Vector A has a magnitude of 8 units and is dire... - SS2 Physics Mechanics - Motion in a Plane Question
Vector A has a magnitude of 8 units and is directed at an angle of 30 degrees above the positive x-axis. Vector B has a magnitude of 5 units and is directed at an angle of 60 degrees below the positive x-axis. Calculate the magnitude and direction of the vector sum A + B.
To find the vector sum A + B, we need to add the x-components and the y-components of the vectors separately and then combine them.
For vector A:
Ax = 8 x cos(30°) = 8 x 0.866 = 6.928 units
Ay = 8 x sin(30°) = 8 x 0.5 = 4 units
For vector B:
Bx = 5 x cos(240°) = 5 x (-0.5) = -2.5 units
By = 5 x sin(240°) = 5 x (-0.866) = -4.33 units
Now, add the x-components and y-components separately:
Sum of x-components = Ax + Bx = 6.928 - 2.5 = 4.428 units
Sum of y-components = Ay + By = 4 - 4.33 = -0.33 units
To find the magnitude of the vector sum, use the Pythagorean theorem:
Magnitude = ((Sum of x-components)2 + (Sum of y-components)2)
Magnitude = ((4.428)2 + (-0.33)2) = (19.62 + 0.1089) = (19.7289) ≈ 4.442 units
To find the direction of the vector sum, use the inverse tangent function:
Direction = tan(-1)((Sum of y-components) / (Sum of x-components))
Direction = tan(-1)(-0.33 / 4.428) ≈ -4.3 degrees
Therefore, the magnitude of the vector sum A + B is approximately 4.442 units, and its direction is approximately -4.3 degrees below the positive x-axis.
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