Question Number 224229 by mr W last updated on 27/Aug/25
Commented by mr W last updated on 28/Aug/25
$${a}\:{loop}\:{of}\:{rope}\:{with}\:{total}\:{length}\:{L} \\ $$
Commented by mr W last updated on 21/Oct/25
![V, A = velocity and acceleration of wedge v, a =relative velocity and acceleration of block along the wedge let λ=(M/m) v=(dx/dt) a=(dv/dt)=v(dv/dx) MV=m(v cos α−V) MA=m(a cos α−A) ⇒A=((ma cos α)/(M+m))=((a cos α)/(λ+1)) mA sin α=mg cos α−N ⇒N=m(g cos α−A sin α) m(a−A cos α)=mg sin α−f f=μN=kxN m(a−A cos α)=mg sin α−kxm(g cos α−A sin α) a−A(cos α+kx sin α)=g(sin α−kx cos α) a−((a cos α)/(λ+1))(cos α+kx sin α)=g(sin α−kx cos α) [λ+sin α(sin α−kx cos α)]a=(λ+1)g(sin α−kx cos α) ⇒a=(((λ+1)g(sin α−kx cos α))/(λ+sin α(sin α−kx cos α))) ⇒A=((g cos α (sin α−kx cos α))/(λ+sin α(sin α−kx cos α))) v(dv/dx)=(((λ+1)g(sin α−kx cos α))/(λ+sin α(sin α−kx cos α))) vdv=(((λ+1)g(sin α−kx cos α)dx)/(λ+sin α(sin α−kx cos α))) ∫_0 ^v vdv=(((λ+1)g)/(sin α))∫_0 ^x (((sin α−kx cos α)dx)/((λ/(sin α))+(sin α−kx cos α))) ∫_0 ^v vdv=−(((λ+1)g)/(k sin α cos α))∫_0 ^x (((sin α−kx cos α)d(sin α−kx cos α))/((λ/(sin α))+(sin α−kx cos α))) (v^2 /2)=−(((λ+1)g)/(k sin α cos α))∫_(sin α) ^(sin α−kx cos α) ((udu)/((λ/(sin α))+u)) (v^2 /2)=−(((λ+1)g)/(k sin α cos α))[u−(λ/(sin α)) ln ((λ/(sin α))+u)]_(sin α) ^(sin α−kx cos α) (v^2 /2)=(((λ+1)g)/(k sin α cos α))[kx cos α+(λ/(sin α)) ln (((λ/(sin α))+sin α−kx cos α)/((λ/(sin α))+sin α))] (v^2 /2)=(((λ+1)g)/(k sin^2 α cos α))[kx sin α cos α+λ ln (1−((kx sin α cos α)/(λ+sin^2 α)))] ⇒v=(√((2(λ+1)g)/(k sin^2 α cos α))) (√(kx sin α cos α+λ ln (1−((kx sin α cos α)/(λ+sin^2 α))))) (dx/dt)=(√((2(λ+1)g)/(k sin^2 α cos α))) (√(kx sin α cos α+λ ln (1−((kx sin α cos α)/(λ+sin^2 α))))) ⇒t=sin α(√((k cos α)/(2(λ+1)g)))∫_0 ^x (dx/( (√(kx sin α cos α+λ ln (1−((kx sin α cos α)/(λ+sin^2 α)))))))](https://www.tinkutara.com/question/Q225337.png)
$${V},\:{A}\:=\:{velocity}\:{and}\:{acceleration}\:{of}\:{wedge} \\ $$$${v},\:{a}\:={relative}\:{velocity}\:{and}\:{acceleration}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{of}\:{block}\:{along}\:{the}\:{wedge} \\ $$$${let}\:\lambda=\frac{{M}}{{m}} \\ $$$${v}=\frac{{dx}}{{dt}} \\ $$$${a}=\frac{{dv}}{{dt}}={v}\frac{{dv}}{{dx}} \\ $$$${MV}={m}\left({v}\:\mathrm{cos}\:\alpha−{V}\right) \\ $$$${MA}={m}\left({a}\:\mathrm{cos}\:\alpha−{A}\right) \\ $$$$\Rightarrow{A}=\frac{{ma}\:\mathrm{cos}\:\alpha}{{M}+{m}}=\frac{{a}\:\mathrm{cos}\:\alpha}{\lambda+\mathrm{1}} \\ $$$${mA}\:\mathrm{sin}\:\alpha={mg}\:\mathrm{cos}\:\alpha−{N} \\ $$$$\Rightarrow{N}={m}\left({g}\:\mathrm{cos}\:\alpha−{A}\:\mathrm{sin}\:\alpha\right) \\ $$$${m}\left({a}−{A}\:\mathrm{cos}\:\alpha\right)={mg}\:\mathrm{sin}\:\alpha−{f} \\ $$$${f}=\mu{N}={kxN} \\ $$$${m}\left({a}−{A}\:\mathrm{cos}\:\alpha\right)={mg}\:\mathrm{sin}\:\alpha−{kxm}\left({g}\:\mathrm{cos}\:\alpha−{A}\:\mathrm{sin}\:\alpha\right) \\ $$$${a}−{A}\left(\mathrm{cos}\:\alpha+{kx}\:\mathrm{sin}\:\alpha\right)={g}\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right) \\ $$$${a}−\frac{{a}\:\mathrm{cos}\:\alpha}{\lambda+\mathrm{1}}\left(\mathrm{cos}\:\alpha+{kx}\:\mathrm{sin}\:\alpha\right)={g}\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right) \\ $$$$\left[\lambda+\mathrm{sin}\:\alpha\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)\right]{a}=\left(\lambda+\mathrm{1}\right){g}\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right) \\ $$$$\Rightarrow{a}=\frac{\left(\lambda+\mathrm{1}\right){g}\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)}{\lambda+\mathrm{sin}\:\alpha\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)} \\ $$$$\Rightarrow{A}=\frac{{g}\:\mathrm{cos}\:\alpha\:\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)}{\lambda+\mathrm{sin}\:\alpha\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)} \\ $$$${v}\frac{{dv}}{{dx}}=\frac{\left(\lambda+\mathrm{1}\right){g}\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)}{\lambda+\mathrm{sin}\:\alpha\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)} \\ $$$${vdv}=\frac{\left(\lambda+\mathrm{1}\right){g}\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right){dx}}{\lambda+\mathrm{sin}\:\alpha\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)} \\ $$$$\int_{\mathrm{0}} ^{{v}} {vdv}=\frac{\left(\lambda+\mathrm{1}\right){g}}{\mathrm{sin}\:\alpha}\int_{\mathrm{0}} ^{{x}} \frac{\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right){dx}}{\frac{\lambda}{\mathrm{sin}\:\alpha}+\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)} \\ $$$$\int_{\mathrm{0}} ^{{v}} {vdv}=−\frac{\left(\lambda+\mathrm{1}\right){g}}{{k}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}\int_{\mathrm{0}} ^{{x}} \frac{\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right){d}\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)}{\frac{\lambda}{\mathrm{sin}\:\alpha}+\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)} \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=−\frac{\left(\lambda+\mathrm{1}\right){g}}{{k}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}\int_{\mathrm{sin}\:\alpha} ^{\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha} \frac{{udu}}{\frac{\lambda}{\mathrm{sin}\:\alpha}+{u}} \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=−\frac{\left(\lambda+\mathrm{1}\right){g}}{{k}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}\left[{u}−\frac{\lambda}{\mathrm{sin}\:\alpha}\:\mathrm{ln}\:\left(\frac{\lambda}{\mathrm{sin}\:\alpha}+{u}\right)\right]_{\mathrm{sin}\:\alpha} ^{\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha} \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=\frac{\left(\lambda+\mathrm{1}\right){g}}{{k}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}\left[{kx}\:\mathrm{cos}\:\alpha+\frac{\lambda}{\mathrm{sin}\:\alpha}\:\mathrm{ln}\:\frac{\frac{\lambda}{\mathrm{sin}\:\alpha}+\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha}{\frac{\lambda}{\mathrm{sin}\:\alpha}+\mathrm{sin}\:\alpha}\right] \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=\frac{\left(\lambda+\mathrm{1}\right){g}}{{k}\:\mathrm{sin}^{\mathrm{2}} \:\alpha\:\mathrm{cos}\:\alpha}\left[{kx}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha+\lambda\:\mathrm{ln}\:\left(\mathrm{1}−\frac{{kx}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\lambda+\mathrm{sin}^{\mathrm{2}} \:\alpha}\right)\right] \\ $$$$\Rightarrow{v}=\sqrt{\frac{\mathrm{2}\left(\lambda+\mathrm{1}\right){g}}{{k}\:\mathrm{sin}^{\mathrm{2}} \:\alpha\:\mathrm{cos}\:\alpha}}\:\sqrt{{kx}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha+\lambda\:\mathrm{ln}\:\left(\mathrm{1}−\frac{{kx}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\lambda+\mathrm{sin}^{\mathrm{2}} \:\alpha}\right)} \\ $$$$\frac{{dx}}{{dt}}=\sqrt{\frac{\mathrm{2}\left(\lambda+\mathrm{1}\right){g}}{{k}\:\mathrm{sin}^{\mathrm{2}} \:\alpha\:\mathrm{cos}\:\alpha}}\:\sqrt{{kx}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha+\lambda\:\mathrm{ln}\:\left(\mathrm{1}−\frac{{kx}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\lambda+\mathrm{sin}^{\mathrm{2}} \:\alpha}\right)} \\ $$$$\Rightarrow{t}=\mathrm{sin}\:\alpha\sqrt{\frac{{k}\:\mathrm{cos}\:\alpha}{\mathrm{2}\left(\lambda+\mathrm{1}\right){g}}}\int_{\mathrm{0}} ^{{x}} \frac{{dx}}{\:\sqrt{{kx}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha+\lambda\:\mathrm{ln}\:\left(\mathrm{1}−\frac{{kx}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\lambda+\mathrm{sin}^{\mathrm{2}} \:\alpha}\right)}} \\ $$
Answered by mr W last updated on 30/Aug/25
![T=tension in rope at A and B T cos θ_1 =a_1 ρg T cos θ_2 =a_2 ρg (a_1 /(cos θ_1 ))=(a_2 /(cos θ_2 ))=(T/(ρg))=k, say sinh (b/(2a_1 ))=tan θ_1 =t_1 (b/(2a_1 ))=(b/(2k cos θ_1 ))=sinh^(−1) t_1 =ln (t_1 +(√(1+t_1 ^2 ))) (b/(2k))=λ=cos θ_1 sinh^(−1) t_1 =((ln (t_1 +(√(1+t_1 ^2 ))))/( (√(1+t_1 ^2 )))) similarly λ=((ln (t_2 +(√(1+t_2 ^2 ))))/( (√(1+t_2 ^2 )))) ⇒λ=((ln (t_1 +(√(1+t_1 ^2 ))))/( (√(1+t_1 ^2 ))))=((ln (t_2 +(√(1+t_2 ^2 ))))/( (√(1+t_2 ^2 )))) ...(i) s_1 =a_1 sinh (b/(2a_1 ))=a_1 tan θ_1 =k cos θ_1 tan θ_1 =k sin θ_1 similarly s_2 =k sin θ_2 2(s_1 +s_2 )=L sin θ_1 +sin θ_2 =(L/(2k))=((λL)/b)=μλ with μ=(L/b) (t_1 /( (√(1+t_1 ^2 ))))+(t_2 /( (√(1+t_2 ^2 ))))=((μln (t_2 +(√(1+t_2 ^2 ))))/( (√(1+t_2 ^2 )))) (t_1 /( (√(1+t_1 ^2 ))))=((μln (t_2 +(√(1+t_2 ^2 )))−t_2 )/( (√(1+t_2 ^2 )))) ⇒t_1 =(√((1/(1−(([μln (t_2 +(√(1+t_2 ^2 )))−t_2 ]^2 )/( 1+t_2 ^2 ))))−1)) ...(ii) we can solve (i) and (ii) for t_1 and t_2 . a_1 =k cos θ_1 =(b/(2ln (t_1 +(√(1+t_1 ^2 ))))) a_2 =k cos θ_2 =(b/(2ln (t_2 +(√(1+t_2 ^2 )))))](https://www.tinkutara.com/question/Q224230.png)
$${T}={tension}\:{in}\:{rope}\:{at}\:{A}\:{and}\:{B} \\ $$$${T}\:\mathrm{cos}\:\theta_{\mathrm{1}} ={a}_{\mathrm{1}} \rho{g} \\ $$$${T}\:\mathrm{cos}\:\theta_{\mathrm{2}} ={a}_{\mathrm{2}} \rho{g} \\ $$$$\frac{{a}_{\mathrm{1}} }{\mathrm{cos}\:\theta_{\mathrm{1}} }=\frac{{a}_{\mathrm{2}} }{\mathrm{cos}\:\theta_{\mathrm{2}} }=\frac{{T}}{\rho{g}}={k},\:{say} \\ $$$$\mathrm{sinh}\:\frac{{b}}{\mathrm{2}{a}_{\mathrm{1}} }=\mathrm{tan}\:\theta_{\mathrm{1}} ={t}_{\mathrm{1}} \\ $$$$\frac{{b}}{\mathrm{2}{a}_{\mathrm{1}} }=\frac{{b}}{\mathrm{2}{k}\:\mathrm{cos}\:\theta_{\mathrm{1}} }=\mathrm{sinh}^{−\mathrm{1}} \:{t}_{\mathrm{1}} =\mathrm{ln}\:\left({t}_{\mathrm{1}} +\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }\right) \\ $$$$\frac{{b}}{\mathrm{2}{k}}=\lambda=\mathrm{cos}\:\theta_{\mathrm{1}} \mathrm{sinh}^{−\mathrm{1}} \:{t}_{\mathrm{1}} =\frac{\mathrm{ln}\:\left({t}_{\mathrm{1}} +\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }} \\ $$$${similarly}\:\lambda=\frac{\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{ln}\:\left({t}_{\mathrm{1}} +\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }}=\frac{\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }}\:\:\:…\left({i}\right) \\ $$$${s}_{\mathrm{1}} ={a}_{\mathrm{1}} \mathrm{sinh}\:\frac{{b}}{\mathrm{2}{a}_{\mathrm{1}} }={a}_{\mathrm{1}} \:\mathrm{tan}\:\theta_{\mathrm{1}} ={k}\:\mathrm{cos}\:\theta_{\mathrm{1}} \:\mathrm{tan}\:\theta_{\mathrm{1}} ={k}\:\mathrm{sin}\:\theta_{\mathrm{1}} \\ $$$${similarly}\:{s}_{\mathrm{2}} ={k}\:\mathrm{sin}\:\theta_{\mathrm{2}} \\ $$$$\mathrm{2}\left({s}_{\mathrm{1}} +{s}_{\mathrm{2}} \right)={L} \\ $$$$\mathrm{sin}\:\theta_{\mathrm{1}} +\mathrm{sin}\:\theta_{\mathrm{2}} =\frac{{L}}{\mathrm{2}{k}}=\frac{\lambda{L}}{{b}}=\mu\lambda\:{with}\:\mu=\frac{{L}}{{b}} \\ $$$$\frac{{t}_{\mathrm{1}} }{\:\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }}+\frac{{t}_{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }}=\frac{\mu\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }} \\ $$$$\frac{{t}_{\mathrm{1}} }{\:\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }}=\frac{\mu\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)−{t}_{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }} \\ $$$$\Rightarrow{t}_{\mathrm{1}} =\sqrt{\frac{\mathrm{1}}{\mathrm{1}−\frac{\left[\mu\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)−{t}_{\mathrm{2}} \right]^{\mathrm{2}} }{\:\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }}−\mathrm{1}}\:\:\:…\left({ii}\right) \\ $$$${we}\:{can}\:{solve}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{for}\:{t}_{\mathrm{1}} \:{and}\:{t}_{\mathrm{2}} . \\ $$$${a}_{\mathrm{1}} ={k}\:\mathrm{cos}\:\theta_{\mathrm{1}} =\frac{{b}}{\mathrm{2ln}\:\left({t}_{\mathrm{1}} +\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }\right)} \\ $$$${a}_{\mathrm{2}} ={k}\:\mathrm{cos}\:\theta_{\mathrm{2}} =\frac{{b}}{\mathrm{2ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)} \\ $$
Commented by mr W last updated on 28/Aug/25
Commented by mr W last updated on 28/Aug/25
Commented by mr W last updated on 28/Aug/25
Commented by mr W last updated on 28/Aug/25
$${valid}\:{range}\:{of}\:{t}_{\mathrm{2}} \:\left({with}\:{t}_{\mathrm{2}} \geqslant{t}_{\mathrm{1}} \right): \\ $$$${t}_{\mathrm{1}} ={t}_{\mathrm{2}} ={t} \\ $$$$\frac{{t}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}=\frac{\mu\mathrm{ln}\:\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)−{t}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }} \\ $$$$\mathrm{2}{t}=\mu\mathrm{ln}\:\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\frac{{t}_{\mathrm{2}} }{\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)}\geqslant\frac{\mu}{\mathrm{2}} \\ $$$${s}_{\mathrm{2}} <\frac{{L}−{b}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\theta_{\mathrm{2}} <\frac{{L}−{b}}{\mathrm{2}{k}}=\left(\mu−\mathrm{1}\right)\lambda \\ $$$$\frac{{t}_{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }}<\left(\mu−\mathrm{1}\right)\lambda=\left(\mu−\mathrm{1}\right)\frac{\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }} \\ $$$$\Rightarrow\frac{{t}_{\mathrm{2}} }{\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)}<\mu−\mathrm{1} \\ $$$${i}.{e}.\:{the}\:{valid}\:{range}\:{of}\:{t}_{\mathrm{2}} \:{is} \\ $$$$\frac{\mu}{\mathrm{2}}\leqslant\frac{{t}_{\mathrm{2}} }{\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)}<\mu−\mathrm{1} \\ $$