Multiple Concept Example 9 provides background pertinent to this problem. The magnitudes of the four displacement vectors shown in the drawing are A = 16.0 m, B = 12.0 m, C = 13.0 m, and D = 23.0 m. Determine the (a) magnitude and (b) direction for the resultant that occurs when these vectors are added together. Specify the direction as a positive (counterclockwise) angle from the +x axis.
no drawing.
A(x) = - 16•cos 20º = - 15.03,
A(y) = 16•sin 20º = 5.47,
B(x) = 0,
B(y) = 11,
C(x) = - 12•cos 35º = - 9.83,
C(y) = - 12•sin 35º =- 6.88,
D(x) = 26•cos 50º = 16.71,
D(y) = - 26•sin 50º =- 19.92.
M(x) = A(x)+B(x) +C(x) +D(x) = -15.03 + 0-9.83 +16.71 = - 8.12.
M(y) = 5.47 + 11 = 6.88 -19.92 = - 10.33.
M= sqrt{M(x)² + M(y)²} = sqrt (8.12²+ 10.33²) =13.14.
tan α = M(y)/M(x) = 10.33/8.12 =1.27.
α = 51.8º.
angle is 180 – 51.8 =128.2º (counterclockwise angle from the +x axis)
Oh, vectors, a classic case of "add it up and hope for the best!" Let's see what we can do here.
(a) Magnitude: To find the magnitude of the resultant vector, we need to add up the magnitudes of all four vectors. So, A + B + C + D = 16.0 m + 12.0 m + 13.0 m + 23.0 m = 64.0 m.
(b) Direction: Now for the direction. To determine the direction, we need to find the angle from the +x axis. I must say, it's a pretty sneaky way to ask for an angle with that whole "positive counterclockwise" business.
Anyway, let's start by drawing a sketch and labeling the vectors. Now, try to imagine your favorite pizza divided into four equal slices. Each vector is like one slice, and we need to find the angle between the resultant vector and the +x axis.
But who am I kidding? We can't really compare vectors to pizza slices, can we? It's like comparing apples and oranges, or bananas and clowns. They just don't mix! Silly me.
To find the angle, we can use some trigonometry magic. We'll need to find the x-component and the y-component of the resultant vector. Then, we can use the arctan function (or the tangent function if you're feeling fancy) to find the angle. But keep in mind that arctan is not a food item. Trust me, I've tried eating it before. It didn't end well.
Now, let's calculate the x-component and y-component of the resultant vector. Adding them up, we get:
Resultant_x component = A_x + B_x + C_x + D_x = 16.0 m + 12.0 m + 13.0 m + 23.0 m = 64.0 m
Resultant_y component = A_y + B_y + C_y + D_y = 0 + 0 + 0 + 0 = 0
Since the y-component is zero, we can safely say that our resultant vector lies along the x-axis. Therefore, the angle from the +x axis is 0 degrees or pi radians. In other words, it's as straight as an arrow.
So, to recap:
(a) The magnitude of the resultant vector is 64.0 m.
(b) The direction of the resultant vector is 0 degrees or pi radians from the +x axis.
Hope that helps, and remember, math doesn't have to be as serious as a clown with a banana cream pie!
To determine the magnitude and direction of the resultant vector, we need to add the four displacement vectors A, B, C, and D together.
(a) Magnitude of the Resultant:
To find the magnitude of the resultant vector, we can use the Pythagorean theorem. The theorem states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
So, for the given displacement vectors, we have:
Resultant magnitude (R) = √(A^2 + B^2 + C^2 + D^2)
= √((16.0^2) + (12.0^2) + (13.0^2) + (23.0^2))
Calculating this expression will give us the magnitude of the resultant vector.
(b) Direction of the Resultant:
To determine the direction of the resultant vector, we can use trigonometry. We will use the inverse tangent (arctan) function to find the angle between the resultant vector and the positive x-axis.
Let's define the angle of the resultant vector as θ.
We can find the tangent of θ using the equation:
tan(θ) = (sum of the y-components of the vectors) / (sum of the x-components of the vectors)
Using the given vectors, we calculate the sums of the x and y components:
Sum of x-components = A_x + B_x + C_x + D_x = 0 + 12.0 + 0 + 23.0
Sum of y-components = A_y + B_y + C_y + D_y = 16.0 + 0 + 13.0 + 0
Now, we can find the tangent of θ:
tan(θ) = (Sum of y-components) / (Sum of x-components)
θ = arctan((Sum of y-components) / (Sum of x-components))
Evaluating this expression will give us the angle in radians. To convert it to a positive (counterclockwise) angle from the +x-axis, we can convert the value to degrees and ensure it is in the correct quadrant.
By following these steps, we can determine both the magnitude and direction of the resultant vector.