Vectors 275
a 1 2 a 2
a 1 1 a 2
2.6 m/s 2
1268
a 1
2a 2
a 2
a 1
1.5 m/s 2
1458
Figure 29.35
Vertical component of a 1 + a 2 ,
V = 1.5 sin90 ◦ + 2.6 sin 145 ◦ = 2.99
From Figure 29.36, the magnitude of a 1 + a 2 ,
R =
(−2.13) 2 + 2.99 2 = 3.67 m/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 = 3.67 m/s
2 at 125.47 ◦
R
0
2.13
2.99
␣
Figure 29.36
Horizontal component of a 1 − a 2
= 1.5 cos 90
◦
− 2.6 cos 145
◦
= 2.13
Vertical component of a 1 − a 2
= 1.5 sin90 ◦ − 2.6 sin 145 ◦ = 0
Magnitude of a 1 − a 2 =
√
2.13 2 + 0 2
= 2.13 m/s
2
Direction of a 1 − a 2 = tan −1
0
2.13
= 0 ◦
Thus, a 1 − a 2 = 2.13 m/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
2H
1H
1V
2V
Figure 29.37
The horizontal component of v 1 − v 2 + v 3
= (22 cos 140 ◦ ) − (40 cos 190 ◦ ) + (15 cos 290 ◦ )
= (−16.85) − (−39.39) + (5.13) = 27.67 units
The vertical component of v 1 −v 2 +v 3
= (22 sin 140 ◦ ) − (40 sin 190 ◦ ) + (15 sin 290 ◦ )
= (14.14) − (−6.95) + (−14.10) = 6.99 units
The magnitude of the resultant,
R =
√
27.67 2 + 6.99 2 = 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 of v 2 − v 1 − v 3
= (40 cos 190 ◦ ) − (22 cos 140 ◦ ) − (15 cos 290 ◦ )
= (−39.39) − (−16.85) − (5.13) =−27.67 units
The vertical component of v 2 − v 1 − v 3
= (40 sin 190 ◦ ) − (22 sin 140 ◦ ) − (15 sin 290 ◦ )
= (−6.95) − (14.14) − (−14.10) = −6.99 units
From Figure 29.38, the magnitude of the resultant, 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 ◦
a 1 2 a 2
a 1 1 a 2
2.6 m/s 2
1268
a 1
2a 2
a 2
a 1
1.5 m/s 2
1458
Figure 29.35
Vertical component of a 1 + a 2 ,
V = 1.5 sin90 ◦ + 2.6 sin 145 ◦ = 2.99
From Figure 29.36, the magnitude of a 1 + a 2 ,
R =
(−2.13) 2 + 2.99 2 = 3.67 m/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 = 3.67 m/s
2 at 125.47 ◦
R
0
2.13
2.99
␣
Figure 29.36
Horizontal component of a 1 − a 2
= 1.5 cos 90
◦
− 2.6 cos 145
◦
= 2.13
Vertical component of a 1 − a 2
= 1.5 sin90 ◦ − 2.6 sin 145 ◦ = 0
Magnitude of a 1 − a 2 =
√
2.13 2 + 0 2
= 2.13 m/s
2
Direction of a 1 − a 2 = tan −1
0
2.13
= 0 ◦
Thus, a 1 − a 2 = 2.13 m/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
2H
1H
1V
2V
Figure 29.37
The horizontal component of v 1 − v 2 + v 3
= (22 cos 140 ◦ ) − (40 cos 190 ◦ ) + (15 cos 290 ◦ )
= (−16.85) − (−39.39) + (5.13) = 27.67 units
The vertical component of v 1 −v 2 +v 3
= (22 sin 140 ◦ ) − (40 sin 190 ◦ ) + (15 sin 290 ◦ )
= (14.14) − (−6.95) + (−14.10) = 6.99 units
The magnitude of the resultant,
R =
√
27.67 2 + 6.99 2 = 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 of v 2 − v 1 − v 3
= (40 cos 190 ◦ ) − (22 cos 140 ◦ ) − (15 cos 290 ◦ )
= (−39.39) − (−16.85) − (5.13) =−27.67 units
The vertical component of v 2 − v 1 − v 3
= (40 sin 190 ◦ ) − (22 sin 140 ◦ ) − (15 sin 290 ◦ )
= (−6.95) − (14.14) − (−14.10) = −6.99 units
From Figure 29.38, the magnitude of the resultant, 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 ◦