Three vectors are specified as A is 5 m at 45o north of east, B is 7 m at 60o east of south, and C is 4 m at 30o west to south. Find the magnitude and direction of their sum.
A = 5m[45o], CCW.
B = 7m[330o], CCW
C = 4m[240o], CCW.
X=5*Cos45 + 7*Cos330 + 4*Cos240 = 7.60 m
Y=5*sin45 + 7*sin330 + 4*sin240=-3.43 m
Tan A = Y/X = -3.43/7.60 = -0.45113
A = -24.28o = 24.28o S of E = 335.7o, CCW.
Resultant = X/CosA = 7.60/Cos335.7 = 8.34m[335.7]
Well, it seems like these vectors are quite directionally challenged! Let me crunch some numbers and find their sum for you.
To add these vectors, we need to break them down into their horizontal and vertical components.
For Vector A, the horizontal component is 5 m * cos(45°), and the vertical component is 5 m * sin(45°).
For Vector B, the horizontal component is 7 m * sin(60°), and the vertical component is 7 m * cos(60°).
For Vector C, the horizontal component is 4 m * sin(30°), and the vertical component is 4 m * cos(30°).
Now let's find the horizontal and vertical components for each vector:
Vector A:
Horizontal component: 5 m * cos(45°) = 5 m * 0.707 = 3.54 m (approx.)
Vertical component: 5 m * sin(45°) = 5 m * 0.707 = 3.54 m (approx.)
Vector B:
Horizontal component: 7 m * sin(60°) = 7 m * 0.866 = 6.06 m (approx.)
Vertical component: 7 m * cos(60°) = 7 m * 0.5 = 3.5 m (approx.)
Vector C:
Horizontal component: 4 m * sin(30°) = 4 m * 0.5 = 2 m
Vertical component: 4 m * cos(30°) = 4 m * 0.866 = 3.46 m (approx.)
Now let's add up the horizontal and vertical components separately:
Horizontal component: 3.54 m + 6.06 m - 2 m = 7.6 m (approx.)
Vertical component: 3.54 m + 3.5 m - 3.46 m = 3.58 m (approx.)
Finally, we can find the magnitude and direction of their sum using the Pythagorean theorem and trigonometry:
Magnitude: sqrt((7.6 m)^2 + (3.58 m)^2) ≈ 8.23 m
Direction: tan^(-1)(3.58 m / 7.6 m) ≈ 26.3°
So, the magnitude of their sum is approximately 8.23 m, and its direction is approximately 26.3°. Happy vector crunching!
To find the magnitude and direction of the sum of vectors A, B, and C, we first need to determine their components and then add them.
Let's break down each vector into their respective x-axis and y-axis components.
Vector A: 5 m at 45° north of east
The x-component of A is A_x = 5 m * cos(45°) = 5 m * √2/2 = 5√2/2 m
The y-component of A is A_y = 5 m * sin(45°) = 5 m * √2/2 = 5√2/2 m
Vector B: 7 m at 60° east of south
The x-component of B is B_x = -7 m * sin(60°) = -7 m * √3/2 = -7√3/2 m
The y-component of B is B_y = -7 m * cos(60°) = -7 m * 1/2 = -7/2 m
Vector C: 4 m at 30° west to south
The x-component of C is C_x = 4 m * cos(30°) = 4 m * √3/2 = 2√3 m
The y-component of C is C_y = 4 m * sin(30°) = 4 m * 1/2 = 2 m
Now we can add up the components to find the resulting vectors in the x-axis and y-axis.
