Question-228227

Question Number 228227 by ajfour last updated on 03/Apr/26

Commented by ajfour last updated on 03/Apr/26

A rod of length L in a cubic room, side s, one end of rod at O while to end P are attached two strings PA=a, PB=b Find P(p,q,h) in terms of a,b,s,L

$${A}\:{rod}\:{of}\:{length}\:{L}\:{in}\:{a}\:{cubic}\:{room}, \\ $$$${side}\:{s},\:\:{one} \\ $$$${end}\:{of}\:{rod}\:{at}\:{O}\:{while}\:{to}\:{end}\:{P}\:{are}\: \\ $$$${attached}\:{two}\:{strings}\:{PA}={a},\: \\ $$$${PB}={b} \\ $$$${Find}\:{P}\left({p},{q},{h}\right)\:{in}\:{terms}\:{of}\:{a},{b},{s},{L} \\ $$$$ \\ $$

Commented by ajfour last updated on 03/Apr/26

Commented by ajfour last updated on 03/Apr/26

Commented by ajfour last updated on 05/Apr/26

Answered by TonyCWX last updated on 04/Apr/26

p^2 + q^2 + h^2 = L^2 ... (1) (s−p)^2 + (0−q)^2 + (s−h)^2 = a^2 ... (2) (0−p)^2 + (s−q)^2 + (s−h)^2 = b^2 ... (3) (2) − (3): (s−p)^2 − (s−q)^2 + q^2 − p^2 = a^2 − b^2 s^2 − 2sp + p^2 − s^2 + 2sq − q^2 + q^2 − p^2 = a^2 − b^2 2s(q − p) = a^2 − b^2 q − p = ((a^2 + b^2 )/(2s)) From (3), p^2 + s^2 − 2sq + q^2 + s^2 − 2sh + h^2 = b^2 2s^2 − 2sq − 2sh + L^2 = b^2 2s^2 − 2s(q + h) = b^2 − L^2 −2s(q + h) = b^2 − L^2 − 2s^2 q = ((2s^2 + L^2 − b^2 )/(2s)) − h p = ((2s^2 + L^2 − a^2 − 2b^2 )/(2s)) − h q = ξ − h determinant (((ξ = ((2s^2 + L^2 − b^2 )/(2s))))) p^2 + (ξ − h)^2 + h^2 = L^2 2h^2 − 2ξh + p^2 + ξ^2 − L^2 = 0 h = ((2ξ ± (√(8L^2 − 8p^2 − 4ξ^2 )))/4) = ((ξ ± (√(2L^2 − 2p^2 − ξ^2 )))/2) h > 0 ⇒ h = ((ξ + (√(2L^2 − 2p^2 − ξ^2 )))/2) h = ((((2s^2 + L^2 − b^2 )/(2s)) + (√(2L^2 − 2(((2s^2 + L^2 − a^2 − 2b^2 )/(2s)))^2 − (((2s^2 + L^2 − b^2 )/(2s)))^2 )))/2) h = ((((2s^2 + L^2 − b^2 )/(2s)) + (√( ((8L^2 s^2 − 2(2s^2 +L^2 − a^2 − 2b^2 )^2 − (2s^2 + L^2 − b^2 )^2 )/(4s^2 )))))/2) h = ((2s^2 + L^2 − b^2 + (√(8L^2 s^2 − 2(2s^2 + L^2 − a^2 − 2b^2 )^2 + (2s^2 + L^2 − b^2 )^2 )))/(4s)) p = ((2s^2 + L^2 − 2a^2 − 3b^2 − (√(8L^2 s^2 − 2(2s^2 + L^2 − a^2 − 2b^2 )^2 + (2s^2 + L^2 − b^2 )^2 )))/(4s)) q = ((2s^2 + L^2 − b^2 − (√(8L^2 s^2 − 2(2s^2 + L^2 − a^2 − 2b^2 )^2 + (2s^2 + L^2 − b^2 )^2 )))/(4s)) determinant (((P(p, q, h) = (((2s^2 + L^2 − 2a^2 − 3b^2 − (√(8L^2 s^2 − 2(2s^2 + L^2 − a^2 − 2b^2 )^2 + (2s^2 + L^2 − b^2 )^2 )))/(4s)), ((2s^2 + L^2 − b^2 − (√(8L^2 s^2 − 2(2s^2 + L^2 − a^2 − 2b^2 )^2 + (2s^2 + L^2 − b^2 )^2 )))/(4s)), ((2s^2 + L^2 − b^2 + (√(8L^2 s^2 − 2(2s^2 + L^2 − a^2 − 2b^2 )^2 + (2s^2 + L^2 − b^2 )^2 )))/(4s))))))

