Question

Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of south. The resultant vector A+ B is given by

Answer 1

The magnitude of the resultant vector when vector A is 5 units towards west and vector B is 3 units towards south is calculated using the Pythagorean theorem, resulting in approximately 5.83 units.

Step-by-step explanation:

To calculate the magnitude of the resultant vector when vector A is 5 units towards west and vector B is 3 units towards south, we use the Pythagorean theorem as these vectors are perpendicular to each other.

The resultant vector (R) can be found by squaring the magnitudes of the two vectors, adding them together, and then taking the square root:

R = √(A² + B²)

Thus

R = √(5² + 3²)

R = √(25 + 9)

R = √34

R = approximately 5.83 units

Therefore, the magnitude of the resultant vector R is about 5.83 units.

Answer 2

The resultant vector
`\vec R = \vec A+\vec B` is given by
`\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]`.

Step-by-step explanation:

Let
`\vec A = 6\cdot (\cos 30^(\circ)\,\hat{i}+\sin 30^(\circ)\,\hat{j})` and
`\vec B = 4\cdot (-\sin 30^(\circ)\,\hat{i}-\cos 30^(\circ)\,\hat{j})`, both measured in meters. The resultant vector
`\vec R` is calculated by sum of components. That is:

`\vec R = \vec A+\vec B` (Eq. 1)

`\vec R = 6\cdot (\cos 30^(\circ)\,\hat{i}+\sin 30^(\circ)\,\hat{j})+4\cdot (-\sin 30^(\circ)\,\hat{i}-\cos 30^(\circ)\,\hat{j})`

`\vec R = (6\cdot \cos 30^(\circ)-4\cdot \sin 30^(\circ))\,\hat{i}+(6\cdot \sin 30^(\circ)-4\cdot \cos 30^(\circ))\,\hat{j}`

`\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]`

The resultant vector
`\vec R = \vec A+\vec B` is given by
`\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]`.