Sum of x-components: Total_x = A_x + B_x + C_x = (5√2/2) + (-7√3/2) + (2√3) = (-7√3 + 4√2) / 2 ≈ -0.34 m
Sum of y-components: Total_y = A_y + B_y + C_y = (5√2/2) + (-7/2) + 2 = (5√2 - 7 + 4) / 2 ≈ 1.93 m
The magnitude of the sum vector is given by the Pythagorean theorem:
Magnitude = √(Total_x^2 + Total_y^2) = √((-0.34)^2 + (1.93)^2) ≈ √(0.1156 + 3.7249) ≈ √3.8405 ≈ 1.96 m
To find the direction of the sum vector, we can use the inverse tangent function:
θ = tan^(-1)(Total_y / Total_x) = tan^(-1)(1.93 / -0.34) ≈ -1.411 radians
Since the sum vector is in the fourth quadrant, we need to add 180° or π radians to the angle:
θ = -1.411 + π ≈ 1.731 radians
Therefore, the magnitude of the sum vector is approximately 1.96 m, and its direction is approximately 1.731 radians.
To find the magnitude and direction of the sum of the three vectors (A, B, and C), we need to first add the vectors together to find the resultant vector.
Step 1: Resolve each vector into its horizontal (x) and vertical (y) components.
Let's find the horizontal and vertical components of each vector:
Vector A:
- Magnitude: 5 m
- Angle: 45° north of east
To resolve this vector, we can use trigonometry. The horizontal component (Ax) is given by:
Ax = A * cos(θ)
Ax = 5 m * cos(45°)
Using the trigonometric identity, cos(45°) = √2 / 2, we can calculate Ax:
Ax = 5 m * (√2 / 2)
Ax = 5√2 / 2 m
The vertical component (Ay) is given by:
Ay = A * sin(θ)
Ay = 5 m * sin(45°)
Using the trigonometric identity, sin(45°) = √2 / 2, we can calculate Ay:
Ay = 5 m * (√2 / 2)
Ay = 5√2 / 2 m
So, the components of vector A are:
Ax = 5√2 / 2 m (horizontal component)
Ay = 5√2 / 2 m (vertical component)
Vector B:
- Magnitude: 7 m
- Angle: 60° east of south
To resolve this vector, we can use trigonometry. The horizontal component (Bx) is given by:
Bx = B * cos(θ)
Bx = 7 m * cos(60°)
Using the trigonometric identity, cos(60°) = 1 / 2, we can calculate Bx:
Bx = 7 m * (1 / 2)
Bx = 7 / 2 m
The vertical component (By) is given by:
By = B * sin(θ)
By = 7 m * sin(60°)
Using the trigonometric identity, sin(60°) = √3 / 2, we can calculate By:
By = 7 m * (√3 / 2)
By = 7√3 / 2 m
So, the components of vector B are:
Bx = 7 / 2 m (horizontal component)
By = 7√3 / 2 m (vertical component)
Vector C:
- Magnitude: 4 m
- Angle: 30° west to south
To resolve this vector, we can use trigonometry. The horizontal component (Cx) is given by:
Cx = C * cos(θ)
Cx = 4 m * cos(30°)
Using the trigonometric identity, cos(30°) = √3 / 2, we can calculate Cx:
Cx = 4 m * (√3 / 2)
Cx = 2√3 m
The vertical component (Cy) is given by:
Cy = C * sin(θ)
Cy = 4 m * sin(30°)
Using the trigonometric identity, sin(30°) = 1 / 2, we can calculate Cy:
Cy = 4 m * (1 / 2)
Cy = 2 m
So, the components of vector C are:
Cx = 2√3 m (horizontal component)
Cy = 2 m (vertical component)
Step 2: Add the horizontal and vertical components separately to find the resultant components.
To find the resultant components, add the corresponding horizontal and vertical components of the three vectors:
Rx = Ax + Bx + Cx
Ry = Ay + By + Cy
Substituting the values we calculated earlier, we have:
Rx = (5√2 / 2) m + (7 / 2) m + (2√3) m
Ry = (5√2 / 2) m + (7√3 / 2) m + 2 m
Step 3: Calculate the magnitude and direction of the resultant vector.
The magnitude of the resultant vector (R) can be calculated using the Pythagorean theorem:
R = √(Rx^2 + Ry^2)
The direction of the resultant vector (θ) can be found using the inverse tangent:
θ = tan^(-1)(Ry / Rx)
Substituting the values we calculated earlier, we can find the magnitude and direction of the sum of vectors A, B, and C.