$${p}^{\mathrm{2}} \:+\:{q}^{\mathrm{2}} \:+\:{h}^{\mathrm{2}} \:=\:{L}^{\mathrm{2}} \:…\:\left(\mathrm{1}\right) \\ $$$$\left({s}−{p}\right)^{\mathrm{2}} \:+\:\left(\mathrm{0}−{q}\right)^{\mathrm{2}} \:+\:\left({s}−{h}\right)^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:…\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{0}−{p}\right)^{\mathrm{2}} \:+\:\left({s}−{q}\right)^{\mathrm{2}} \:+\:\left({s}−{h}\right)^{\mathrm{2}} \:=\:{b}^{\mathrm{2}} \:…\:\left(\mathrm{3}\right) \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:−\:\left(\mathrm{3}\right): \\ $$$$\left({s}−{p}\right)^{\mathrm{2}} \:−\:\left({s}−{q}\right)^{\mathrm{2}} \:\:+\:{q}^{\mathrm{2}} −\:{p}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \\ $$$${s}^{\mathrm{2}} \:−\:\mathrm{2}{sp}\:+\:{p}^{\mathrm{2}} \:−\:{s}^{\mathrm{2}} \:+\:\mathrm{2}{sq}\:−\:{q}^{\mathrm{2}} \:+\:{q}^{\mathrm{2}} \:−\:{p}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \\ $$$$\mathrm{2}{s}\left({q}\:−\:{p}\right)\:=\:{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \\ $$$${q}\:−\:{p}\:=\:\frac{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }{\mathrm{2}{s}} \\ $$$$ \\ $$$$\mathrm{From}\:\left(\mathrm{3}\right), \\ $$$${p}^{\mathrm{2}} \:+\:{s}^{\mathrm{2}} \:−\:\mathrm{2}{sq}\:+\:{q}^{\mathrm{2}} \:+\:{s}^{\mathrm{2}} \:−\:\mathrm{2}{sh}\:+\:{h}^{\mathrm{2}} \:=\:{b}^{\mathrm{2}} \\ $$$$\mathrm{2}{s}^{\mathrm{2}} \:−\:\mathrm{2}{sq}\:−\:\mathrm{2}{sh}\:+\:{L}^{\mathrm{2}} \:=\:{b}^{\mathrm{2}} \\ $$$$\mathrm{2}{s}^{\mathrm{2}} \:−\:\mathrm{2}{s}\left({q}\:+\:{h}\right)\:=\:{b}^{\mathrm{2}} \:−\:{L}^{\mathrm{2}} \\ $$$$−\mathrm{2}{s}\left({q}\:+\:{h}\right)\:=\:{b}^{\mathrm{2}} \:−\:{L}^{\mathrm{2}} \:−\:\mathrm{2}{s}^{\mathrm{2}} \\ $$$${q}\:=\:\frac{\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }{\mathrm{2}{s}}\:−\:{h} \\ $$$${p}\:=\:\frac{\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{a}^{\mathrm{2}} \:−\:\mathrm{2}{b}^{\mathrm{2}} }{\mathrm{2}{s}}\:−\:{h} \\ $$$${q}\:=\:\xi\:−\:{h}\:\begin{array}{|c|}{\xi\:=\:\frac{\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }{\mathrm{2}{s}}}\\\hline\end{array} \\ $$$$ \\ $$$${p}^{\mathrm{2}} \:+\:\left(\xi\:−\:{h}\right)^{\mathrm{2}} \:+\:{h}^{\mathrm{2}} \:=\:{L}^{\mathrm{2}} \\ $$$$\mathrm{2}{h}^{\mathrm{2}} \:−\:\mathrm{2}\xi{h}\:+\:{p}^{\mathrm{2}} \:+\:\xi^{\mathrm{2}} \:−\:{L}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$${h}\:=\:\frac{\mathrm{2}\xi\:\pm\:\sqrt{\mathrm{8}{L}^{\mathrm{2}} \:−\:\mathrm{8}{p}^{\mathrm{2}} \:−\:\mathrm{4}\xi^{\mathrm{2}} }}{\mathrm{4}}\:=\:\frac{\xi\:\pm\:\sqrt{\mathrm{2}{L}^{\mathrm{2}} \:−\:\mathrm{2}{p}^{\mathrm{2}} \:−\:\xi^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${h}\:>\:\mathrm{0}\:\Rightarrow\:{h}\:=\:\frac{\xi\:+\:\sqrt{\mathrm{2}{L}^{\mathrm{2}} \:−\:\mathrm{2}{p}^{\mathrm{2}} \:−\:\xi^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${h}\:=\:\frac{\frac{\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }{\mathrm{2}{s}}\:+\:\sqrt{\mathrm{2}{L}^{\mathrm{2}} \:−\:\mathrm{2}\left(\frac{\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{a}^{\mathrm{2}} \:−\:\mathrm{2}{b}^{\mathrm{2}} }{\mathrm{2}{s}}\right)^{\mathrm{2}} \:−\:\left(\frac{\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }{\mathrm{2}{s}}\right)^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${h}\:=\:\frac{\frac{\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }{\mathrm{2}{s}}\:+\:\sqrt{\:\frac{\mathrm{8}{L}^{\mathrm{2}} {s}^{\mathrm{2}} \:−\:\mathrm{2}\left(\mathrm{2}{s}^{\mathrm{2}} \:+{L}^{\mathrm{2}} \:−\:{a}^{\mathrm{2}} \:−\:\mathrm{2}{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\:\left(\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}{s}^{\mathrm{2}} }}}{\mathrm{2}} \\ $$$${h}\:=\:\frac{\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \:+\:\sqrt{\mathrm{8}{L}^{\mathrm{2}} {s}^{\mathrm{2}} \:−\:\mathrm{2}\left(\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{a}^{\mathrm{2}} \:−\:\mathrm{2}{b}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\left(\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{4}{s}} \\ $$$$ \\ $$$${p}\:=\:\frac{\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:\mathrm{2}{a}^{\mathrm{2}} \:−\:\mathrm{3}{b}^{\mathrm{2}} \:−\:\sqrt{\mathrm{8}{L}^{\mathrm{2}} {s}^{\mathrm{2}} \:−\:\mathrm{2}\left(\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{a}^{\mathrm{2}} \:−\:\mathrm{2}{b}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\left(\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{4}{s}} \\ $$$${q}\:=\:\frac{\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} −\:\sqrt{\mathrm{8}{L}^{\mathrm{2}} {s}^{\mathrm{2}} \:−\:\mathrm{2}\left(\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{a}^{\mathrm{2}} \:−\:\mathrm{2}{b}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\left(\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{4}{s}} \\ $$$$ \\ $$$$\begin{array}{|c|}{{P}\left({p},\:{q},\:{h}\right)\:=\:\left(\frac{\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:\mathrm{2}{a}^{\mathrm{2}} \:−\:\mathrm{3}{b}^{\mathrm{2}} \:−\:\sqrt{\mathrm{8}{L}^{\mathrm{2}} {s}^{\mathrm{2}} \:−\:\mathrm{2}\left(\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{a}^{\mathrm{2}} \:−\:\mathrm{2}{b}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\left(\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{4}{s}},\:\frac{\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \:−\:\sqrt{\mathrm{8}{L}^{\mathrm{2}} {s}^{\mathrm{2}} \:−\:\mathrm{2}\left(\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{a}^{\mathrm{2}} \:−\:\mathrm{2}{b}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\left(\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{4}{s}},\:\frac{\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \:+\:\sqrt{\mathrm{8}{L}^{\mathrm{2}} {s}^{\mathrm{2}} \:−\:\mathrm{2}\left(\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{a}^{\mathrm{2}} \:−\:\mathrm{2}{b}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\left(\mathrm{2}{s}^{\mathrm{2}} \:+\:{L}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{4}{s}}\right)}\\\hline\end{array} \\ $$

