Basic Engineering Mathematics, Fifth Edition

Vectors 275

a 12 a 2

a 11 a 2

2.6m/s^2

1268

a 1

2 a 2

a 2

a 1

1.5m/s^2

1458

Figure 29.35

Vertical component ofa 1 +a 2 ,

V= 1 .5sin90◦+ 2 .6sin145◦= 2. 99

From Figure 29.36, the magnitude ofa 1 +a 2 ,

R=

√

(− 2. 13 )^2 + 2. 992 = 3 .67m/s^2

In Figure 29.36,α=tan−^1

(

2. 99

2. 13

)

= 54. 53 ◦and

θ= 180 ◦− 54. 53 ◦= 125. 47 ◦

Thus,a 1 +a 2 = 3 .67m/s^2 at 125. 47 ◦

R

2.13 0

2.99





Figure 29.36

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

(a)v 1 −v 2 +v 3 and (b)v 2 −v 1 −v 3 when

v 1 =22 units at 140◦,v 2 =40 units at 190◦and

v 3 =15 units at 290◦

(a) The vectors are shown in Figure 29.37.

15

40

22

1408

1908

2908

2 H 1 H

1 V

2 V

Figure 29.37

The horizontal component ofv 1 −v 2 +v 3

=(22cos140◦)−(40cos190◦)+(15cos290◦)

=(− 16. 85 )−(− 39. 39 )+( 5. 13 )=27.67 units

The vertical component ofv 1 −v 2 +v 3

=(22sin140◦)−(40sin190◦)+(15sin290◦)

=( 14. 14 )−(− 6. 95 )+(− 14. 10 )=6.99 units

The magnitude of the resultant,

R=

√

27. 672 + 6. 992 =28.54 units

The direction of the resultant

R=tan−^1

(

6. 99

27. 67

)

= 14. 18 ◦

Thus,v 1 −v 2 +v 3 = 28 .54 units at 14. 18 ◦

(b) The horizontal component ofv 2 −v 1 −v 3

=(40cos190◦)−(22cos140◦)−(15cos290◦)

=(− 39. 39 )−(− 16. 85 )−( 5. 13 )=−27.67 units

The vertical component ofv 2 −v 1 −v 3

=(40sin190◦)−(22sin140◦)−(15sin290◦)

=(− 6. 95 )−( 14. 14 )−(− 14. 10 )=−6.99 units

From Figure 29.38, the magnitude of the resul-

tant,R=

√

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

and α=tan−^1

(

6. 99

27. 67

)

= 14. 18 ◦, from which,

θ= 180 ◦+ 14. 18 ◦= 194. 18 ◦