Horizontal and vertical components.

In summary: Cartesian Coordinate graph and found my line at 81m east of north. i then found my vertical component and it was 79.898. finally, i calculated the horizontal component.
  • #1
yyttr2
46
0
I am doing physics homework and I came upon a problem that I had forgotten how to do, and i do not believe my answer is correct. I have tried to Google this but failed, can you please tell me if i have done this correct?
the question:
"4. Give the horizontal and vertical components of:
a.70m/s West
b. 42m North"

At first I was wondering how do i solve the horizontal and vertical components of a single variable? So i combined them and hoped they were the same problem.
203px-DiagramaPotenciasWPde.jpg

were the 42m is at the right and 70m/s is the base.
I then found the unknown angle showed in the picture (ignore all the symbols up there)

[tex]\phi[/tex]=arctan[tex]\frac{42}{70}[/tex]

[tex]\phi[/tex]=80.54[tex]\circ[/tex]

and I found the length of the hypotenuses.

hy=[tex]\sqrt{(l1)^2+(l2)^2}[/tex]

hy=[tex]\sqrt{(42)^2+(70)^2}[/tex]

hy=[tex]\sqrt{6664}[/tex]

hy=81.6

I then graphed the hypotenuse on a Cartesian Coordinate graph were x= east -x= west
y=north and -y=south
and drew my line 81m at 80.54[tex]\circ[/tex] east of north

and then found my vertical component.

[tex]\frac{sin\phi}{1}[/tex]=[tex]\frac{op}{hy}[/tex]

op(1)=sin[tex]\phi[/tex]hy

op=sin80.54[tex]\circ[/tex]81
op=79.898
vertical component= 79.898

and next my horizontal component.
hy=[tex]\sqrt{(l1)^2+(l2)^2}[/tex]

[tex]hy^{2}[/tex]=[tex]l1^{2}[/tex]+[tex]l2^{2}[/tex]

[tex]hy^{2}[/tex]-[tex]Vc^{2}[/tex]=[tex]Hc^{2}[/tex]

[tex]81^{2}[/tex]-[tex]79.898^{2}[/tex]=[tex]Hc^{2}[/tex]

Hc=[tex]\sqrt{81^2-79.898^2}[/tex]

Hc=13.4

If i have done this wrong please explain to me how, If i have done them right please explain to me which units I should use for the Hc and Vc and why.
 
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  • #2
The question, as you have posed it, makes no sense.

Even though it makes no sense, I'll play ball. If we assume, as you did in your solution, that the north-south direction is "vertical" and the east-west direction is "horizontal", then you already HAVE the horizontal and vertical components of the velocity, so why would you need to try and calculate them?

If you calculated the hypotenuse and then used trig to calculate the horizontal and vertical components, *then you should have gotten back what you started with.*

The reason you didn't get back what you started with is because your computation of the angle was wrong.

arctan(42 / 70) = 30.9637565 degrees
 
  • #3
:frown: It makes perfect sense
thank you.
 

What are horizontal and vertical components?

Horizontal and vertical components are used to describe the direction and magnitude of a vector. The horizontal component represents the vector's direction in the x-axis, while the vertical component represents the vector's direction in the y-axis.

How do you calculate horizontal and vertical components?

To calculate the horizontal and vertical components of a vector, you can use trigonometric functions. The horizontal component can be found by multiplying the magnitude of the vector by the cosine of the angle between the vector and the x-axis. The vertical component can be found by multiplying the magnitude of the vector by the sine of the angle.

Why are horizontal and vertical components important?

Horizontal and vertical components are important because they allow us to break down a vector into its individual parts. This can be helpful in solving physics and engineering problems, as well as understanding the motion of objects.

What is the relationship between horizontal and vertical components?

The horizontal and vertical components of a vector are always perpendicular to each other. This means that if you draw a right triangle with the vector as the hypotenuse, the horizontal and vertical components will be the adjacent and opposite sides, respectively.

Can horizontal and vertical components be negative?

Yes, horizontal and vertical components can be negative. The direction of a vector is determined by its angle, so if the angle is negative, then the horizontal or vertical component will also be negative. This simply means that the vector is pointing in the opposite direction of the positive component.

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