Adding vectors algebraically to find displacement

In summary, Kevin was trying to solve for the length of vector c, which is perpendicular to the x-axis and has a magnitude of 190. However, the angles for c are wrong and the solution is therefore incorrect.
  • #1
KevinFuerst
4
0
Hi everyone,
I am taking a physics course this year and I find myself already looking for help unfortunately. I think I understand the basics of adding vectors using Ax = A cosθ , Ay = A sinθ, etc. I am getting confused with the 3rd leg of the trip though. Since it is due west wouldn't that mean there is no vertical displacement? How can work that into my equation?

Homework Statement


attachment.php?attachmentid=50133&stc=1&d=1345744862.jpg

Homework Equations


Ax = A cosθ
Ay = A sinθ
...
R = √Rx2 + Ry2

The Attempt at a Solution


I have tried using the previous formulas to solve it but i get an answer of ≈320km. the back of the book says it should be 245km. could someone walk me though this? Thank you!
 

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  • #2
Hey Kevin ,
In the picture I have uploaded you can see
that
(*note the symbols in bold are vectors)

r =a + b

, subsequently
R= c+r
 

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  • #3
What you want is [itex]\vec{R} = \vec{a} + \vec{b} + \vec{c}. [/itex]
Write out the vector equations for each of [itex] \vec{a}, \vec{b}, \vec{c} [/itex] and then add them together. This will give an eqn for [itex] \vec{R} [/itex]

Using [itex] |\vec{R}| = \sqrt{R_x^2 + R_y^2} [/itex] will give you it's length.
Can you find the angle from this?
 
  • #4
thanks for the quick response but I am not quite following you guys. Still new to this.
Here is my work, maybe you can show me what i am doing wrong that way. Once i get the right distance i'll have no problem finding the angle. I under stand how to do that part.

Ax = A cos(30) = 175km (.866) = 151.55km
Ay = A sin(30) = 175 (.5) = 87.5km
Bx = B sin(20) =150km (.342) = -51.3km
By = B cos(20) = 150 (.939) = 140.96km
Cx = C cos(110) = -190km (-.342) = 64.98km
Cy = C sin(110) = 190 (.939) = 178.54km

√165.232 + 4072 = √192951.4 ≈ 439km
 
  • #5
Hey kelvin ,
As the diagram and the question says , the vector C is parallel to X axis , which mean there is no change in the y component of total displacement .
On the side note , What does the figure mean by 110 degrees ? what is it .
 
  • #6
Try solving
Cx as 190
and Cy as 0
 
  • #7
Cx = C cos(110) = -190km (-.342) = 64.98km
Cy = C sin(110) = 190 (.939) = 178.54km

You have the components correct for vectors [itex] \vec{a} [/itex] and [itex] \vec{b},[/itex]however the angles are wrong for the components of [itex] \vec{c} [/itex]

What angle does [itex] \vec{c} [/itex] make with the positive [itex] x [/itex] direction?
There is no vertical component to [itex] \vec{c} [/itex]
 
  • #8
kushan said:
Try solving
Cx as 190
and Cy as 0
thank you so much. i understand that now. it will help me on a previos one i thought i had right too.

CAF123 said:
You have the components correct for vectors [itex] \vec{a} [/itex] and [itex] \vec{b},[/itex]however the angles are wrong for the components of [itex] \vec{c} [/itex]

What angle does [itex] \vec{c} [/itex] make with the positive [itex] x [/itex] direction?
There is no vertical component to [itex] \vec{c} [/itex]

vector c is at 0 or 180 degrees isn't it?
 
  • #9
180 you can call it .
 
  • #10
Vector [itex]\vec{c} [/itex] is at angle of 180o with respect to the positive [itex] x [/itex] axis. Therefore the vector eqn for [itex]\vec{c} [/itex] is,
[itex] \vec{c} = (190cos180)\hat x = -190\hat x [/itex]
 
  • #11
oh okay! that's where that 110 came from. i was trying to figure out the angle so i decided to try the interior angle of ΔABC

i had a very similar problem there i found the interior angle like that and coincidently the resultant was extremely close either way. I have now corrected both and i can't thank you two enough.
 

What is the concept of adding vectors algebraically?

Adding vectors algebraically is the process of combining two or more vectors to find the total displacement or resultant vector. It involves using mathematical operations such as addition, subtraction, and multiplication to find the magnitude and direction of the resultant vector.

What are the steps involved in adding vectors algebraically?

The steps for adding vectors algebraically are as follows:
1. Represent the vectors using their magnitude and direction.
2. Add the horizontal components of the vectors together and then the vertical components together.
3. Use the Pythagorean theorem to find the magnitude of the resultant vector.
4. Use trigonometric functions to find the direction of the resultant vector.

Can vectors be added in any order?

No, the order in which vectors are added can affect the final resultant vector. It is important to add the vectors in the correct order, typically starting with the first vector and then adding subsequent vectors one by one. This ensures that the direction of each vector is taken into account and the final result is accurate.

What is the difference between adding vectors graphically and algebraically?

Adding vectors graphically involves physically drawing the vectors and using a scale to determine the magnitude and direction of the resultant vector. On the other hand, adding vectors algebraically involves using mathematical operations to find the magnitude and direction of the resultant vector. While both methods result in the same answer, adding vectors algebraically is often more accurate and efficient.

Can vectors be subtracted using the same method as addition?

Yes, vectors can be subtracted using the same method as addition. Subtraction of vectors involves adding the negative of the vector being subtracted. For example, to subtract vector B from vector A, we would add vector A and the negative of vector B (represented as -B). The resultant vector will have a magnitude and direction that represents the difference between the two vectors.

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