If A=250g at an angle of 0 degrees and B=150g at an angle of 120 degrees, what will C be? both magnitude and direction.
use law of cosines
or, break into components and add x and y parts
A + B = 250g[0o] + 150g[120o].
X = 250 + 150*Cos120 = 175 Grams.
Y = 150*sin120 = 130 grams.
175 + 130i = 218g[36.6o] CCW.
Tan A = Y/X.
The magnitude and direction of C for equilibrium = 218g[36.6+180] =
218g[216.6] CCW.
To find the magnitude and direction of vector C, we need to add vectors A and B using vector addition. To perform vector addition, we break down each vector into its horizontal (x) and vertical (y) components.
For vector A:
Magnitude (A): 250g
Angle (θA): 0 degrees
For vector B:
Magnitude (B): 150g
Angle (θB): 120 degrees
First, we need to find the x and y components of each vector.
The x-component can be obtained by using the formula: x = magnitude * cos(angle)
The y-component can be obtained by using the formula: y = magnitude * sin(angle)
For vector A:
Ax = 250g * cos(0 degrees) = 250g * 1 = 250g
Ay = 250g * sin(0 degrees) = 0g
For vector B:
Bx = 150g * cos(120 degrees) = 150g * (-0.5) = -75g
By = 150g * sin(120 degrees) = 150g * (√3/2) ≈ 129.9g
Next, we add the x-components and y-components separately to obtain the resultant vector Cx and Cy.
Cx = Ax + Bx
Cy = Ay + By
Cx = 250g + (-75g) = 175g
Cy = 0g + 129.9g ≈ 129.9g
Now, we have the x and y components of vector C, which are Cx = 175g and Cy ≈ 129.9g, respectively.
To find the magnitude of vector C, we use the Pythagorean theorem:
Magnitude (C) = √(Cx^2 + Cy^2)
Magnitude (C) = √((175g)^2 + (129.9g)^2) ≈ √30625g^2 + 16880.01g^2 ≈ √47505g^2 ≈ 217.97g
To find the direction of vector C, we use the inverse tangent function:
Angle (θC) = atan(Cy / Cx)
Angle (θC) = atan(129.9g / 175g) ≈ atan(0.7429) ≈ 36.68 degrees
Therefore, the magnitude of vector C is approximately 217.97g, and the direction is approximately 36.68 degrees.