Chapter 3, Problem 54GP

To determine

The magnitude of displacement of the plumber, angles with x- axis in horizontal and vertical planes.

Answer to Problem 54GP

The displacement in the component is $\left(50\text{m},-25\text{m},-10\text{m}\right)$ , magnitude of the displacement is $56.79\text{m}$ , the angle of the x-axis in horizontal plane is $-26.57°$ and angle of the x-axis in vertical plane is $-11.31°$ .

Explanation of Solution

Given info:

The displacement along the east is, $d_x=50\text{m}$ .

The displacement along the south is, $d_y=-25\text{m}$ .

The displacement in the down direction is, $d_z=-10\text{m}$

Formula Used:

The displacement in the component is given as,

  $d=\left(d_x,d_y,d_z\right)$

The magnitude of the displacement is given as,

  $d=\sqrt{d_x^2+d_y^2+d_z^2}$

The angle of the x-axis in horizontal plane is,

  $\tan \alpha =\frac{d_y}{d_x}$

The angle of the x-axis in vertical plane is,

  $\tan \beta =\frac{d_z}{d_x}$

Calculation:

Substitute all the values in the above expression.

  $d=\left(50\text{m},-25\text{m},-10\text{m}\right)$

Substitute all the values in the above expression.

  $\begin{array}{c}d=\sqrt{{\left(50\text{m}\right)}^2+{\left(-25\text{m}\right)}^2+{\left(-10\text{m}\right)}^2}\\ =56.79\text{m}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}\tan \alpha =\frac{-25\text{m}}{50\text{m}}\\ \alpha =-26.57°\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}\tan \beta =\frac{-10\text{m}}{50\text{m}}\\ \beta =-11.31°\end{array}$

Conclusion:

Thus, the displacement in the component is $\left(50\text{m},-25\text{m},-10\text{m}\right)$ , magnitude of the displacement is $56.79\text{m}$ , the angle of the x-axis in horizontal plane is $-26.57°$ and angle of the x-axis in vertical plane is $-11.31°$ .