Question

Find the magnitude of the sum

Answer 1

Answer:

Approximately $3.01\; \rm m$.

Explanation:

Decompose each vector into the sum of two vectors: a horizontal one (parallel to the arrow that points to the right) and a vertical one (parallel the arrow that points upwards.)

Vector $\sf A$ is horizontal and is at an angle of $0^\circ$ with the horizon.  

• Horizontal component of vector $\sf A$: to the right, with a length of $5.00\; \rm m \cdot \cos\left(0^\circ \right) = 5.00\; \rm m$.
• Vertical component of vector $\sf A$: $5.00\; \rm m \cdot \sin\left(0^\circ \right) = 0\; \rm m$.

Vector $\sf B$ is at an angle of $30^\circ$ below the horizon.

• Horizontal component of vector $\sf B$: to the right, with a length of $\displaystyle 6.00\; \rm m \cdot \cos\left(30^\circ \right) = (6.00\; \rm m) \times \frac{\sqrt{3}}{2}\approx 5.19615\; \rm m$.
• Vertical component of vector $\sf B$: downwards, with a length of$\displaystyle 6.00\; \rm m \cdot \sin\left(30^\circ \right) = 6.00\; \rm m \times \frac{1}{2} = 3.00\; \rm m$.

Calculate the sum of vector $\sf A$ and vector $\sf B$.

The horizontal component of vector $\sf A$ and vector $\sf B$ are opposite to one another. Therefore, the length of the horizontal component of $\sf (A + B)$ would be the difference between the length of the horizontal components of vector $\sf A\!$ and of vector $\sf B\!$:

$\displaystyle (6.00\; \rm m) \times \frac{\sqrt{3}}{2} - 5.00\; \rm m \approx 0.196152\; \rm m$.

The length of the vertical component of vector $\sf A$ is $0\; \rm m$. Therefore, the length of the vertical component of $\sf (A + B)$ would be equal to the length of the vertical component of vector $\sf B$, $\displaystyle 6.00\; \rm m \times \frac{1}{2} = 3.00\; \rm m$.

Therefore, the length of the horizontal and vertical component of $\sf (A + B)$ are approximately $0.196152\; \rm m$ and $3.00\; \rm m$, respectively. The length of vector $\sf (A + B)\!$ would be approximately:

$\displaystyle \sqrt{(0.196152\; \rm m)^{2} + (3.00\; \rm m)^{2}} \approx 3.01\; \rm m$.