Chapter 4, Problem 4.1P

Interpretation Introduction

Interpretation: The velocity component in a particular direction is to be calculated.

Concept introduction: Velocity vector is a sum of the two velocity vectors perpendicular to each other.

  

  $\begin{array}{l}{\text{V = V}}_{\text{x}}{\text{cosθ + V}}_{\text{y}}\text{sinθ }\mathrm{.......}\text{ (1)}\\ {\text{V}}_{\text{x}}\text{= Velocity component Parallel to the direction}\\ {\text{V}}_{\text{y}}\text{= Velocity component Perpendicular to the direction}\end{array}$

  $\text{θ =}$ Angle between the 2 vectors, i.e. the angle between hypotenuse and base of the above figure.

The component in a particular direction is the dot product of the velocity vector and unit vector.

Answer to Problem 4.1P

The component of velocity is 1.66 m/s.

Explanation of Solution

The velocity vector given is,

  ${\text{v = 10e}}_{\text{x}}{\text{+7xe}}_{\text{y}}$ ........... (2)

At point (2,2) the equation (2) can be rewritten as,

  ${\text{v = 10e}}_{\text{x}}{\text{+14e}}_{\text{y}}$

A unit vector in $-30°$ direction is given as,

  

Using equation (1) in the above vector diagram at angle $-30°$ is given as,

  $\begin{array}{l}{\text{e = e}}_{\text{x}}{\text{cos(-30°) + e}}_{\text{x}}\text{sin(-30°) }\\ {\text{e = e}}_{\text{x}}\frac{\sqrt{\text{3}}}{\text{2}}{\text{ - e}}_{\text{y}}\frac{\text{1}}{\text{2}}\\ \end{array}$ ........... (3)

Taking the dot product of equation (2) and (3) we get,

  $\begin{array}{l}\text{v}{\text{.e = (10e}}_{\text{x}}{\text{+14e}}_{\text{y}}\text{)}\text{.(}\frac{\sqrt{\text{3}}}{\text{2}}{\text{e}}_{\text{x}}\text{-}\frac{\text{1}}{\text{2}}{\text{e}}_{\text{y}}\text{)}\\ \text{v}\text{.e = 5}\sqrt{\text{3}}\text{ - 7}\\ \text{v}\text{.e = 1}\text{.66 m/s}\end{array}$

Conclusion

Thus, the dot product of the velocity vector and the vector in a particular direction gives the velocity component and its value here is 1.66 m/s.