Commented by TonyCWX last updated on 03/Apr/26

I apologise for my silly mistake. Will edit later.

$$\mathrm{I}\:\mathrm{apologise}\:\mathrm{for}\:\mathrm{my}\:\mathrm{silly}\:\mathrm{mistake}. \\ $$$$\mathrm{Will}\:\mathrm{edit}\:\mathrm{later}. \\ $$

Commented by TonyCWX last updated on 04/Apr/26

I have finished editting. All corrections are shown in blue colour. Please help me recheck.

$$\mathrm{I}\:\mathrm{have}\:\mathrm{finished}\:\mathrm{editting}. \\ $$$$\mathrm{All}\:\mathrm{corrections}\:\mathrm{are}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{blue}\:\mathrm{colour}. \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{recheck}. \\ $$

Answered by fantastic2 last updated on 03/Apr/26

coordinate configuration (x,y,z) q^2 +p^2 +h^2 =L^2 ...i A(0,s,s) & P(q,p,h) q^2 +(s−p)^2 +(s−h)^2 =a^2 ...ii B(s,0,s) &P(q,p,h) (s−q)^2 +p^2 +(s−h)^2 =b^2 ... iii expanding ii a^2 =p^2 +q^2 +h^2 +2s^2 −2s(p+h) ⇒p+h=((L^2 +2s^2 −a^2 )/(2s)) ...iv similarly by expanding iii ⇒q+h=((L^2 +2s^2 −b^2 )/(2s)) ...v now iv−v⇒p−q=((b^2 −a^2 )/(2s)) ... vi from iv ⇒p= −h ( =((L^2 +2s^2 −a^2 )/(2s))) from v⇒q= −h( =((L^2 +2s^2 −b^2 )/(2s))) now it becomes ( −h)^2 +( −h)^2 +h^2 =L^2 ⇒3h^2 −2h( + )+(^2 +^2 −L^2 )=0 h=((2( + )±(√(4( + )^2 −12(^2 +^2 −L^2 ))))/6)>0 h=((2( + )+(√(4( + )^2 −12(^2 +^2 −L^2 ))))/6) p= −((2( + )±(√(4( + )^2 −12(^2 +^2 −L^2 ))))/6) q= −((2( + )±(√(4( + )^2 −12(^2 +^2 −L^2 ))))/6) ∴P( −((2( + )±(√(4( + )^2 −12(^2 +^2 −L^2 ))))/6), −((2( + )±(√(4( + )^2 −12(^2 +^2 −L^2 ))))/6),((2( + )+(√(4( + )^2 −12(^2 +^2 −L^2 ))))/6)) where =((L^2 +2s^2 −a^2 )/(2s)) & =((L^2 +2s^2 −b^2 )/(2s))

$${coordinate}\:{configuration}\:\left({x},{y},{z}\right) \\ $$$${q}^{\mathrm{2}} +{p}^{\mathrm{2}} +{h}^{\mathrm{2}} ={L}^{\mathrm{2}} \:\:…{i} \\ $$$${A}\left(\mathrm{0},\boldsymbol{{s}},\boldsymbol{{s}}\right)\:\&\:{P}\left({q},{p},{h}\right) \\ $$$${q}^{\mathrm{2}} +\left({s}−{p}\right)^{\mathrm{2}} +\left({s}−{h}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \:…{ii} \\ $$$${B}\left(\boldsymbol{{s}},\mathrm{0},\boldsymbol{{s}}\right)\:\&{P}\left({q},{p},{h}\right) \\ $$$$\left({s}−{q}\right)^{\mathrm{2}} +{p}^{\mathrm{2}} +\left({s}−{h}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} \:…\:{iii} \\ $$$${expanding}\:{ii} \\ $$$${a}^{\mathrm{2}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{h}^{\mathrm{2}} +\mathrm{2}{s}^{\mathrm{2}} −\mathrm{2}{s}\left({p}+{h}\right) \\ $$$$\Rightarrow{p}+{h}=\frac{{L}^{\mathrm{2}} +\mathrm{2}{s}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{s}}\:…{iv} \\ $$$${similarly}\:{by}\:{expanding}\:{iii} \\ $$$$\Rightarrow{q}+{h}=\frac{{L}^{\mathrm{2}} +\mathrm{2}{s}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{s}}\:…{v} \\ $$$${now}\:{iv}−{v}\Rightarrow{p}−{q}=\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{s}}\:…\:{vi} \\ $$$${from}\:{iv}\:\Rightarrow{p}=\:−{h}\:\left(\:=\frac{{L}^{\mathrm{2}} +\mathrm{2}{s}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{s}}\right) \\ $$$${from}\:{v}\Rightarrow{q}=\:−{h}\left(\:=\frac{{L}^{\mathrm{2}} +\mathrm{2}{s}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{s}}\right) \\ $$$${now}\:{it}\:{becomes} \\ $$$$\left(\:−{h}\right)^{\mathrm{2}} +\left(\:−{h}\right)^{\mathrm{2}} +{h}^{\mathrm{2}} ={L}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}{h}^{\mathrm{2}} −\mathrm{2}{h}\left(\:+\:\right)+\left(\:^{\mathrm{2}} +\:^{\mathrm{2}} −{L}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${h}=\frac{\mathrm{2}\left(\:+\:\right)\pm\sqrt{\mathrm{4}\left(\:+\:\right)^{\mathrm{2}} −\mathrm{12}\left(\:^{\mathrm{2}} +\:^{\mathrm{2}} −{L}^{\mathrm{2}} \right)}}{\mathrm{6}}>\mathrm{0} \\ $$$${h}=\frac{\mathrm{2}\left(\:+\:\right)+\sqrt{\mathrm{4}\left(\:+\:\right)^{\mathrm{2}} −\mathrm{12}\left(\:^{\mathrm{2}} +\:^{\mathrm{2}} −{L}^{\mathrm{2}} \right)}}{\mathrm{6}} \\ $$$${p}=\:−\frac{\mathrm{2}\left(\:+\:\right)\pm\sqrt{\mathrm{4}\left(\:+\:\right)^{\mathrm{2}} −\mathrm{12}\left(\:^{\mathrm{2}} +\:^{\mathrm{2}} −{L}^{\mathrm{2}} \right)}}{\mathrm{6}} \\ $$$${q}=\:−\frac{\mathrm{2}\left(\:+\:\right)\pm\sqrt{\mathrm{4}\left(\:+\:\right)^{\mathrm{2}} −\mathrm{12}\left(\:^{\mathrm{2}} +\:^{\mathrm{2}} −{L}^{\mathrm{2}} \right)}}{\mathrm{6}} \\ $$$$\therefore{P}\left(\:−\frac{\mathrm{2}\left(\:+\:\right)\pm\sqrt{\mathrm{4}\left(\:+\:\right)^{\mathrm{2}} −\mathrm{12}\left(\:^{\mathrm{2}} +\:^{\mathrm{2}} −{L}^{\mathrm{2}} \right)}}{\mathrm{6}},\:−\frac{\mathrm{2}\left(\:+\:\right)\pm\sqrt{\mathrm{4}\left(\:+\:\right)^{\mathrm{2}} −\mathrm{12}\left(\:^{\mathrm{2}} +\:^{\mathrm{2}} −{L}^{\mathrm{2}} \right)}}{\mathrm{6}},\frac{\mathrm{2}\left(\:+\:\right)+\sqrt{\mathrm{4}\left(\:+\:\right)^{\mathrm{2}} −\mathrm{12}\left(\:^{\mathrm{2}} +\:^{\mathrm{2}} −{L}^{\mathrm{2}} \right)}}{\mathrm{6}}\right) \\ $$$${where}\:\:=\frac{{L}^{\mathrm{2}} +\mathrm{2}{s}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{s}}\:\&\:\:=\frac{{L}^{\mathrm{2}} +\mathrm{2}{s}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{s}} \\ $$

