Question

A vector $\overrightarrow{\mathrm{A}}$ has a magnitude of $40.0\mathrm{m}$ and points in a direction ${20.0}^{\circ }$ below the positive $x$ axis. A second vector, $\overrightarrow{\mathbf{B}}$, has a magnitude of $75.0\mathrm{m}$ and points in a direction ${50.0}^{\circ }$ above the positive $x$ axis. Sketch the vectors $\overrightarrow{A},\overrightarrow{B}$, and $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$.

Short answer

Vector C has a magnitude of 96.00 m and is at a 27.09° angle above the x-axis.

Step-by-step solution

Decompose Vector A into Components

First, we will find the individual components of the vector $\overrightarrow{\mathrm{A}}$. Given the magnitude is $40.0\mathrm{m}$ and the direction is ${20.0}^{\circ }$ below the positive $x$-axis, this angle can be taken as negative. - $A_x=A\cdot \cos (-{20.0}^{\circ })=40.0\cdot \cos (-{20.0}^{\circ })$- $A_y=A\cdot \sin (-{20.0}^{\circ })=40.0\cdot \sin (-{20.0}^{\circ })$Calculating these:- $A_x\approx 40.0\cdot 0.9397=37.59\mathrm{m}$- $A_y\approx 40.0\cdot (-0.3420)=-13.68\mathrm{m}$

Decompose Vector B into Components

Next, we calculate the components of $\overrightarrow{\mathbf{B}}$. The magnitude is $75.0\mathrm{m}$, and the angle is ${50.0}^{\circ }$ above the $x$-axis.- $B_x=B\cdot \cos ({50.0}^{\circ })=75.0\cdot \cos ({50.0}^{\circ })$- $B_y=B\cdot \sin ({50.0}^{\circ })=75.0\cdot \sin ({50.0}^{\circ })$Calculating these:- $B_x\approx 75.0\cdot 0.6428=48.21\mathrm{m}$- $B_y\approx 75.0\cdot 0.7660=57.45\mathrm{m}$

Find Components of Vector C

Now we will find the components of $\overrightarrow{\mathbf{C}}=\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}$ by adding the corresponding components of $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$.- $C_x=A_x+B_x=37.59+48.21\approx 85.80\mathrm{m}$- $C_y=A_y+B_y=-13.68+57.45\approx 43.77\mathrm{m}$

Calculate Magnitude and Direction of Vector C

The magnitude of $\overrightarrow{\mathbf{C}}$ is calculated using the Pythagorean theorem:$$C=\sqrt{C_x^2+C_y^2}=\sqrt{(85.80{)}^2+(43.77{)}^2}\approx 96.00\mathrm{m}$$The direction (angle $\theta$ from the positive $x$-axis) can be found using the tangent function:$$\theta ={\tan }^{-1}\left(\frac{C_y}{C_x}\right)={\tan }^{-1}\left(\frac{43.77}{85.80}\right)\approx {27.09}^{\circ }$$

Sketch the Vectors

Draw vectors $\overrightarrow{A}$, $\overrightarrow{B}$, and $\overrightarrow{C}$ on a coordinate system:- $\overrightarrow{A}$ should be drawn at a ${20}^{\circ }$ angle below the positive $x$-axis, extending to a length that represents $40\mathrm{m}$.- $\overrightarrow{B}$ should be drawn at a ${50}^{\circ }$ angle above the positive $x$-axis, extended to a length for $75\mathrm{m}$.- $\overrightarrow{C}$ is the resultant vector starting from the origin and ending at the tip of $\overrightarrow{B}$ when $\overrightarrow{A}$ is positioned head to tail. The angle with the $x$-axis is ${27.09}^{\circ }$ and magnitude $96.00\mathrm{m}$.

Key concepts

Magnitude

In the world of vectors, magnitude refers to the "length" or "size" of the vector. It's a measure of how much of the quantity the vector represents. For instance, in physics, if a vector represents a force, its magnitude is how strong that force is measured typically in units like meters (m) for displacement or newtons (N) for force.

To calculate the magnitude of a vector, we use the Pythagorean theorem for components on the xy-plane:

• For vector $\overrightarrow{\mathrm{C}}$, we find it by $$C=\sqrt{C_x^2+C_y^2}\approx 96.00\mathrm{m}$$

The method involves squaring each component (x and y), adding them together, and then taking the square root of this sum.
It's like finding the hypotenuse of a right-angled triangle. This is a universal method for any vector if components are known.

Remember, the magnitude is always a non-negative number, as it represents a length. Even if the vector points in a negative direction, like below the x-axis, the magnitude remains positive.

Direction

A vector isn’t just about size; direction is equally important, indicating where the vector points. Direction can be given in degrees or radians, often measured from a reference line like the positive x-axis.

For example, the direction of vector $\overrightarrow{\mathrm{A}}$ in the exercise is "20° below the positive x-axis," and $\overrightarrow{\mathrm{B}}$ is "50° above the positive x-axis." These directions specify the orientation of each vector, allowing accurate representation in diagrams.
To determine the direction of a resultant vector, like $\overrightarrow{\mathrm{C}}$, the tangent inverse function is handy:

• $$\theta ={\tan }^{-1}\left(\frac{C_y}{C_x}\right)\approx {27.09}^{\circ }$$

This angle $\theta$ tells us how vector $\overrightarrow{\mathrm{C}}$ is tilted from the positive x-line. Mastery of direction involves being able to interpret and calculate these angles based on given information.

Direction, just like magnitude, is crucial for understanding how vectors interact or combine.

Component Vectors

Component vectors are the building blocks that help break down a vector into more manageable parts along common axes, usually the x and y axes. Imagine taking a walk that includes moving forward and sideward. We could break this movement into two components: forward (x-axis) and sideward (y-axis).

For instance, vector $\overrightarrow{\mathrm{A}}$ with a magnitude of 40.0 m can be split into components $A_x$ and $A_y$ using trigonometric functions:

• $A_x=40.0\cdot \cos (-{20.0}^{\circ })$
• $A_y=40.0\cdot \sin (-{20.0}^{\circ })$

Similarly, vector $\overrightarrow{\mathrm{B}}$ with a magnitude of 75.0 m can also be split:

• $B_x=75.0\cdot \cos ({50.0}^{\circ })$
• $B_y=75.0\cdot \sin ({50.0}^{\circ })$

These components allow for easier vector addition, as we simply sum the x-components together and the y-components together to find the resultant vector, $\overrightarrow{\mathrm{C}}$. Components transform complex vector problems into simpler arithmetic, aiding in both calculation and conceptual understanding.