Chapter 6.5, Problem 50E

(a)

To determine

Plot the vectors that represent the two forces in the complex plane.

Answer to Problem 50E

  $f_{1}y$ is vertical component of force 1, $f_{1}x$ is horizontal component of force 1 and f1 is the force 1

And $f_{2}y$ is vertical component of force 2, $f_{2}x$ is horizontal component of force 2 and f2 is the force 2

And $fr$ is the resultant force

Explanation of Solution

Given:

On the complex plane

Force 1 had horizontal component of 5 Newton

And vertical component of 3 Newton

Force 2 has horizontal component of 4 Newton

And vertical component of 2 Newton

On the complex plane

Force 1 had horizontal component of 5 Newton

And vertical component of 3 Newton

Force 2 has horizontal component of 4 Newton

And vertical component of 2 Newton

Draw complex plane

Here, $f_{1}y$ is vertical component of force 1, $f_{1}x$ is horizontal component of force 1 and f1 is the force 1

And $f_{2}y$ is vertical component of force 2, $f_{2}x$ is horizontal component of force 2 and f2 is the force 2

And $fr$ is the resultant force

(b)

To determine

Determine the horizontal and vertical components of the resultant force acting on the point using the complex plane.

Answer to Problem 50E

The resultant force is of $\sqrt{2}$ Newton acting on the point

Explanation of Solution

Given:

On the complex plane

Force 1 had horizontal component of 5 Newton

And vertical component of 3 Newton

Force 2 has horizontal component of 4 Newton

And vertical component of 2 Newton

From the figure

The resultant force can be given by the distance between two complex numbers

Distance formula is given by

  $d=\sqrt{{\left(s-a\right)}^{\begin{array}{l}2\\ \end{array}}+{\left(t-b\right)}^2}$

Where d is the distance between the complex numbers

And a,b,s,t can be find as

Compare (5,3) with the standard form of complex number $a+bi$ which in point form is $(a,b)$

  $a=5,b=3$ Similarly, compare (4,2) with $s+it$ which in point form is $\left(s,t\right)$

  $s=4,t=2$

Now put these values in the distance formula

  $\begin{array}{cc}d& =\sqrt{{(4-5)}^2+{(2-3)}^2}\\ & =\sqrt{{(-1)}^2+{(-1)}^2}\\ & =\sqrt{1+1}\\ & =\sqrt{2}\end{array}$

Therefore, the resultant force is of $\sqrt{2}$ Newton acting on the point