Chapter 14.2, Problem 8E

To determine

To find:The magnitude and bearing of the additional force $F$ that will keep the object at rest.

Answer to Problem 8E

The magnitude and bearing of the additional force $F$$20\cdot 61\text{N}$ and $39\cdot {09}^{\circ }$ respectively.

Explanation of Solution

Given information:

There is given a force $F_1=6\text{N}$ , bearing ${300}^{\circ }$ and another force $F_2=15\text{N}$ , bearing ${30}^{\circ }$ .

The magnitude of the resultant force can be calculated as,

  $\begin{array}{c}{F}_1=6\cos {300}^{\circ }+6\sin {300}^{\circ }\\ =6\left(0\cdot 5\right)i+6\left(-0\cdot 86\right)j\\ =3i+5j\end{array}$

And,

  $\begin{array}{c}{F}_2=15\cos {30}^{\circ }+15\sin {30}^{\circ }\\ =15\left(0\cdot 86\right)i+15\left(0\cdot 5\right)j\\ =13i+8j\end{array}$

Therefore,

The additional force is given by,

  $\begin{array}{c}{F}_A=\left(3i+5j\right)+\left(13i+8j\right)\\ =16i+13j\end{array}$

  $F_A=\left(16,13\right)$

Now the magnitude can be calculated as,

  $\begin{array}{c}{D}_R=\sqrt{{\left(16-0\right)}^2+{\left(13-0\right)}^2}\\ =\sqrt{256+169}\\ =\sqrt{425}\\ =20\cdot 61\text{N}\end{array}$

Now, the bearing can be calculated as,

  $\begin{array}{c}\tan \theta =\frac{13}{16}\\ \theta ={\tan }^{-1}\left(0\cdot 8125\right)\\ =39\cdot {09}^{\circ }\end{array}$

Therefore, the magnitude and bearing of the additional force $F$$20\cdot 61\text{N}$ and $39\cdot {09}^{\circ }$ respectively.