Higher Engineering Mathematics, Sixth Edition

Vectors 261

R

2 2.13 0

2.99





Figure 24.36

Horizontal component ofa 1 −a 2

= 1 .5cos90◦− 2 .6cos145◦=2.13

Vertical component ofa 1 −a 2

= 1 .5sin90◦− 2 .6sin145◦= 0

Magnitude ofa 1 −a 2 =

√

2. 132 + 02

=2.13m/s^2

Direction ofa 1 −a 2 =tan−^1

(

0

2. 13

)

= 0 ◦

Thus, a 1 −a 2 =2.13m/s^2 at 0◦

Problem 12. Calculate the resultant of (i)

v 1 −v 2 +v 3 and (ii)v 2 −v 1 −v 3 whenv 1 = 22

units at 140◦,v 2 =40 units at 190◦andv 3 = 15

units at 290◦.

(i) The vectors are shown in Fig. 24.37.

15

40

22

1408

1908

2 H 2908 1 H

1 V

2 V

Figure 24.37

The horizontal component of

v 1 −v 2 +v 3 =(22cos140◦)−(40cos190◦)

+(15cos290◦)

=(− 16. 85 )−(− 39. 39 )+( 5. 13 )

=27.67units

The vertical component of

v 1 −v 2 +v 3 =(22sin140◦)−(40sin190◦)

+(15sin290◦)

=( 14. 14 )−(− 6. 95 )+(− 14. 10 )

=6.99units

The magnitude of the resultant,

R=

√

27. 672 + 6. 992 =28.54units

The direction of the resultantR=tan−^1

(

6. 99

27. 67

)

= 14. 18 ◦

Thus,v 1 −v 2 +v 3 =28.54 units at 14. 18 ◦

Usingcomplex numbers,

v 1 −v 2 +v 3 = 22 ∠ 140 ◦− 40 ∠ 190 ◦+ 15 ∠ 290 ◦

=(− 16. 853 +j 14. 141 )

−(− 39. 392 −j 6. 946 )

+( 5. 130 −j 14. 095 )

= 27. 669 +j 6. 992 =28.54∠ 14. 18 ◦

(ii) The horizontal component of

v 2 −v 1 −v 3 =(40cos190◦)−(22cos140◦)

−(15cos290◦)

=(− 39. 39 )−(− 16. 85 )−( 5. 13 )

=−27.67 units

The vertical component of

v 2 −v 1 −v 3 =(40sin190◦)−(22sin140◦)

−(15sin290◦)

=(− 6. 95 )−( 14. 14 )−(− 14. 10 )

=−6.99 units

From Fig. 24.38 the magnitude of the resultant,

R=

√

(− 27. 67 )^2 +(− 6. 99 )^2 =28.54 units

andα=tan−^1

(

6. 99

27. 67

)

= 14. 18 ◦, from which,

θ= 180 ◦+ 14. 18 ◦= 194. 18 ◦