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Question Number 181651 by mr W last updated on 28/Nov/22

Commented by Acem last updated on 05/Dec/22

$D i d y o u b o t h b e l i e v e h i m w i t h a l l y o u r m i n d s ?$ $H e j u s t t r y i n g t o m a k e y o u f e e l r e m o r s e a n d$ $g i v e t h e g o a t e t e r n a l r e s t a n d t h e n i n v i t e h i m$ $t o a b a r b e c u e$
Commented by mr W last updated on 28/Nov/22

$a g o a t i s t i e d w i t h a 40 m l o n g r o p e$ $o n a s i l o a t a h e i g h t o f 2 m . t h e r a d i u s$ $o f t h e s i l o i s 10 m . f i n d t h e a r e a$ $w h e r e t h e g o a t c a n g r a z e a r o u n d t h e$ $s i l o .$
Commented by Frix last updated on 28/Nov/22

$What about the Animal Welfare ?!$
Commented by mr W last updated on 28/Nov/22

$a r e y o u s e e i n g t h a t t h e g o a t i s$ $s u f f e r i n g ?$
Commented by Frix last updated on 28/Nov/22
Ask the goat... ��
Commented by mr W last updated on 29/Nov/22

$w h e n y o u k n o w t h a t t h e g o a t i s$ $s u f f e r i n g, t h e n e i t h e r y o u h a v e a s k e d$ $t h e g o a t o r . . .$ $o t h e r w i s e h o w c o u l d y o u k n o w t h a t ?$
Commented by MJS_new last updated on 30/Nov/22

$my {grandmother}^{'} s goat would have eaten$ $the rope . . .$
	Answered by a.lgnaoui last updated on 28/Nov/22

	$△ OAB \sin θ = \frac{1}{20} \Rightarrow \cos θ = \sqrt{\frac{399}{400}}$ $Area = A_{2} - A_{1}$ $A_{2} = π {(10 + 40 \sqrt{\frac{399}{400}})}^{2} = 7838, 2717$ $A_{1} = 100 π = 314, 1592$ $A r e a = A_{2} - A_{1} = 7524, 1124$
	Commented by a.lgnaoui last updated on 28/Nov/22

	Commented by mr W last updated on 28/Nov/22

	$w r o n g!$ $t h e r o p e i s f i x e d o n t h e s i l o .$
	Commented by mr W last updated on 28/Nov/22

	Answered by Acem last updated on 28/Nov/22

	Commented by Acem last updated on 28/Nov/22

	$D e e m t h a t t h e c i l o a s a w h e e l i s s p i n n i n g w i t h o u t$ $r o l l i n g, a n d h e r e i t h i n k t h a t t h e p l o t t e r i s t h e$ $s e g m e n t A P^{'}$ $N o w w e n e e d a f u n c t i o n i n t e r m s$ $o f t h e v a r i a b l e θ$ $I m p o r t a n t N o t e, i {d i d n}^{'} t n o t i c e t h e f i x e d p o i n t$ $w h i c h h a s 2 m h e i g h t, i w i l l m o d i f y t h e g r a p h$ $a n d s o m e v a l u e s l a t e r$
	Answered by mahdipoor last updated on 28/Nov/22


	Commented by mahdipoor last updated on 28/Nov/22

	$i c a n t s e n d i m a g e : ($
	Commented by mahdipoor last updated on 28/Nov/22
	$so F.S. is: π[((f^2 ((π/2)))/2)−R^2 +2∣∫_θ_1 ^( θ_2 ) g(θ)dθ∣] ⇒⇒ f(θ)=(√(L^2 −H^2 ))−R((π/2)−θ) g(θ)=(R^2 +f^2 (θ))(−1+(d/dθ)[tan^(−1) (((f(θ))/R))]) { ((θ_2 =(π/2))),((0>θ_1 ≥((−π)/2)=Answer {Rcos(θ)+f(θ)sinθ=0})) :} H=2m R=10m L=40m$
	$s o F . S . i s :$ $π [\frac{f^{2} (\frac{π}{2})}{2} - R^{2} + 2 ∣ \int_{θ_{1}}^{θ_{2}} g (θ) d θ ∣]$ $\Rightarrow\Rightarrow$ $f (θ) = \sqrt{L^{2} - H^{2}} - R (\frac{π}{2} - θ)$ $g (θ) = (R^{2} + f^{2} (θ)) (- 1 + \frac{d}{d θ} [{t a n}^{- 1} (\frac{f (θ)}{R})])$ ${\begin{cases} θ_{2} = \frac{π}{2} \\ 0 > θ_{1} ⩾ \frac{- π}{2} = A n s w e r {R c o s (θ) + f (θ) s i n θ = 0} \end{cases}$ $H = 2 m R = 10 m L = 40 m$
	Answered by mr W last updated on 28/Nov/22

