Question

Forces of 7.6N at 38 degrees and 11.8N at 143 degrees act at a point.Calculate the magnitude and direction of their resultant.

Answer 1

Approximately
`12.3\; {\rm N}` at approximately
`106^(\circ)`.

Step-by-step explanation:

Assume the two given directions are measured with respect to the positive
`x` axis.

If a vector of magnitude
`\| a\|` is at an angle of
`\theta` from the positive
`x` axis, this vector can be written in the component form as:

`\begin{aligned}\| a\|\begin{bmatrix}\cos(\theta) \\ \sin(\theta)\end{bmatrix}\end{aligned}`;

Or equivalently:

`\begin{aligned}\begin{bmatrix}\|a\|\, \cos(\theta) \\ \|a\|\, \sin(\theta)\end{bmatrix}\end{aligned}`.

For example, the
`7.6\; {\rm N}` force is a vector with magnitude
`7.6\; {\rm N}` at a direction of
`38^(\circ)` from the positive
`x` axis. This vector can be represented as:

`\begin{aligned} 7.6\, \begin{bmatrix}\cos(38^(\circ)) \\ \sin(38^(\circ))\end{bmatrix} &= \begin{bmatrix}7.6\, \cos(38^(\circ)) \\ 7.6\, \sin(38^(\circ))\end{bmatrix} \approx \begin{bmatrix}5.9889 \\ 4.6790 \end{bmatrix}\end{aligned}`.

Similarly, the
`11.8\; {\rm N}` vector can be represented as:

`\begin{aligned}11.8\, \begin{bmatrix}\cos(143^(\circ)) \\ \sin(143^(\circ))\end{bmatrix} &= \begin{bmatrix}11.8\, \cos(143^(\circ)) \\ 11.8\, \sin(143^(\circ))\end{bmatrix} \approx \begin{bmatrix}-9.4239 \\ 7.1014 \end{bmatrix}\end{aligned}`.

To find the sum of the two vectors, take the sum of each component separately:

`\begin{aligned}& \begin{bmatrix}5.9889 \\ 4.6790 \end{bmatrix} + \begin{bmatrix}-9.4239 \\ 7.1014\end{bmatrix} \\ =\; & \begin{bmatrix}5.9889 + (-9.4239)\\ 4.6790 + 7.1014\end{bmatrix} \\ \approx\; & \begin{bmatrix}-3.4350 \\ 11.780\end{bmatrix} \end{aligned}`.

Apply the Pythagorean Theorem to find the magnitude of this vector sum:

`\displaystyle \sqrt{(-3.4350)^(2) + (11.780)^(2)} \approx 12.271`.

Note that the first component (
`x`-component) of this vector is negative, such that this vector would point to the left of the vertical axis. Since the second component (
`y`-component) of this vector is positive, this vector would point above the horizontal axis. Hence, the direction of this vector (relative to the positive
`x\!`-axis) would be an angle between
`90^(\circ)` and
`180^(\circ)`.

Divide the
`x`-component of this vector by its magnitude to find the cosine of the angle between this vector and the positive
`x\!`-axis. Apply the inverse cosine function to find this angle:

`\begin{aligned}\arccos \left((-3.4350)/(12.271)\right) \approx 106^(\circ)\end{aligned}`.