Commented by fantastic2 last updated on 03/Apr/26

notice how we dont need p−q=((b^2 −a^2 )/(2s))

$${notice}\:{how}\:{we}\:{dont}\:{need}\:\:{p}−{q}=\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{s}} \\ $$

Answered by mr W last updated on 03/Apr/26

p^2 +q^2 +h^2 =l^2 ...(i) (s−p)^2 +(s−h)^2 +q^2 =a^2 2s^2 −2sp−2sh+p^2 +q^2 +h^2 =a^2 ⇒p=s+((l^2 −a^2 )/(2s))−h=A−h ...(ii) (s−q)^2 +(s−h)^2 +p^2 =b^2 2s^2 −2sq−2sh+p^2 +q^2 +h^2 =b^2 ⇒q=s+((l^2 −b^2 )/(2s))−h=B−h ...(iii) with A=s+((l^2 −a^2 )/(2s)), B=s+((l^2 −b^2 )/(2s)) (A−h)^2 +(B−h)^2 +h^2 =l^2 3h^2 −2(A+B)h+A^2 +B^2 −l^2 =0 h=(1/3)[A+B±(√((A+B)^2 −3A^2 −3B^2 +3l^2 ))] ⇒h=(1/3)[A+B±(√(3l^2 −2(A^2 +B^2 −AB)))]

$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{h}^{\mathrm{2}} ={l}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$$\left({s}−{p}\right)^{\mathrm{2}} +\left({s}−{h}\right)^{\mathrm{2}} +{q}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\mathrm{2}{s}^{\mathrm{2}} −\mathrm{2}{sp}−\mathrm{2}{sh}+{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{h}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\Rightarrow{p}={s}+\frac{{l}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{s}}−{h}={A}−{h}\:\:\:…\left({ii}\right) \\ $$$$\left({s}−{q}\right)^{\mathrm{2}} +\left({s}−{h}\right)^{\mathrm{2}} +{p}^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\mathrm{2}{s}^{\mathrm{2}} −\mathrm{2}{sq}−\mathrm{2}{sh}+{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{h}^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\Rightarrow{q}={s}+\frac{{l}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{s}}−{h}={B}−{h}\:\:\:…\left({iii}\right) \\ $$$${with}\:{A}={s}+\frac{{l}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{s}},\:{B}={s}+\frac{{l}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{s}} \\ $$$$\left({A}−{h}\right)^{\mathrm{2}} +\left({B}−{h}\right)^{\mathrm{2}} +{h}^{\mathrm{2}} ={l}^{\mathrm{2}} \\ $$$$\mathrm{3}{h}^{\mathrm{2}} −\mathrm{2}\left({A}+{B}\right){h}+{A}^{\mathrm{2}} +{B}^{\mathrm{2}} −{l}^{\mathrm{2}} =\mathrm{0} \\ $$$${h}=\frac{\mathrm{1}}{\mathrm{3}}\left[{A}+{B}\pm\sqrt{\left({A}+{B}\right)^{\mathrm{2}} −\mathrm{3}{A}^{\mathrm{2}} −\mathrm{3}{B}^{\mathrm{2}} +\mathrm{3}{l}^{\mathrm{2}} }\right] \\ $$$$\Rightarrow{h}=\frac{\mathrm{1}}{\mathrm{3}}\left[{A}+{B}\pm\sqrt{\mathrm{3}{l}^{\mathrm{2}} −\mathrm{2}\left({A}^{\mathrm{2}} +{B}^{\mathrm{2}} −{AB}\right)}\right] \\ $$

