Add the following vectors using the Component law.
a. 9 m/s [N30°E] and 2 m/s [N 60°E].
b. 3 N in a direction 15° south of west and 4 N in a direction 12° east of south.
N30°E---- > 60° in standard notation
N 60°E --- > 30°
so .... I read
9 m/s [N30°E] and 2 m/s [N 60°E]
as
(9cos60°, 9sin60°) + (2cos30°, 2sin30°)
= (9(1/2), 9(√3/2) ) + (2(√3/2), 2(1/2) )
= ....
do b) in the same way
N and E +ve
measure angles from E (0 deg) clockwise
assuming 3 N and 4 N are forces in Newtons
W (180 deg) + 15 deg S (clockwise) = 195 deg
S (270 deg) + 12 deg E (clockwise) = 282 deg
Rx = 3 cos 195 + 4 cos 282
Ry = 3 sin 195 + 4 sin 282
@Reiny, is the correct way of approaching or nah?
(3cos12°, 3sin12°) + (4cos15°, 4sin15°)
To add the given vectors using the Component law, we need to break down each vector into its horizontal and vertical components. Then, we can sum the horizontal components and the vertical components separately.
a. 9 m/s [N30°E] and 2 m/s [N 60°E]:
First, let's find the horizontal and vertical components of each vector.
For the vector 9 m/s [N30°E]:
- The horizontal component is 9 m/s * cos(30°) = 7.794 m/s towards the east.
- The vertical component is 9 m/s * sin(30°) = 4.5 m/s towards the north.
For the vector 2 m/s [N60°E]:
- The horizontal component is 2 m/s * cos(60°) = 1 m/s towards the east.
- The vertical component is 2 m/s * sin(60°) = 1.732 m/s towards the north.
Now, we can add the horizontal components and the vertical components separately.
Horizontal component: 7.794 m/s + 1 m/s = 8.794 m/s towards the east.
Vertical component: 4.5 m/s + 1.732 m/s = 6.232 m/s towards the north.
Finally, we can combine the horizontal and vertical components to find the resultant vector. To do this, we will use the Pythagorean theorem and trigonometry.
The magnitude of the resultant vector is sqrt((8.794 m/s)^2 + (6.232 m/s)^2) = 10.873 m/s.
The angle of the resultant vector can be found using the inverse tangent function:
angle = tan^(-1)(6.232 m/s / 8.794 m/s) = 35.89°.
Thus, the resultant vector is approximately 10.873 m/s at an angle of 35.89° (measured counterclockwise from the east direction).
b. 3 N in a direction 15° south of west and 4 N in a direction 12° east of south:
Similarly, we'll find the horizontal and vertical components of each vector.
For the vector 3 N in a direction 15° south of west:
- The horizontal component is 3 N * cos(15°) = 2.879 N towards the west.
- The vertical component is 3 N * sin(15°) = 0.749 N in the southern direction.
For the vector 4 N in a direction 12° east of south:
- The horizontal component is 4 N * cos(12°) = 3.894 N towards the east.
- The vertical component is 4 N * sin(12°) = 0.834 N in the southern direction.
Adding the horizontal components:
2.879 N (west) + 3.894 N (east) = -1.015 N
Adding the vertical components:
0.749 N (south) + 0.834 N (south) = 1.583 N (south)
The resultant vector has a magnitude of sqrt((-1.015 N)^2 + (1.583 N)^2) = 1.865 N.
The angle of the resultant vector can be found using the inverse tangent function:
angle = tan^(-1)(1.583 N / -1.015 N) ≈ -58.62°.
Note: The angle is measured counterclockwise from the west direction in this case since the resultant vector is directed toward the south and slightly west.
Therefore, the resultant vector is approximately 1.865 N at an angle of -58.62° (measured counterclockwise from the west direction).