Finding magnitude of vector without direction (kite problem)

In summary, a child is flying a kite with a mass of 6.8 kg on a windy day. The tension from the string on the kite has a magnitude of 17.0 N and makes an angle of 32.6° from the vertical, while the acceleration of the kite has a magnitude of 7.72 m/s2 and makes an angle of 37.6° from the vertical. The only forces acting on the kite are its weight, the tension from the string, and a force from the wind. Using Newton's second law, we can set up equations in component form and solve for the unknown force from the wind and the angle between the wind and the vertical axis. The correct answer, found by
  • #1
Lorelei42
2
0

Homework Statement


A child is flying a large kite of mass 6.8 kg on a windy day. At the moment shown the tension from the string on the kite has a magnitude of 17.0 N and makes an angle of
char12.png
= 32.6° from the vertical, and the acceleration of the kite has a magnitude of ak = 7.72 m/s2 and makes an angle of
char1E.png
=37
char3A.png
6° from the vertical as shown in the figure below. The only forces felt by the kite are its own weight, the tension from the string, and a force from the wind.
(picture of problem, shows angles)
upload_2018-2-7_15-40-48.png

Homework Equations


Newton's second Law F(net)=m*a

The Attempt at a Solution



Knowns[/B]: m(kite)=6.8kg, θ(1)=32.6°(vertical angle between y-axis and string), Tension=T(yk)=17N, Φ=37.6°(angle between direction of acceleration and y-axis), a(k)=7.72m/s/s, a(g)=9.81m/s/s
Unknowns: F(wk), θ(2)(angle between wind and y-axis)(see FBD if unclear)

Free Body Diagram:
Free body diagram, apologies for poor handwriting/picture quality)
upload_2018-2-7_15-57-13.png


Equations in Component Form:
X: a*sinΦ*m(k)=F(wk)*sin(θ2)-T(yk)*sin(θ1)
Solved for F(wk) becomes: F(wk)=(a*sinΦ*m(k)+T(yk)*sin(θ1))/sin(θ2)

Y:a*cosΦ*m(k)=F(wk)*cos(θ2)-T(yk)*cos(θ1)-m*g
Solved for F(wk)=(a*cosΦ*m(k)+T(yk)*cos(θ1)+mg)/(cos(θ2))

Solve for θ2
Since both equations are set equal to F(wk) they can be set equal to each other:
(a*cosΦ*m(k)+T(yk)*cos(θ1)+mg)/(cos(θ2))=(a*sinΦ*m(k)+T(yk)*sin(θ1))/sin(θ2)
Add in the given knowns
(7.72*cos(37.6)*6.8+17*cos(32.6)+6.8*9.81)/(cos(θ2))=(7.72*sin(37.6)*6.8+17*sin(32.6))/sin(θ2)
Simplify
122.554/(cos(θ2)=41.189/sin(θ2)
Solve for (θ2)
sin(θ2)/cos(θ2)=41.189/122.554
tan(θ2)=0.336
(θ2)=arctan(.336)=18.57°

Substitute (θ2) back into equation to find F(wk)

xeq. F(wk)=122.554/cos(θ2)=122.554/cos(18.570=129.29N
yeq. F(wk)=41.189/sin(θ2)=41.189/sin(18.57)=129.29N

But apparently this is not the correct answer, so somewhere I went wrong in my process. Any friendly people care to show me where that "somewhere" was.
 

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  • #2
Nevermind, I solved it.

For anyone else who has a similar problem I recommend using a tilted coordinate axis where the positive x direction is parallel to the direction of acceleration.
 

1. What is the purpose of finding the magnitude of a vector without direction?

The magnitude of a vector without direction, also known as the absolute value or length of a vector, allows us to understand the size or strength of the vector without considering its direction. This is useful in various scientific and mathematical applications, such as analyzing forces or motion.

2. How do you find the magnitude of a vector without direction?

To find the magnitude of a vector without direction, we use the Pythagorean theorem. This involves squaring the x and y components of the vector, adding them together, and then taking the square root of the sum. This gives us the length of the vector regardless of its direction.

3. What is the difference between magnitude with and without direction?

The magnitude of a vector with direction is a measure of both the size and direction of the vector. On the other hand, the magnitude of a vector without direction only considers the size of the vector, disregarding its direction. This means that two vectors with the same magnitude but different directions will have different magnitudes when direction is taken into account.

4. Can a vector have a negative magnitude without direction?

No, the magnitude of a vector without direction is always a positive value. This is because the Pythagorean theorem involves squaring the components of the vector, which results in positive values. Therefore, the square root of the sum will also be positive.

5. How is finding the magnitude of a vector without direction useful in real-world situations?

Finding the magnitude of a vector without direction is useful in various real-world situations, such as in navigation or engineering. For example, knowing the magnitude of a force acting on a structure can help engineers determine the strength and stability of the structure. In navigation, it can help determine the distance traveled and the speed of an object without considering its direction.

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