Find the resultants of the following sets of vectors using the algebraic process. Check your answers by finding the resultant graphically as well.
Set 1:
Vector A: 3500 m @ 60 degrees;
Vector B: 5000 m @ 20 degrees S of W.
Set 2:
Vector A: 500 m/s @ 30 degrees;
Vector B: 400 m/s @ 25 degrees N of W;
Vector C: 300 m/s @ 15 degrees S of E.
Set 3:
Vector A: 250 N @ -45 degrees;
Vector B: 150 N @ 50 degrees .
NOTE: You should be familiar with subtraction of vectors also. For example, for any two vectors A and B, know how to find A - B, B - A, or B - 2A. The test will also include graphical addition/subtraction of vectors.
Set 1:
Vector A: 3500 m @ 60 degrees;
Vector B: 5000 m @ 20 degrees S of W.
Solution: First, sketch the vectors so you know whether the
components should be negative or positive. Then find the x and y
components of each vector using
the equations Ax = Acos(theta); Ay = Asin(theta). Give each component its correct sign.
In this problem,
Ax = 3500cos60 = 1750 m. It is positive because it goes to the right.
Ay = 3500sin60 = 3031 m, and is positive because its direction is up.
Bx = -5000cos20 = -4698 m and is negative because its direction is to the left.
By = -5000sin20 = -1710 m and is negative because its direction is down.
Now add the x components to get the x component of the resultant:
Rx = Ax + Bx
Rx = 1750 + -4698 = -2948 m.
Then add the y components to get the y component of the resultant:
Ry = Ay + By
Ry = 3031 + -1710 = 1321 m.
You now have two vectors, Rx and Ry. Sketch them head to tail (paying attention to their signs) to
determine what quadrant you are in. In this case, because Rx is negative and Ry
is positive, we are in the second quadrant.
We find the resultant using the Pythagorean Theorem: R2 = Rx2 + Ry2. In this problem,
R = 3230 m and we need to find its direction. Theta = arc tan 1321/2948.
So theta = 24 degrees, and because we are in the second quadrant, we say 24 degrees
N of W, or 156 degrees, or 24 degrees above the negative x axis, etc.
Set 2:
Vector A: 500 m/s @ 30 degrees;
Vector B: 400 m/s @ 25 degrees N of W;
Vector C: 300 m/s @ 15 degrees S of E.
In this problem,
Ax = 500cos30 = 433 m/s. It is positive because it goes to the right.
Ay = 500sin30 = 250 m/s, and is positive because its direction is up.
Bx = -400cos25 = -363 m/s and is negative because its direction is to the left.
By = 400sin25 = 169 m/s and is positive because its direction is up.
Cx = 300cos15 = 290 m/s and is positive because its direction is to the right.
Cy = -300sin15 = -78 m/s and is negative because its direction is down.
Now add the x components to get the x component of the resultant:
Rx = Ax + Bx + Cx
Rx = 433 + -363 + 290 = 360 m/s
Then add the y components to get the y component of the resultant:
Ry = Ay + By + Cy
Ry = 250 + 169 + -78 = 341 m/s
You now have two vectors, Rx and Ry. Sketch them head to tail (paying attention to their signs) to
determine what quadrant you are in. In this case, because Rx and Ry
are positive, we are in the first quadrant.
We find the resultant using the Pythagorean Theorem: R2 = Rx2 + Ry2. In this problem,
R = 496 m/s and we need to find its direction. Theta = arc tan 341/360.
So theta = 43.4 degrees, and because we are in the first quadrant, we say 43.4 degrees.
We could also say 43.4 degrees
N of W, or 43.4 degrees above the positive x axis, etc. (but probably wouldn't).
Set 3:
Vector A: 250 N @ -45 degrees;
Vector B: 150 N @ 50 degrees .
In this problem,
Ax = 250cos45 = 177 N. It is positive because it goes to the right.
Ay = -250sin45 = -177 N, and is negative because its direction is down.
Bx = 150cos50 = 96 N and is positive because its direction is to the right.
By = 150sin50 = 115 N and is positive because its direction is up.
Now add the x components to get the x component of the resultant:
Rx = Ax + Bx
Rx = 177 + 96 = 273 N.
Then add the y components to get the y component of the resultant:
Ry = Ay + By
Ry = -177 + 115 = -62 N.
You now have two vectors, Rx and Ry. Sketch them head to tail (paying attention to their signs) to
determine what quadrant you are in. In this case, because Rx is positive and Ry
is negative, we are in the fourth quadrant.
We find the resultant using the Pythagorean Theorem: R2 = Rx2 + Ry2. In this problem,
R = 280 N and we need to find its direction. Theta = arc tan 62/273.
So theta = 12.8 degrees, and because we are in the fourth quadrant, we say -12.8 degrees or
12.8 degrees S of E, or 12.8 degrees below the positive x axis or 347.2 degrees.
See also, Vector Resolution.