Commented by ajfour last updated on 03/Apr/26

$${yes}\:{sir},\:{i}\:{did}\:{same}\:{way}. \\ $$$${Now}\:{we}\:{can}\:{proceed}\:{to}\:{find} \\ $$$${T}_{{A}} /{mg}\:\:\:{and}\:{T}_{{B}} /{mg}. \\ $$

Commented by mr W last updated on 04/Apr/26

Commented by mr W last updated on 04/Apr/26

Commented by mr W last updated on 05/Apr/26

((GT)/(s−GT))=(q/p) ⇒GT=((qs)/(p+q)) CG=(((√2)qs)/(p+q)) ((OG)/(GT))=((√(p^2 +q^2 ))/q) ⇒OG=((s(√(p^2 +q^2 )))/(p+q)) QG=(((√(p^2 +q^2 ))s)/(p+q))−(√(p^2 +q^2 ))=(√(p^2 +q^2 ))((s/(p+q))−1) tan φ=((QG)/(s−h))=((√(p^2 +q^2 ))/(s−h))((s/(p+q))−1) tan θ=(h/( (√(p^2 +q^2 )))) ((sin (θ+(π/2)−ϕ))/(sin ϕ))=((sin ((π/2)−θ−φ))/(sin φ)) ((cos (ϕ−θ))/(sin ϕ))=((cos (θ+φ))/(sin φ)) (1/(tan ϕ))=(1/(tan φ))−2 tan θ ⇒tan ϕ=(1/((1/(tan φ))−2 tan θ))=((√(p^2 +q^2 ))/((((p+q)(s−h))/(s−p−q))−2h)) (T/(mg))=((sin ϕ)/(sin (φ+ϕ)))=(1/(((sin φ)/(tan ϕ))+cos φ)) ⇒T=((mg)/((((tan φ)/(tan ϕ))+1)cos φ)) AH=CG=(((√2)qs)/(p+q)) cos γ=((a^2 +2s^2 −b^2 )/(2(√2)as)) cos δ=((b^2 +2s^2 −a^2 )/(2(√2)bs)) ((sin (γ+α))/(sin α))=(a/(AH))=((a(p+q))/( (√2)qs)) ((sin γ)/(tan α))+cos γ=((a(p+q))/( (√2)qs)) ⇒tan α=((sin γ)/(((a(p+q))/( (√2)qs))−cos γ)) similarly ⇒tan β=((sin δ)/(((b(p+q))/( (√2)ps))−cos δ)) (T_1 /(sin β))=(T_2 /(sin α))=(T/(sin (α+β)))=((mg)/(sin (α+β)(((tan φ)/(tan ϕ))+1)cos φ)) ⇒T_1 =((mg sin β)/(sin (α+β)(((tan φ)/(tan ϕ))+1)cos φ)) ⇒T_2 =((mg sin α)/(sin (α+β)(((tan φ)/(tan ϕ))+1)cos φ))

