Add the following vectors using trigonometry (i.e. cosine and sine laws). 9 N on a heading of [S2°W] and 11 N on a heading of [N31°W].?
To add the given vectors using trigonometry, we need to break down each vector into its x and y components.
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.
To add the given vectors, we first need to break them down into their horizontal and vertical components. Since trigonometric functions deal with right triangles, we can use the cosine and sine laws to find these components.
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].
[S2°W] ---> 268°
[N31°W] ---> 121° in standard trig notation
Resultant = 9(cos268° , sin268°) + 11(cos 121°, sin 121°)
= .....
All angles are measured CCW from +X-axis
Fr = 9N[268 ]+ 11N[121o].
X = 9*Cos268 + 11*Cos121 =
Y = 9*sin268 + 11*sin121 =
Fr = sqrt(X^2 + Y^2).
TanA = Y/X.