What is the magnitude and angle of displacement given three vectors?

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Homework Help Overview

The discussion revolves around calculating the magnitude and angle of displacement given three vectors. The subject area includes vector resolution and trigonometry.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the breakdown of vector components, particularly focusing on the correct resolution of Vector C. Questions arise regarding the accuracy of the initial calculations and the proper method for determining the components of the vectors.

Discussion Status

The discussion is ongoing, with participants providing feedback on the breakdown of vector components and questioning assumptions about the angles involved. Some guidance has been offered regarding visualizing the vectors and constructing triangles for better understanding.

Contextual Notes

There are indications of missing information in the calculations, such as a negative sign in the y-direction. Participants are also exploring the implications of angles and their respective components in relation to the horizontal axis.

laladude
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Homework Statement


1.jpg

Compute the magnitude of the displacement (\vec{}R and the angle with respect to the x-axis.


Homework Equations





The Attempt at a Solution



A = 10m * sin30° \hat{}y+ 10m * cos30° \hat{}x

\vec{}A = 5m \hat{}y + 8.66m \hat{}x

B = 15m

C = 20m * sin80° \hat{}x + 20m * cos80° \hat{}y

\vec{}R = \hat{}A + \hat{}B + \hat{}C
\hat{}R = √(43.36m)2 + (1.58m)2 = 64.34m

tanθ = 43.36/1.58
= 87°

Is this correct? Please help!
 
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laladude said:
C = 20m * sin80° \hat{}x + 20m * cos80° \hat{}y
Your breakdown of Vector C isn't quite right.
 
NascentOxygen said:
Your breakdown of Vector C isn't quite right.

Thanks.. Would it just be 20m * cos80° in the y-direction and none in the x-dir?
 
laladude said:
Thanks.. Would it just be 20m * cos80° in the y-direction and none in the x-dir?
No, not if that's 80 degrees to the horizontal. Unless a vector lies exactly along the x (or y)[/color] axis, it must resolve into both x and y components.
 
NascentOxygen said:
No, not if that's 80 degrees to the horizontal. Unless a vector lies exactly along the x (or y)[/color] axis, it must resolve into both x and y components.

I noticed I'm missing a negative in the y direction, but how will I correctly break it into its components? I can make a triangle from the dotted line to the bottom arrow of C, but I'm not so sure of what to use.
 
To make it easier to visualize, in your rough working (which only you see) draw the angle so that it's more like 70 degrees than 80 so you have more space on the paper to work with. Then construct that triangle you spoke of, and determine the lengths of its vert and horiz sides.

There is more amiss than the missing - sign.
 
NascentOxygen said:
To make it easier to visualize, in your rough working (which only you see) draw the angle so that it's more like 70 degrees than 80 so you have more space on the paper to work with. Then construct that triangle you spoke of, and determine the lengths of its vert and horiz sides.

There is more amiss than the missing - sign.

draw the triangle on the inside of the picture? I am so losttt :cry:
 
Vector C is almost vertical, so when you break it into its components, the longest of these components is going to be the vertical one. Check that you have this correct.
 

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