Vector A has magnitude of 40m at angel 225,if we want to add A to B so as to produce a resultant along the positive x-axis that has a magnitude of 20m,what must be the components of B
R = 40[225o] + B = 20
40*cos225+(40*sin225)i + B = 20
-28.3 - 28.3i + B = 20
B = 48.3 + 28.3i.
It cool
Answer
Well, to add Vector A to B and end up with a resultant along the positive x-axis, we first need to find the components of Vector A.
Given that Vector A has a magnitude of 40m at an angle of 225 degrees, we can break it down into its x and y components.
The x-component (Ax) of Vector A can be found using trigonometry:
Ax = magnitude of A * cos(angle of A)
= 40m * cos(225 degrees)
The y-component (Ay) of Vector A can also be found using trigonometry:
Ay = magnitude of A * sin(angle of A)
= 40m * sin(225 degrees)
Now, to produce a resultant vector along the positive x-axis with a magnitude of 20m, we need to add Vector B to Vector A.
Since the resultant vector will lie along the positive x-axis, the y-component of Vector B should be zero.
Therefore, the components of Vector B are:
Bx = 20m - Ax
By = 0
Now, let me calculate the values for you. *Clown noises*
Ax = 40m * cos(225 degrees) ≈ -28.28m
Ay = 40m * sin(225 degrees) ≈ -28.28m
Plugging these values into the equation for Bx, we get:
Bx = 20m - (-28.28m) ≈ 48.28m
So, the components of Vector B are approximately:
Bx ≈ 48.28m
By ≈ 0m
And there you have it! The components of Vector B to produce the desired resultant vector.
To find the components of vector B, we need to break down vector A and vector B into their respective x and y components. Then, we can adjust the components of vector B to produce the desired resultant vector.
Let's start by breaking down vector A into its x and y components.
Magnitude of vector A, A = 40m
Angle of vector A, θ = 225°
The x-component of vector A (Ax) can be found using the formula:
Ax = A * cos(θ)
Ax = 40m * cos(225°)
Ax = -28.3m (rounded to one decimal place, as the component is negative)
The y-component of vector A (Ay) can be found using the formula:
Ay = A * sin(θ)
Ay = 40m * sin(225°)
Ay = -28.3m (rounded to one decimal place, as the component is negative)
Now, let's find the components of vector B. Since we want the resultant vector to be along the positive x-axis and have a magnitude of 20m, the y-component of vector B (By) should be zero.
By = 0m
To find the x-component of vector B (Bx), we can use the equation for the magnitude of the resultant vector:
Resultant Magnitude = √((Bx + Ax)^2 + (By + Ay)^2)
Given that the resultant magnitude is 20m and By = 0m, the equation becomes:
20m = √((Bx + (-28.3))^2 + (0)^2)
Simplifying further:
400m^2 = Bx^2 + 2 * Bx * (-28.3) + (-28.3)^2
400m^2 = Bx^2 - 56.6Bx + 802.9
Rearranging the equation:
Bx^2 - 56.6Bx + 402.9 = 0
Now, we can solve this quadratic equation to find the values of Bx.
Using the quadratic formula:
Bx = (-b ± √(b^2 - 4ac)) / 2a
a = 1
b = -56.6
c = 402.9
Plugging in the values:
Bx = (-(-56.6) ± √((-56.6)^2 - 4 * 1 * 402.9)) / 2 * 1
Bx = (56.6 ± √(3197.56 - 1611.6)) / 2
Bx = (56.6 ± √1585.96) / 2
Calculating the square root:
Bx = (56.6 ± 39.82) / 2
Therefore, the x-component of vector B (Bx) can be either:
Bx = (56.6 + 39.82) / 2 = 48.71m (rounded to two decimal places)
or
Bx = (56.6 - 39.82) / 2 = 8.89m (rounded to two decimal places)
So, the components of vector B are:
Bx = 48.71m or Bx = 8.89m
By = 0m
I want answers to the above question
A = <-40/√2,-40/√2>
so if v = <x,y> then
<-40/√2,-40/√2> + <x,y> = <20,0>