Higher Engineering Mathematics, Sixth Edition

114 Higher Engineering Mathematics

448

488

30.0 m

D

C

B

A

Figure 11.36

For triangleABC, using Pythagoras’ theorem:

BC^2 =AB^2 +AC^2

(

DC

tan44◦

) 2

=( 30. 0 )^2 +

(

DC

tan48◦

) 2

DC^2

(

1

tan^244 ◦

−

1

tan^248 ◦

)

= 30. 02

DC^2 ( 1. 072323 − 0. 810727 )= 30. 02

DC^2 =

30. 02

0. 261596

= 3440. 4

Hence, height of aerial,

DC=

√

3440.4=58.65m

Problem 35. A crank mechanism of a petrol

engine is shown in Fig. 11.37. ArmOAis 10.0cm

long and rotates clockwise about O. The connecting

rodABis 30.0cm long and endBis constrained to

move horizontally.

30.0 cm

A

B

O

10.0 cm

508

Figure 11.37

(a) For the position shown in Fig. 11.37 determine

the angle between the connecting rodABand

the horizontal and the length ofOB.

(b) How far doesBmove when angleAOB

changes from 50◦to 120◦?

(a) Applying the sine rule:

AB

sin50◦

=

AO

sinB

from which,

sinB=

AOsin50◦

AB

=

10 .0sin50◦

30. 0

= 0. 2553

Hence B=sin−^10. 2553 = 14 ◦ 47 ′ (or 165◦ 13 ′,

which is impossible in this case).

Hence the connecting rodABmakes an angle

of 14◦ 47 ′with the horizontal.

AngleOAB= 180 ◦− 50 ◦− 14 ◦ 47 ′= 115 ◦ 13 ′.

Applying the sine rule:

30. 0

sin50◦

=

OB

sin115◦ 13 ′

from which,

OB=

30 .0sin115◦ 13 ′

sin50◦

=35.43cm

(b) Figure11.38showstheinitial andfinalpositionsof

thecrank mechanism.In triangleOA′B′, applying

the sine rule:

30. 0

sin120◦

=

10. 0

sinA′B′O

from which,

sinA′B′O=

10 .0sin120◦

30. 0

= 0. 2887

A A

BB O

50 

30.0 cm

10.0 cm

120 

Figure 11.38

Hence A′B′O=sin−^10. 2887 = 16 ◦ 47 ′ (or 163◦ 13 ′

which is impossible in this case).