Q74P - University Physics with Modern Physics

Question

 A fence post is $52.0m$ from where you are standing, in a direction $37.0°$north of east. A second fence post is due south from you. How far are you from the second post if the distance between the two posts is $68.0m$ ?

Step-by-Step Solution

Answer

I am standing at a separation of $22.6m$ from the second post.

Significance of the triangle law of vector addition for evaluating the distance

This law illustrates that when two triangle’s side represents two vectors, then the other side represents the resultant vector.

The resultant vector in the triangle gives the distance between two posts which we have to find out.

Identification of the given data

The given data can be listed below as:

• The fence post is situated at a separation of $52.0m$,$37.0°$ north of east.

• Two posts are situated at a separation of  $68.0m$.

Determination of the distance from the second post

The free-body diagram of the triangle using the reference frame, having $+x$direction to the right and $+y$ direction to the top, has been illustrated below

Analysing the above figure, according to the triangle law of vector addition, the distance h can be measured as:

$a^2=h^2+b^2-2hb\cos \theta$

Here, h and b are the sides of the triangle and a is the latent height of the triangle

${\left(68m\right)}^2=h^2+{\left(52m\right)}^2-2\times 52m\times h\cos \theta$

The above equation can be broken down as-

${\left(68.0\right)}^2={\left(52.0\right)}^2+h^2-2\left(52.0\right)h\cos \left(37°+90°\right)$

$h^2-104\left(\cos 127°\right)h-1920=0$

Furthermore, the quadratic equation can be expressed as:

$$\begin{gathered} h=\frac{104\cos 127°\pm \sqrt{{\left[10816\right]}^2-4\left(1\left(-1920\right)\right)}}{2\left(1\right)} \\ =\frac{-62.58\pm 42370091.99+7680}{2} \\ =\frac{-62.58\pm 42377771.99}{2} \\ =\left\{-85.1,22.6\right\} \end{gathered}$$

As the distance must be positive, hence, the distance is $h=22.6m$

Thus, I am about $22.6m$ away from the second post.