What are the components and sum of two vectors?

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In summary, Vector V1 is 7.59 units long and points along the negative x axis. Vector V2 is 4.34 units long and points at +55.0° to the positive x axis. The sum of V1 and V2 is 6.22 units.
  • #1
physicsss
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Vector V1 is 7.59 units long and points along the negative x axis. Vector V2 is 4.34 units long and points at +55.0° to the positive x axis.

(a) What are the x and y components of each vector?
I got:
V1x = -7.59
V1y = 0
V2x = 2.49
V2y = 3.56

(b) Determine the sum V1 + V2 (magnitude and angle).
I got:
6.22

All of the above is correct when I submitted it, but the angle was wrong. I've tried entering -34.9 and 34.9, but they are all wrong. Any ideas?
:confused:
 
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  • #2
Gosh, I hate these "submit to a computer" things! So often they are "wrong" just because you have too many significant figures or some picky thing like that.

But in this case, you answer is simply "completely" wrong.

The problem is that arctan, like all of the inverse trig functions is "multi-valued"
Adding the two vectors in component form, you get <-4.8,3.56> and the angle is given by arctan(-3.56/4.8)= arctan(-.742). You calculator can only give one answer and so gives the "principal" value, -36.6 degrees.
Since we are measuring angles from the positive x-axis, that would correspond to a vector pointing down and to the right: <4.8,-3.56> (Obviously, the arctan has no way of "knowing" if the negative is from the numerator or denominator).
But the correct vector is <-4.8, 3.56> which points up and to the right: you need to add 180 degrees. The correct angle is 180- 36.6= 143.4 degrees.
 
  • #3
How did you get -4.8?
 
  • #4
What would the angle be if the x-component is -193.65 and the y-component is -12.941? anyone?
 
  • #5
physicsss said:
What would the angle be if the x-component is -193.65 and the y-component is -12.941? anyone?
[tex]tan\theta=\frac{193.65}{12.941}[/tex]
[tex]\theta=86.177^0[/tex]
don't insert the negative sign, just treat it as a normal triangle with certain lengths which are always positive.
so, you can say that the vector makes an angle of –(90+86.177)=-176.177 degree from the positive x-axis; or
(360-176.177)=183.823 degree from the positive x-axis.
 

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  • #6
physicsss said:
How did you get -4.8?

By adding, just as the problem said to do:
(-7.59, 0)+ (2.49, 3.56)= (-7.59+ 2.49, 0+ 3.56)= (-5.1, 3.56)

Hmm, blasted calculator doesn't know what it's doing!
 
  • #7
Leong said:
[tex]tan\theta=\frac{193.65}{12.941}[/tex]
[tex]\theta=86.177^0[/tex]
don't insert the negative sign, just treat it as a normal triangle with certain lengths which are always positive.
so, you can say that the vector makes an angle of –(90+86.177)=-176.177 degree from the positive x-axis; or
(360-176.177)=183.823 degree from the positive x-axis.

OR put in the negative sign: [itex]tan\theta= \frac{-193.65}{-12.941}[/itex] and then observe that the negatives cancel!
 
  • #8
Um...tan x=y/x...should it be tan^-1(12.941/193.65)?
 
  • #9
There are two ways to find the angle:
1. Draw the vector, find the angle you want by treating it like a right triangle using trig like what i have done.
2. Use the vector notation, [tex]\vec{a}= x\vec{i} + y\vec{j}[/tex]
Then, [tex]tan\ \theta = \frac{y}{x}[/tex]

I use option #2 here.

[tex]\vec{a}=-193.65\vec{i}-12.941\vec{j}[/tex]
[tex]tan\ \theta = \frac{-12.941}{-193.65}[/tex]
Your calculator will give [tex]\theta=3.8232^0[/tex]
See HallofIvy's point in post #2?
Your calculator will give the principal value to you.
But you know that [tex]\vec{a}[/tex] lies in the 3rd quadrant. so you should add 180 degree to the answer you have just got; i.e. 183.82 degree; just like what we have got using option #1 method.

and when you use the formula to find the angle by method #2. the angle is always relative to the positive x-axis.find the principal value, identify the quadrant where the vector lies. add some appropriate degree and get the answer.
 
  • #10
arent vectors fun :smile: , hehe.
 

What are vectors and how are they used in science?

Vectors are mathematical quantities that have both magnitude and direction. In science, they are commonly used to represent physical quantities such as force, velocity, and acceleration. They are often drawn as arrows in diagrams to show the direction and magnitude of a physical quantity.

How are vectors different from scalars?

Scalars are quantities that have only magnitude, while vectors have both magnitude and direction. For example, speed is a scalar quantity as it only tells us the magnitude of an object's motion, while velocity is a vector quantity as it tells us both the magnitude and direction of an object's motion.

What are the different types of vectors?

There are several types of vectors, including displacement vectors, velocity vectors, force vectors, and electric field vectors. Each type represents a different physical quantity and has its own rules for addition, subtraction, and multiplication.

Can vectors be added and subtracted?

Yes, vectors can be added and subtracted using the rules of vector addition and subtraction. These rules take into account both the magnitude and direction of the vectors. When adding or subtracting vectors, it is important to ensure that they are in the same units and are represented by the same coordinate system.

How are vectors used in scientific experiments?

Vectors are used in scientific experiments to accurately represent and analyze physical quantities such as force, velocity, and acceleration. They can be used to predict the motion of objects, calculate the net force acting on an object, and determine the direction and magnitude of an electric field. Vectors are also essential in the fields of engineering, physics, and mathematics.

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