[S2°W] ---> 268°
[N31°W] ---> 121° in standard trig notation
Resultant = 9(cos268° , sin268°) + 11(cos 121°, sin 121°)
= .....
[N31°W] ---> 121° in standard trig notation
Resultant = 9(cos268° , sin268°) + 11(cos 121°, sin 121°)
= .....
Fr = 9N[268 ]+ 11N[121o].
X = 9*Cos268 + 11*Cos121 =
Y = 9*sin268 + 11*sin121 =
Fr = sqrt(X^2 + Y^2).
TanA = Y/X.
Let's start with the first vector of magnitude 9 N on a heading of S2°W.
Step 1: Determine the x and y components using trigonometry.
The angle of S2°W can be represented as 182° (180° + 2°) counterclockwise from the positive x-axis.
To find the x component (horizontally), we can use the cosine law:
cos(182°) = adj/hypotenuse
adj = cos(182°) * 9 N
To find the y component (vertically), we can use the sine law:
sin(182°) = opp/hypotenuse
opp = sin(182°) * 9 N
Step 2: Calculate the x and y components.
adj = cos(182°) * 9 N ≈ -8.97 N (rounded to two decimal places)
opp = sin(182°) * 9 N ≈ -0.31 N (rounded to two decimal places)
So, the first vector can be represented as approximately (-8.97 N, -0.31 N).
Let's move on to the second vector of magnitude 11 N on a heading of N31°W.
Step 1: Determine the x and y components using trigonometry.
The angle of N31°W can be represented as 331° (360° - 31°) counterclockwise from the positive x-axis.
To find the x component (horizontally), we can use the cosine law:
cos(331°) = adj/hypotenuse
adj = cos(331°) * 11 N
To find the y component (vertically), we can use the sine law:
sin(331°) = opp/hypotenuse
opp = sin(331°) * 11 N
Step 2: Calculate the x and y components.
adj = cos(331°) * 11 N ≈ -9.47 N (rounded to two decimal places)
opp = sin(331°) * 11 N ≈ -5.79 N (rounded to two decimal places)
So, the second vector can be represented as approximately (-9.47 N, -5.79 N).
Now, let's add the x and y components of both vectors to find the resultant vector.
The x component of the resultant vector will be (-8.97 N) + (-9.47 N) = -18.44 N.
The y component of the resultant vector will be (-0.31 N) + (-5.79 N) = -6.1 N.
Therefore, the resultant vector can be represented as approximately (-18.44 N, -6.1 N) when added using trigonometry with cosine and sine laws.
Let's start with the first vector: 9 N on a heading of [S2°W].
Step 1: Determine the horizontal and vertical components.
The horizontal component is given by:
horizontal component = magnitude × cos(angle)
Here, the angle is 2° West of South, which means we subtract 2° from 180° (since South is 180°). So, the angle is 178°.
horizontal component = 9 N × cos(178°)
Similarly, the vertical component is given by:
vertical component = magnitude × sin(angle)
vertical component = 9 N × sin(178°)
Now, let's calculate these components.
horizontal component = 9 N × cos(178°) ≈ -8.991 N (round to 3 decimal places)
vertical component = 9 N × sin(178°) ≈ -0.272 N (round to 3 decimal places)
Therefore, the first vector can be represented as approximately -8.991 N in the horizontal direction and -0.272 N in the vertical direction.
Now, let's move on to the second vector: 11 N on a heading of [N31°W].
Step 2: Determine the horizontal and vertical components.
The angle is 31° West of North. To convert it to a right triangle angle, add 90° to get 121°.
horizontal component = 11 N × cos(121°)
vertical component = 11 N × sin(121°)
Let's calculate these components.
horizontal component = 11 N × cos(121°) ≈ -5.015 N (round to 3 decimal places)
vertical component = 11 N × sin(121°) ≈ 9.662 N (round to 3 decimal places)
Therefore, the second vector can be represented as approximately -5.015 N in the horizontal direction and 9.662 N in the vertical direction.
Step 3: Add the horizontal and vertical components separately.
Resultant horizontal component = sum of horizontal components of both vectors
Resultant vertical component = sum of vertical components of both vectors
Resultant horizontal component = -8.991 N + (-5.015 N) ≈ -14.006 N (round to 3 decimal places)
Resultant vertical component = -0.272 N + 9.662 N ≈ 9.390 N (round to 3 decimal places)
Finally, the resultant vector is approximately -14.006 N in the horizontal direction and 9.390 N in the vertical direction.
Step 4: Calculate the magnitude and direction of the resultant vector.
To find the magnitude of the resultant vector, use the Pythagorean theorem:
magnitude = √(resultant horizontal component^2 + resultant vertical component^2)
magnitude = √((-14.006 N)^2 + (9.390 N)^2) ≈ 16.735 N (round to 3 decimal places)
To find the direction of the resultant vector, use the inverse tangent:
angle = arctan(resultant vertical component / resultant horizontal component)
angle = arctan(9.390 N / -14.006 N) ≈ -31.77° (round to 2 decimal places)
Therefore, the resultant vector is approximately 16.735 N on a heading of [S31.77°W].