Answer:
The elongation rate is 1.17 ft/s
Explanation:
The expression is:
-(L - x)θsinθ + x*cosθ = 0
x = (L+x)θ * tanθ
CD² = 2b²(1 - cos(θ + δ))
2CC = 2b²θsin(θ + δ)
[tex]O=\frac{CC*C\sqrt{2^{2}(1-cos(O+Y) } }{b^{2}sin(O+Y) }[/tex]
Where O = θ and Y = δ
[tex]O=\frac{\sqrt{2} }{b} (\frac{\sqrt{1-cos(O+Y)} }{\sqrt{1-cos^{2}(O+Y) } } )C[/tex]
[tex]x=(L+x)tanO\frac{\sqrt{2} C}{b\sqrt{1+cos(O+Y)} }[/tex]
Where
C = 1 ft/s
θ = 28°
L = 13 ft
h = 3.2 ft
b = 9 ft
x = 5 ft
δ = sin⁻¹(h/b) = sin⁻¹(3.2/9) = 20.83°
Replacing:
[tex]x=\frac{(13+5)*tan(28)*\sqrt{2}*1 }{9*\sqrt{1+cos(28+20.83)} } =1.17ft/s[/tex]
