Chapter 1, Problem 1.48P

The pole OB is subjected to the 6004b force at B. Determine (a) the rectangular components of the force; and (b) the angles between the force vector and the coordinate axes.

To determine

(a)

The rectangular components of the force.

Answer to Problem 1.48P

The rectangular components of the force are,

  $\begin{array}{l}{F}_x=-109.09\\ F_y=327.273\\ F_z=-490.901\end{array}$

Explanation of Solution

Given Information:

  

Calculation:

The force is directed from point B towards point A. The coordinates of these points are,

  $\begin{array}{l}A\left(-20,60,0\right)\text{ ft}\\ B\left(0,0,90\right)\text{ ft}\end{array}$

The position vector of A relative to B ,

  $\begin{array}{l}\overrightarrow{BA}=\left(-20-0\right)\text{i}+\left(60-0\right)\text{j}+\left(0-90\right)\text{k}\\ =-20\text{i}+60\text{j}-90\text{k}\end{array}$

The unit vector in the direction of this position vector,

  $\begin{array}{l}\lambda =\frac{\overrightarrow{BA}}{\left|\overrightarrow{BA}\right|}=\frac{-20\text{i}+60\text{j}-90\text{k}}{\sqrt{{20}^2+{60}^2+{90}^2}}\\ =\frac{1}{110}\left(-20\text{i}+60\text{j}-90\text{k}\right)\\ =-\frac{2}{11}\text{i}+\frac{6}{11}\text{j}-\frac{9}{11}\text{k}\end{array}$

So, the force vector,

  $\begin{array}{l}\text{F}=600\left(-\frac{2}{11}\text{i}+\frac{6}{11}\text{j}-\frac{9}{11}\text{k}\right)\\ =-109.09\text{i}+327.273\text{j}-490.901\text{k}\end{array}$

The rectangular components,

  $\begin{array}{l}{F}_x=-109.09\\ F_y=327.273\\ F_z=-490.901\end{array}$

Conclusion:

The rectangular components of force are calculated by given coordinates of points into the vector equations.

To determine

(b)

The angle between force vector and coordinate axes.

Answer to Problem 1.48P

The angle between force vector and coordinate axes are,

  $\begin{array}{l}{\theta }_x=100.48°\\ {\theta }_y=56.94°\\ {\theta }_z=144.9°\end{array}$

Explanation of Solution

Given Information:

  

The rectangular components of the force are,

  $\begin{array}{l}{F}_x=-109.09\\ F_y=327.273\\ F_z=-490.901\end{array}$

Calculation:

The rectangular components are given.

  $\begin{array}{l}{F}_x=F\cos {\theta }_x=-109.09\\ \Rightarrow {\theta }_x={\cos }^{-1}\left(\frac{-109.09}{F}\right)={\cos }^{-1}\left(\frac{-109.09}{600}\right)\\ =100.48°\\ F_y=F\cos {\theta }_y=327.273\\ \Rightarrow {\theta }_y={\cos }^{-1}\left(\frac{327.273}{F}\right)={\cos }^{-1}\left(\frac{327.273}{600}\right)\\ =56.94°\\ F_z=F\cos {\theta }_z=-490.901\\ \Rightarrow {\theta }_z={\cos }^{-1}\left(\frac{-490.901}{F}\right)={\cos }^{-1}\left(\frac{-490.901}{600}\right)\\ =144.9°\end{array}$

Conclusion:

The angle between force vector and coordinate axes are calculated by putting rectangular components into the trigonometric equations.