Need help understanding vectors.?

Question

Vector A has a magnitude of 8.80 units and makes an angle of 51.0° counter-clockwise from the positive x-axis. Vector B has a magnitude of 8.00 units and is directed along the negative x-axis.

Find the magnitude of A + B:
Find the direction of A + B counterclockwise from the positive X axis:

Any help on setting it up would be great!

Answer 1

The tricky bit here is with doing it on the web without diagrams: drawing the diagram is 90% of the battle with vectors.



So draw a set of axes in your book, and make 'em like a big plus sign, so you have both positive and negative on both the x and y axes.



Now, B is pretty easy. It starts at the origin (0,0) and goes left 8 units along the negative side of the x-axis to the point (-8,0). Draw an arrow at that end point to show which way the vector goes.



A is a bit trickier, but not much. It also starts at the origin and goes out toward between where 1 and 2 o'clock would be if you put a clock over your axes. It's 8.8 units long and it is at an angle of 51 degrees above the positive side of the x-axis. Draw an arrow at the top end to show which way the vector goes.



OK, that's your starting situation. Now, to solve it, you need to imagine grabbing the A vector and just sliiiding it left alone the negative x-axis, until its 'tail (non-arrow) end is at the head of the other vector (in this case, the point (-8,0). Its angle stays exactly the same. Draw in the vector A in its new position.



Now, the *resultant* vector, A+B, the answer you're looking for, starts from the origin (0,0) and goes up to the tip of the new A vector you just drew.



If you drew the diagram super carefully, and measured the 51 degrees accurately with a protractor, then you can just get the answer from the graph - magniutde of A+B you measure with a ruler, direction you measure with a protractor.



Otherwise, you would use trig: you know two sides of the triangle you just created, and at least one of the angles, so you should be able to use sine rule and cos rule to find the other sides and angles.



All the best with it!

Answer 2

Here is another method called "Superposition." A very useful technique, learn it.



Vector A is less than 90 degrees so it is in the first quadrant ====> Both x and y components are positive



We have a right triangle with angle 51 degrees and the hypotenuse is 8.8 units. We need to solve for the 2 other sides using trig.



cos 51 = x/8.8

x = 5.54



sin 51 = y/8.8

y = 6.84



Vector B is easy



x = - 8.0

y = 0



Add up the components



X = 5.54+ (- 8.0) = - 2.46

Y = 6.84 + 0 = 6.84



X is negative and Y is positive ===> Our RESULTANT is in the 2nd quadrant



That is our new triangle, Those are the x and y components for our RESULTANT vector. We need angle and hypotenuse. We measure the angle to the RESULTANT clockwise from the - x axis





Lets do the Pythagorean Theorem for magnitude



-2.46^2 + 6.84^2 = R^2

6.05 + 46.79 = R^2



R^2 = 52.84

R = 7.27 units



angle



tan^-1 = Y/X (I know where the angle is, so I am going to work with + numbers)

tan^-1 = 6.84/2.46 = 2.78

angle = 70.22 degrees above the - x axis

Answer 3

i'm in geometry an in eighth grade this 365 days and we already discovered vectors. they're particularly elementary actualy. they're used for shifting one shape to a distinctive are on the graph using the vector. as an occasion in case you have factor (0,0) and the vector is <2,3> then u basically upload te vector to the factor making the applicable factor (2,3). wish I helped

Answer 4

Do it by components:

Xa = -8; Ya = 0

Xb = 8*cos51°; Yb = 8*sin51°



Xr = Xa+Xb

Yr = Ya+Yb



Mr = √(Xr²+Yr²) = 6.888 units

Θr = arctan(Yr/Xr) = 115.5°