- #1

Lorelei42

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## 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

(picture of problem, shows angles)

## 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)

**
**

**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)

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.