Engineering Mechanics

(^404) „„„„„ A Textbook of Engineering Mechanics

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


2. In the first case, since the cyclist is riding towards West at 20 km. p.h., therefore cut off
   OA equal to 20 km. to some suitable scale towards East (opposite to West i.e. actual
   direction of the cyclist).


3. At A, draw a line at an angle of 45°^ with OA which represents the relative direction of the
   rain.


4. In the second case, since the man is riding at 12 km.p.h., therefore cut off OB equal to
   12 km. to the scale towards East.


5. At B, draw a line at an angle of 60°^ with OA, which represents the direction of the relative
   velocity of the rain. This is so because the rain meets the man at angle of 30° with the
   vertical. Let the two lines (from A and B) meet at C.


6. Join CO, which gives the actual direction and velocity of the wind. By measurement, we
   find that ∠ α = 3·3° and CO = 19 km.p.h. Ans.

Mathematical check

From C, draw a perpendicular CD to the line OA. Let OD = x and CD = y. From the triangle

ACD,

20 – 20 –

tan 45 or 1

xx

yy

°= = ...(Q tan 45° = 1)

or y = 20 – x ..(i)

Similarly, in triangle BCD

12 –

tan 30

x

y

°= or

12 –

0.577

x

y

=

or

12 –

0.577

x

y= ...(ii)

Equating equations (i) and (ii),

12 –

20 –

0.577

x

x=

11.54 – 0.577 x = 12 – x

0.423 x = 0.46

or

0.46

1.09 km

0.423

x==

Substituting the value of x in equation (i),

y = 20 – x = 20 – 1.09 = 18.91 km

∴

1.09

tan 0.0576

18.91

x

y

α= = =

∴ a = 3.3° Ans.

Now from the triangle OCD,

CO=+=+OD^2222 CD x y

=+=(1.09)^22 (18.91) 18.94 km.p.h.^ Ans.