$$\frac{{GT}}{{s}−{GT}}=\frac{{q}}{{p}} \\ $$$$\Rightarrow{GT}=\frac{{qs}}{{p}+{q}} \\ $$$${CG}=\frac{\sqrt{\mathrm{2}}{qs}}{{p}+{q}} \\ $$$$\frac{{OG}}{{GT}}=\frac{\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}{{q}} \\ $$$$\Rightarrow{OG}=\frac{{s}\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}{{p}+{q}} \\ $$$${QG}=\frac{\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }{s}}{{p}+{q}}−\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }\left(\frac{{s}}{{p}+{q}}−\mathrm{1}\right) \\ $$$$\mathrm{tan}\:\phi=\frac{{QG}}{{s}−{h}}=\frac{\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}{{s}−{h}}\left(\frac{{s}}{{p}+{q}}−\mathrm{1}\right) \\ $$$$\mathrm{tan}\:\theta=\frac{{h}}{\:\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{2}}−\varphi\right)}{\mathrm{sin}\:\varphi}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta−\phi\right)}{\mathrm{sin}\:\phi} \\ $$$$\frac{\mathrm{cos}\:\left(\varphi−\theta\right)}{\mathrm{sin}\:\varphi}=\frac{\mathrm{cos}\:\left(\theta+\phi\right)}{\mathrm{sin}\:\phi} \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\varphi}=\frac{\mathrm{1}}{\mathrm{tan}\:\phi}−\mathrm{2}\:\mathrm{tan}\:\theta \\ $$$$\Rightarrow\mathrm{tan}\:\varphi=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{tan}\:\phi}−\mathrm{2}\:\mathrm{tan}\:\theta}=\frac{\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}{\frac{\left({p}+{q}\right)\left({s}−{h}\right)}{{s}−{p}−{q}}−\mathrm{2}{h}} \\ $$$$\frac{{T}}{{mg}}=\frac{\mathrm{sin}\:\varphi}{\mathrm{sin}\:\left(\phi+\varphi\right)}=\frac{\mathrm{1}}{\frac{\mathrm{sin}\:\phi}{\mathrm{tan}\:\varphi}+\mathrm{cos}\:\phi} \\ $$$$\Rightarrow{T}=\frac{{mg}}{\left(\frac{\mathrm{tan}\:\phi}{\mathrm{tan}\:\varphi}+\mathrm{1}\right)\mathrm{cos}\:\phi} \\ $$$${AH}={CG}=\frac{\sqrt{\mathrm{2}}{qs}}{{p}+{q}} \\ $$$$\mathrm{cos}\:\gamma=\frac{{a}^{\mathrm{2}} +\mathrm{2}{s}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}{as}} \\ $$$$\mathrm{cos}\:\delta=\frac{{b}^{\mathrm{2}} +\mathrm{2}{s}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}{bs}} \\ $$$$\frac{\mathrm{sin}\:\left(\gamma+\alpha\right)}{\mathrm{sin}\:\alpha}=\frac{{a}}{{AH}}=\frac{{a}\left({p}+{q}\right)}{\:\sqrt{\mathrm{2}}{qs}} \\ $$$$\frac{\mathrm{sin}\:\gamma}{\mathrm{tan}\:\alpha}+\mathrm{cos}\:\gamma=\frac{{a}\left({p}+{q}\right)}{\:\sqrt{\mathrm{2}}{qs}} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{sin}\:\gamma}{\frac{{a}\left({p}+{q}\right)}{\:\sqrt{\mathrm{2}}{qs}}−\mathrm{cos}\:\gamma} \\ $$$${similarly} \\ $$$$\Rightarrow\mathrm{tan}\:\beta=\frac{\mathrm{sin}\:\delta}{\frac{{b}\left({p}+{q}\right)}{\:\sqrt{\mathrm{2}}{ps}}−\mathrm{cos}\:\delta} \\ $$$$\frac{{T}_{\mathrm{1}} }{\mathrm{sin}\:\beta}=\frac{{T}_{\mathrm{2}} }{\mathrm{sin}\:\alpha}=\frac{{T}}{\mathrm{sin}\:\left(\alpha+\beta\right)}=\frac{{mg}}{\mathrm{sin}\:\left(\alpha+\beta\right)\left(\frac{\mathrm{tan}\:\phi}{\mathrm{tan}\:\varphi}+\mathrm{1}\right)\mathrm{cos}\:\phi} \\ $$$$\Rightarrow{T}_{\mathrm{1}} =\frac{{mg}\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\left(\alpha+\beta\right)\left(\frac{\mathrm{tan}\:\phi}{\mathrm{tan}\:\varphi}+\mathrm{1}\right)\mathrm{cos}\:\phi} \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\frac{{mg}\:\mathrm{sin}\:\alpha}{\mathrm{sin}\:\left(\alpha+\beta\right)\left(\frac{\mathrm{tan}\:\phi}{\mathrm{tan}\:\varphi}+\mathrm{1}\right)\mathrm{cos}\:\phi} \\ $$

