Home » Classroom » SS2 Physics Mechanics - Motion in a Plane Question

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.

Please share this, thanks:

Add your answer

Notice: Posting irresponsibily can get your account banned!

No responses