at 30° , 141.4N at 45° and 100N at 240°

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F= f1 +f2

F1=100cos30 + 141.4cos45 + 100cos240
=136.6i
F2=100sin30 + 141.4sin45 + 100sin240
=63.4j

R^2 =f^2 +f^2
R= 0.15kN at 25

Tan= y/x

Well, well, well! Looks like we have some coplanar forces playing around. Let's find out their resultant!

First, let's represent each force as a vector. The direction is given by the angle and the magnitude (the length) is given by the value in front of it.

So, we have three forces:
1. 100N at 30°
2. 141.4N at 45°
3. 100N at 240°

Now, let's break these down into their horizontal and vertical components and add them up separately.

For the first force (100N at 30°):
Horizontal component = 100N * cos(30°)
Vertical component = 100N * sin(30°)

For the second force (141.4N at 45°):
Horizontal component = 141.4N * cos(45°)
Vertical component = 141.4N * sin(45°)

For the third force (100N at 240°):
Horizontal component = 100N * cos(240°)
Vertical component = 100N * sin(240°)

Now, let's add up the horizontal and vertical components separately:

Horizontal resultant = Sum of all horizontal components
Vertical resultant = Sum of all vertical components

Finally, we can calculate the magnitude and angle of the resultant vector using the horizontal and vertical resultants:

Resultant magnitude = sqrt((Horizontal resultant)^2 + (Vertical resultant)^2)
Resultant angle = arctan(Vertical resultant / Horizontal resultant)

And voila! There you have it! The mathemagical resultant of those coplanar forces. I hope my clowning around helped you compute it all!

To compute the algebraic resultant of these coplanar forces, we first need to represent the forces as vectors. We can use vector addition to find the resultant.

1. Let's represent the first force of 100N at 30°. We can break it down into its horizontal and vertical components. The horizontal component (Fx1) can be calculated using cosine, and the vertical component (Fy1) can be calculated using sine:

Fx1 = 100N * cos(30°)
= 100N * (√3/2)
= 100√3/2 N

Fy1 = 100N * sin(30°)
= 100N * (1/2)
= 50N

So, the first force can be represented as a vector: F1 = 100√3/2 N i + 50N j

2. Now, let's represent the second force of 141.4N at 45° in a similar manner:

Fx2 = 141.4N * cos(45°)
= 141.4N * (1/√2)
= 100N

Fy2 = 141.4N * sin(45°)
= 141.4N * (1/√2)
= 100N

So, the second force can be represented as a vector: F2 = 100N i + 100N j

3. Finally, let's represent the third force of 100N at 240°:

Fx3 = 100N * cos(240°)
= -100N/2
= -50N

Fy3 = 100N * sin(240°)
= -100N * √3/2
= -100√3/2 N

So, the third force can be represented as a vector: F3 = -50N i + (-100√3/2 N) j

4. Now, we can find the resultant force by adding all the x-components and y-components separately:

Rx = Fx1 + Fx2 + Fx3
= (100√3/2 N) + (100N) + (-50N)
= 100√3/2 N + 50N

Ry = Fy1 + Fy2 + Fy3
= (50N) + (100N) + (-100√3/2 N)
= 150N - 100√3/2 N

Therefore, the resultant force can be represented as a vector: R = (100√3/2 N + 50N) i + (150N - 100√3/2 N) j

To find the magnitude and direction of the resultant force, we can use the Pythagorean theorem and inverse tangent:

Magnitude (R) = √((100√3/2 N + 50N)^2 + (150N - 100√3/2 N)^2)

Direction (θ) = tan^(-1)((150N - 100√3/2 N) / (100√3/2 N + 50N))

You can now compute the values for the magnitude and direction of the resultant force using the above formulas and the given values of the forces.

100cis30° + 141.4cis45° + 100cis240° = 136.6 + 63.38i = 150.6 @ 24.9°