Commented by ajfour last updated on 05/Apr/26

$${Great}\:{work}\:{Sir}.\:{I}\:{shall}\:{try}\:{to}\:{follow}.. \\ $$

Commented by mr W last updated on 07/Apr/26

Methode II (vector) T_1 =(T_1 /a)(s−p, −q, s−h) T_2 =(T_2 /b)(−p, s−q, s−h) N=(N/l)(p, q, h) ΣF_(px) =0: ((T_1 (s−p))/a)+((T_2 (−p))/b)+((Np)/l)=0 ΣF_(py) =0: ((T_1 (−q))/a)+((T_2 (s−q))/b)+((Nq)/l)=0 ΣF_(pz) =0: ((T_1 (s−h))/a)+((T_2 (s−h))/b)+((Nh)/l)−((mg)/2)=0 ⇒(N/l)=((mg)/(2h))−((T_1 (s−h))/(ah))−((T_2 (s−h))/(bh)) ((T_1 (s−p))/a)+((T_2 (−p))/b)+((mgp)/(2h))−((T_1 (s−h)p)/(ah))−((T_2 (s−h)p)/(bh))=0 ⇒((T_1 s(h−p))/(ap))−((T_2 s)/b)+((mg)/2)=0 ...(i) ((T_1 (−q))/a)+((T_2 (s−q))/b)+((mgq)/(2h))−((T_1 (s−h)q)/(ah))−((T_2 (s−h)q)/(bh))=0 ⇒−((T_1 s)/a)+((T_2 s(h−q))/(bq))+((mg)/2)=0 ...(ii) ⇒T_1 =((mgap)/(2s(p+q−h))) ⇒T_2 =((mgbq)/(2s(p+q−h)))

$${Methode}\:{II}\:\left({vector}\right) \\ $$$$\boldsymbol{{T}}_{\mathrm{1}} =\frac{{T}_{\mathrm{1}} }{{a}}\left({s}−{p},\:−{q},\:{s}−{h}\right) \\ $$$$\boldsymbol{{T}}_{\mathrm{2}} =\frac{{T}_{\mathrm{2}} }{{b}}\left(−{p},\:{s}−{q},\:{s}−{h}\right) \\ $$$$\boldsymbol{{N}}=\frac{{N}}{{l}}\left({p},\:{q},\:{h}\right) \\ $$$$\Sigma{F}_{{px}} =\mathrm{0}:\:\frac{{T}_{\mathrm{1}} \left({s}−{p}\right)}{{a}}+\frac{{T}_{\mathrm{2}} \left(−{p}\right)}{{b}}+\frac{{Np}}{{l}}=\mathrm{0} \\ $$$$\Sigma{F}_{{py}} =\mathrm{0}:\:\frac{{T}_{\mathrm{1}} \left(−{q}\right)}{{a}}+\frac{{T}_{\mathrm{2}} \left({s}−{q}\right)}{{b}}+\frac{{Nq}}{{l}}=\mathrm{0} \\ $$$$\Sigma{F}_{{pz}} =\mathrm{0}:\:\frac{{T}_{\mathrm{1}} \left({s}−{h}\right)}{{a}}+\frac{{T}_{\mathrm{2}} \left({s}−{h}\right)}{{b}}+\frac{{Nh}}{{l}}−\frac{{mg}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{N}}{{l}}=\frac{{mg}}{\mathrm{2}{h}}−\frac{{T}_{\mathrm{1}} \left({s}−{h}\right)}{{ah}}−\frac{{T}_{\mathrm{2}} \left({s}−{h}\right)}{{bh}} \\ $$$$\frac{{T}_{\mathrm{1}} \left({s}−{p}\right)}{{a}}+\frac{{T}_{\mathrm{2}} \left(−{p}\right)}{{b}}+\frac{{mgp}}{\mathrm{2}{h}}−\frac{{T}_{\mathrm{1}} \left({s}−{h}\right){p}}{{ah}}−\frac{{T}_{\mathrm{2}} \left({s}−{h}\right){p}}{{bh}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{T}_{\mathrm{1}} {s}\left({h}−{p}\right)}{{ap}}−\frac{{T}_{\mathrm{2}} {s}}{{b}}+\frac{{mg}}{\mathrm{2}}=\mathrm{0}\:\:\:…\left({i}\right) \\ $$$$\frac{{T}_{\mathrm{1}} \left(−{q}\right)}{{a}}+\frac{{T}_{\mathrm{2}} \left({s}−{q}\right)}{{b}}+\frac{{mgq}}{\mathrm{2}{h}}−\frac{{T}_{\mathrm{1}} \left({s}−{h}\right){q}}{{ah}}−\frac{{T}_{\mathrm{2}} \left({s}−{h}\right){q}}{{bh}}=\mathrm{0} \\ $$$$\Rightarrow−\frac{{T}_{\mathrm{1}} {s}}{{a}}+\frac{{T}_{\mathrm{2}} {s}\left({h}−{q}\right)}{{bq}}+\frac{{mg}}{\mathrm{2}}=\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow{T}_{\mathrm{1}} =\frac{{mgap}}{\mathrm{2}{s}\left({p}+{q}−{h}\right)} \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\frac{{mgbq}}{\mathrm{2}{s}\left({p}+{q}−{h}\right)} \\ $$

Commented by ajfour last updated on 07/Apr/26