When two vectors vector A and vector B are drawn from a common point, the angle between them is phi.
If vector A and vector B have the same magnitude, for which value of phi will their vector sum have the same magnitude as vector A or vector B?
When the triangle of forces is an equilateral triangle, i.e. when φ=120°. So Acos(φ/2)+Bcos(φ/2)=A/2+B/2=A=B
To find the value of phi for which the vector sum of vector A and vector B has the same magnitude as vector A or vector B, we can analyze the situation using vector addition.
Let's assume that the common magnitude of vector A and vector B is denoted as "m" for simplicity.
The vector sum, denoted as vector C, can be found by adding vector A and vector B. Mathematically, we can represent this as:
vector C = vector A + vector B
Since vector A and vector B have the same magnitude, their sum will result in a parallelogram where the sides are of equal length.
The magnitude of vector C can be expressed as:
|vector C| = √(Cx^2 + Cy^2)
Where Cx and Cy are the x and y components of vector C, respectively.
To have the same magnitude as vector A or vector B, the expression above should be equal to "m".
So, we have:
√(Cx^2 + Cy^2) = m
To determine the value of phi, we need to consider the components of vector A and vector B.
Let's assume that the angle between the positive x-axis and vector A is θ1, and the angle between the positive x-axis and vector B is θ2.
Considering the magnitudes and angles, we can represent vector A and vector B in terms of their components as:
Vector A: Ax = m * cos(θ1) and Ay = m * sin(θ1)
Vector B: Bx = m * cos(θ2) and By = m * sin(θ2)
Now, let's substitute the x and y components of vector A and vector B into the expression for the magnitude of vector C:
|vector C| = √((Ax + Bx)^2 + (Ay + By)^2)
= √((m * cos(θ1) + m * cos(θ2))^2 + (m * sin(θ1) + m * sin(θ2))^2)
= √(m^2 * (cos^2(θ1) + 2 * cos(θ1) * cos(θ2) + cos^2(θ2)) + m^2 * (sin^2(θ1) + 2 * sin(θ1) * sin(θ2) + sin^2(θ2)))
= √(2 * m^2 * (1 + cos(θ1) * cos(θ2) + sin(θ1) * sin(θ2)))
Simplifying further:
|vector C| = √(2 * m^2 * (1 + cos(θ1 - θ2)))
To have the same magnitude as vector A or vector B, |vector C| must be equal to m:
√(2 * m^2 * (1 + cos(θ1 - θ2))) = m
Squaring both sides of the equation:
2 * m^2 * (1 + cos(θ1 - θ2)) = m^2
2 + 2 * cos(θ1 - θ2) = 1
Simplifying:
cos(θ1 - θ2) = -1/2
Now, to find the value of phi, we need to solve for θ1 - θ2. We know that θ1 - θ2 is the angle between vector A and vector B.
Using the inverse cosine function, we can determine the possible values of θ1 - θ2:
θ1 - θ2 = ±120° + 360° * n (n is an integer)
Finally, phi can be calculated by subtracting θ1 - θ2 from 180°:
phi = 180° - (θ1 - θ2) = 180° - (±120° + 360° * n)
Therefore, for the value of phi where the vector sum of vector A and vector B has the same magnitude as vector A or vector B, phi can be either 60° + 360° * n or 300° + 360° * n, where n is an integer.