	Commented by mr W last updated on 28/Nov/22
	$l=rope length r=radius of silo h=height of anchor point of rope R=(√(l^2 −h^2 )) let λ=(R/r) PB=a (a+rθ)^2 +h^2 =l^2 ⇒a=(√(l^2 −h^2 ))−rθ=R−rθ x_P =r cos θ−a sin θ =r cos θ−(R−rθ) sin θ =r[cos θ−(λ−θ) sin θ] y_P =r sin θ+a cos θ =r sin θ+(R−rθ) cos θ =r[sin θ+(λ−θ) cos θ] at y_P =0: θ_1 −tan θ_1 =λ A_1 =2∫_0 ^R (r−x)dy A_1 =2r^2 ∫_θ_1 ^0 [1−cos θ+(λ−θ) sin θ][−(λ−θ) sin θ]dθ A_1 =2r^2 ∫_0 ^θ_1 (λ−θ) sin θ [1−cos θ+(λ−θ) sin θ]dθ A=((πR^2 )/2)+A_1 −πr^2 ⇒A=πr^2 {(λ^2 /2)−1+(2/π)∫_0 ^θ_1 (λ−θ) sin θ [1 +(λ−θ) sin θ−cos θ]dθ} example: r=10, l=40, h=2 λ=((√(l^2 −h^2 ))/r)=((√(40^2 −2^2 ))/(10))=((√(399))/5) θ_1 ≈2.0446 A≈π×10^2 ×(((399)/(50))−1+7.249688150) ≈4470.4 m^2$
	$l = r o p e l e n g t h$ $r = r a d i u s o f s i l o$ $h = h e i g h t o f a n c h o r p o i n t o f r o p e$ $R = \sqrt{l^{2} - h^{2}}$ $l e t λ = \frac{R}{r}$ $P B = a$ ${(a + r θ)}^{2} + h^{2} = l^{2}$ $\Rightarrow a = \sqrt{l^{2} - h^{2}} - r θ = R - r θ$ $x_{P} = r \cos θ - a \sin θ$ $= r \cos θ - (R - r θ) \sin θ$ $= r [\cos θ - (λ - θ) \sin θ]$ $y_{P} = r \sin θ + a \cos θ$ $= r \sin θ + (R - r θ) \cos θ$ $= r [\sin θ + (λ - θ) \cos θ]$ $a t y_{P} = 0 :$ $θ_{1} - \tan θ_{1} = λ$ $A_{1} = 2 \int_{0}^{R} (r - x) d y$ $A_{1} = 2 r^{2} \int_{θ_{1}}^{0} [1 - \cos θ + (λ - θ) \sin θ] [- (λ - θ) \sin θ] d θ$ $A_{1} = 2 r^{2} \int_{0}^{θ_{1}} (λ - θ) \sin θ [1 - \cos θ + (λ - θ) \sin θ] d θ$ $A = \frac{π R^{2}}{2} + A_{1} - π r^{2}$ $\Rightarrow A = π r^{2} {\frac{λ^{2}}{2} - 1 + \frac{2}{π} \int_{0}^{θ_{1}} (λ - θ) \sin θ [1 + (λ - θ) \sin θ - \cos θ] d θ}$ $e x a m p l e : r = 10, l = 40, h = 2$ $λ = \frac{\sqrt{l^{2} - h^{2}}}{r} = \frac{\sqrt{40^{2} - 2^{2}}}{10} = \frac{\sqrt{399}}{5}$ $θ_{1} \approx 2.0446$ $A \approx π \times 10^{2} \times (\frac{399}{50} - 1 + 7.249688150)$ $\approx 4470.4 m^{2}$
	Commented by mr W last updated on 28/Nov/22

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