4. If F1 stands for a force vector of magnitude 30.0 N and F2 stands for a force vector of magnitude 40.0 N
acting in the directions shown in Figure 3-6, what are the magnitude and direction of the resultant
obtained by the vector addition of these two vectors using the analytical method? Show your work.
no diagram, but just convert each vector to x-y coordinates, and then add them up.
then convert back to polar form.
To find the magnitude and direction of the resultant vector obtained by adding two vectors using the analytical method, follow these steps:
Step 1: Draw a coordinate system and label the axes. In this case, it is not provided in the question, so let's assume the positive x-axis points to the right, and the positive y-axis points upwards.
Step 2: Label the given vectors F1 and F2 with their magnitudes and directions. F1 has a magnitude of 30.0 N and is directed at an angle of 60 degrees from the positive x-axis, while F2 has a magnitude of 40.0 N and is directed at an angle of 150 degrees from the positive x-axis.
Step 3: Decompose the given vectors into their x and y components. To do this, use the following trigonometric relationships:
Fx = F * cos(angle)
Fy = F * sin(angle)
For F1:
Fx1 = 30.0 N * cos(60 degrees) = 30.0 N * 0.5 = 15.0 N (positive x-component)
Fy1 = 30.0 N * sin(60 degrees) = 30.0 N * 0.866 = 25.98 N (positive y-component)
For F2:
Fx2 = 40.0 N * cos(150 degrees) = 40.0 N * -0.866 = -34.64 N (negative x-component)
Fy2 = 40.0 N * sin(150 degrees) = 40.0 N * 0.5 = 20 N (positive y-component)
Step 4: Add the x-components and y-components separately to get the resultant x-component (Rx) and resultant y-component (Ry):
Rx = Fx1 + Fx2 = 15.0 N + (-34.64 N) = -19.64 N (negative x-component)
Ry = Fy1 + Fy2 = 25.98 N + 20 N = 45.98 N (positive y-component)
Step 5: Find the magnitude (R) and direction (θ) of the resultant vector using the Pythagorean theorem and inverse tangent function:
R = sqrt(Rx^2 + Ry^2)
θ = atan(Ry / Rx)
R = sqrt((-19.64 N)^2 + (45.98 N)^2) = sqrt(385.0496 N^2 + 2116.9604 N^2) = sqrt(2502.01 N^2) = 50.02 N (approximately)
θ = atan(45.98 N / -19.64 N) ≈ atan(-2.344)
θ ≈ -67.81 degrees
So, the magnitude of the resultant vector is approximately 50.02 N, and its direction is approximately -67.81 degrees from the positive x-axis.
Therefore, the answer is: The magnitude of the resultant obtained by the vector addition is approximately 50.02 N, and its direction is approximately -67.81 degrees.
To find the magnitude and direction of the resultant obtained by adding the two force vectors using the analytical method, we can use the Pythagorean theorem and trigonometric functions.
Let's denote the angle between F1 and the x-axis as θ1, and the angle between F2 and the x-axis as θ2. In Figure 3-6, θ1 is given as 30° and θ2 is given as 60°.
Step 1: Resolving F1 and F2 into their x and y components:
F1x = F1 * cos(θ1)
F1y = F1 * sin(θ1)
F2x = F2 * cos(θ2)
F2y = F2 * sin(θ2)
Substituting the given values, we get:
F1x = 30.0 N * cos(30°) ≈ 26.0 N
F1y = 30.0 N * sin(30°) ≈ 15.0 N
F2x = 40.0 N * cos(60°) = 20.0 N
F2y = 40.0 N * sin(60°) ≈ 34.6 N
Step 2: Finding the x and y components of the resultant vector (Rx and Ry):
Rx = F1x + F2x
Ry = F1y + F2y
Substituting the values, we get:
Rx = 26.0 N + 20.0 N = 46.0 N
Ry = 15.0 N + 34.6 N = 49.6 N
Step 3: Calculating the magnitude of the resultant vector (R):
R = √(Rx^2 + Ry^2)
Substituting the values, we get:
R = √(46.0 N^2 + 49.6 N^2) = √(2116 N^2 + 2450.16 N^2) = √(4566.16 N^2) ≈ 67.6 N
Step 4: Determining the direction of the resultant vector (θR):
θR = tan^(-1)(Ry / Rx)
Substituting the values, we get:
θR = tan^(-1)(49.6 N / 46.0 N) ≈ tan^(-1)(1.08) ≈ 47.6°
Therefore, the magnitude of the resultant vector is approximately 67.6 N, and the direction is approximately 47.6° (measured counterclockwise from the x-axis).