Engineering Mechanics

Chapter 19 : Relative Velocity „„„„„ 403

1. First of all, draw East, West, North and South lines meeting at O.


2. In the first case, since the man is running eastwards at 6 km. p.h., therefore cut off OA
   equal to 6 km. to some suitable scale on the westwards (opposite to eastwards i.e.actual
   direction of the man).


3. At A, draw a perpendicular CA, which represents the direction of the relative velocity of
   wind.


4. In the second case, since the man doubles his speed, therefore cut off OB equal to 12 km.
   to the scale on the westward.


5. At B, draw a line CB at 45° (i.e. North-East) meeting the vertical line through A at C,
   whichrepresents the direction of the new relative velocity of the wind.


6. Join CO, which gives the actual direction and velocity of the wind. By measurement, we
   find that ∠ θ = ∠ AOC = 45° and CO = 8.5 km.p.h.

Mathematical check

We know that

OA=AB = 6 km. ...(Speed of man)

and CA=AB = 6 km. ... tan 45^1

CA

AB

⎛⎞==

⎜⎟

⎝⎠

o

∴ OA=CA or θ = 45° Ans.
Now in triangle OAC
6
8.49 km. p.h.
sin 45 0.707

CA

CO===

°

Ans.

Example 19.3. When a cyclist is riding towards West at 20 km per hour, he finds the rain
meeting at an angle of 45° with the vertical. When he rides at 12 km per hour, he meets the rain at an
angle of 30° with the vertical. What is the actual velocity, in magnitude and direction, of the rain?

Solution. Given : When velocity is 20 km. p.h., apparent direction, of the rain = 45° with
vertical and when velocity is 12 km. p.h. apparent direction of rain = 30° with vertical.

Fig. 19.4.
Let us draw the relative velocity diagram for both the cases as shown in Fig. 19.4 and as
discussed below :