Question Number 47957 by ajfour last updated on 17/Nov/18
Commented by ajfour last updated on 17/Nov/18
$${If}\:{the}\:{stick}\:{is}\:{on}\:{the}\:{verge}\:{of} \\ $$$${slipping}\:{on}\:{ground},\:{in}\:{the} \\ $$$${situation}\:{shown},\:{find}\:\boldsymbol{\theta},\:\boldsymbol{\varphi},\:\boldsymbol{{N}},\:\boldsymbol{{T}}. \\ $$
Commented by ajfour last updated on 17/Nov/18
$${As}\:{a}\:{special}\:{case}\:{take} \\ $$$$\:\:{m}\:=\:\mathrm{1}.\mathrm{5}\:{kg}\:,\:{h}=\:\mathrm{4}{m}\:,\:{l}=\sqrt{\mathrm{13}}{m}\:, \\ $$$$\:\:{b}\:=\:\frac{\sqrt{\mathrm{10}}}{\mathrm{3}}{m}\:,\:{a}\:=\:\frac{\sqrt{\mathrm{10}}}{\mathrm{6}}{m}\:,\:\mu\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:; \\ $$$$\:{Find}\:{at}\:{least}\:\varphi\:. \\ $$
Commented by mr W last updated on 18/Nov/18
$${thanks}\:{for}\:{creating}\:{this}\:{nice}\:{question}\centerdot \\ $$
Answered by mr W last updated on 17/Nov/18
![l cos ϕ+b sin θ=h ...(i) μN−T sin ϕ=0 ...(ii) N+T cos ϕ=mg ...(iii) Tb(cos ϕ cos θ+sin ϕ sin θ)=mg (((a+b)cos θ)/2) ...(iv) from (i): cos ϕ=((h−b sin θ)/l) ...(v) from (ii) and (iii): ⇒T=((μmg)/(μ cos ϕ+sin ϕ)) ...(vi) ⇒N=((mg sin ϕ)/(μ cos ϕ+sin ϕ)) ...(vii) 2μ(cos ϕ+sin ϕ tan θ)=((a/b)+1)(μ cos ϕ+sin ϕ) ⇒(2μ tan θ −(a/b)−1) tan ϕ=μ((a/b)−1) ⇒(2μ tan θ −(a/b)−1)^2 ((1/(cos^2 ϕ))−1)=μ^2 ((a/b)−1)^2 ⇒(2μ tan θ −(a/b)−1)^2 [(l^2 /((h−b sin θ)^2 ))−1]=μ^2 ((a/b)−1)^2 we get θ from this eqn. and then ϕ from (v) and T, N from (vi) and (vii). with h=4, l=(√(13)),a=((√(10))/6),b=((√(10))/3),μ=(1/3) ⇒((2/3) tan θ −(1/2)−1)^2 [((13×9)/((12−(√(10)) sin θ)^2 ))−1]=(1/9)((1/2)−1)^2 ⇒θ=23.9883° ⇒ϕ=cos^(−1) ((12−(√(10)) sin θ)/(3(√(13)))) =7.8854° or θ=61.5076°, ϕ=31.5200°](https://www.tinkutara.com/question/Q47987.png)
$${l}\:\mathrm{cos}\:\varphi+{b}\:\mathrm{sin}\:\theta={h}\:\:…\left({i}\right) \\ $$$$\mu{N}−{T}\:\mathrm{sin}\:\varphi=\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$${N}+{T}\:\mathrm{cos}\:\varphi={mg}\:\:\:…\left({iii}\right) \\ $$$${Tb}\left(\mathrm{cos}\:\varphi\:\mathrm{cos}\:\theta+\mathrm{sin}\:\varphi\:\mathrm{sin}\:\theta\right)={mg}\:\frac{\left({a}+{b}\right)\mathrm{cos}\:\theta}{\mathrm{2}}\:\:\:…\left({iv}\right) \\ $$$$ \\ $$$${from}\:\left({i}\right): \\ $$$$\mathrm{cos}\:\varphi=\frac{{h}−{b}\:\mathrm{sin}\:\theta}{{l}}\:\:\:…\left({v}\right) \\ $$$${from}\:\left({ii}\right)\:{and}\:\left({iii}\right): \\ $$$$\Rightarrow{T}=\frac{\mu{mg}}{\mu\:\mathrm{cos}\:\varphi+\mathrm{sin}\:\varphi}\:\:\:…\left({vi}\right) \\ $$$$\Rightarrow{N}=\frac{{mg}\:\mathrm{sin}\:\varphi}{\mu\:\mathrm{cos}\:\varphi+\mathrm{sin}\:\varphi}\:\:\:…\left({vii}\right) \\ $$$$\mathrm{2}\mu\left(\mathrm{cos}\:\varphi+\mathrm{sin}\:\varphi\:\mathrm{tan}\:\theta\right)=\left(\frac{{a}}{{b}}+\mathrm{1}\right)\left(\mu\:\mathrm{cos}\:\varphi+\mathrm{sin}\:\varphi\right) \\ $$$$\Rightarrow\left(\mathrm{2}\mu\:\mathrm{tan}\:\theta\:−\frac{{a}}{{b}}−\mathrm{1}\right)\:\mathrm{tan}\:\varphi=\mu\left(\frac{{a}}{{b}}−\mathrm{1}\right) \\ $$$$\Rightarrow\left(\mathrm{2}\mu\:\mathrm{tan}\:\theta\:−\frac{{a}}{{b}}−\mathrm{1}\right)^{\mathrm{2}} \:\left(\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\varphi}−\mathrm{1}\right)=\mu^{\mathrm{2}} \left(\frac{{a}}{{b}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{2}\mu\:\mathrm{tan}\:\theta\:−\frac{{a}}{{b}}−\mathrm{1}\right)^{\mathrm{2}} \:\left[\frac{{l}^{\mathrm{2}} }{\left({h}−{b}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }−\mathrm{1}\right]=\mu^{\mathrm{2}} \left(\frac{{a}}{{b}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$${we}\:{get}\:\theta\:{from}\:{this}\:{eqn}.\:{and}\:{then} \\ $$$$\varphi\:{from}\:\left({v}\right)\:{and}\:{T},\:{N}\:{from}\:\left({vi}\right)\:{and}\:\left({vii}\right). \\ $$$$ \\ $$$${with}\:{h}=\mathrm{4},\:{l}=\sqrt{\mathrm{13}},{a}=\frac{\sqrt{\mathrm{10}}}{\mathrm{6}},{b}=\frac{\sqrt{\mathrm{10}}}{\mathrm{3}},\mu=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\left(\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{tan}\:\theta\:−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} \:\left[\frac{\mathrm{13}×\mathrm{9}}{\left(\mathrm{12}−\sqrt{\mathrm{10}}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }−\mathrm{1}\right]=\frac{\mathrm{1}}{\mathrm{9}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\theta=\mathrm{23}.\mathrm{9883}° \\ $$$$\Rightarrow\varphi=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{12}−\sqrt{\mathrm{10}}\:\mathrm{sin}\:\theta}{\mathrm{3}\sqrt{\mathrm{13}}}\:=\mathrm{7}.\mathrm{8854}° \\ $$$${or}\:\theta=\mathrm{61}.\mathrm{5076}°,\:\varphi=\mathrm{31}.\mathrm{5200}° \\ $$
Answered by mr W last updated on 18/Nov/18
Commented by mr W last updated on 18/Nov/18
![an other way: b sin θ+l cos ϕ=h ...(i) (((a+b) cos (β+θ))/(2 sin β))=(((b−a) cos (ϕ−θ))/(2 sin ϕ)) ...(ii) (((a+b) cos (β+θ))/((b−a) sin β))=((cos θ)/(tan ϕ))+sin θ tan ϕ=((cos θ)/((((a+b) cos (β+θ))/((b−a) sin β))−sin θ)) let λ=((a+b)/(b−a)) ⇒tan ϕ=(1/((λ/μ)−(1+λ)tan θ)) ⇒ϕ=(π/2)−tan^(−1) {(λ/μ)−(1+λ)tan θ} from(i): ⇒(b/h) sin θ+(l/h) sin [tan^(−1) {(λ/μ)−(1+λ) tan θ}]=1 with h=4,l=(√(13)),a=((√(10))/6),b=((√(10))/3),μ=(1/3): ⇒λ=3 ⇒((√(10))/(12)) sin θ+((√(13))/4) sin [tan^(−1) (9−4 tan θ)]=1 ⇒θ=23.9883° or 61.5076° ⇒ϕ=90°−tan^(−1) (9−4 tan θ) =7.8855° or 31.5199°](https://www.tinkutara.com/question/Q47996.png)
$${an}\:{other}\:{way}: \\ $$$$ \\ $$$${b}\:\mathrm{sin}\:\theta+{l}\:\mathrm{cos}\:\varphi={h}\:\:\:…\left({i}\right) \\ $$$$\frac{\left({a}+{b}\right)\:\mathrm{cos}\:\left(\beta+\theta\right)}{\mathrm{2}\:\mathrm{sin}\:\beta}=\frac{\left({b}−{a}\right)\:\mathrm{cos}\:\left(\varphi−\theta\right)}{\mathrm{2}\:\mathrm{sin}\:\varphi}\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$$\frac{\left({a}+{b}\right)\:\mathrm{cos}\:\left(\beta+\theta\right)}{\left({b}−{a}\right)\:\mathrm{sin}\:\beta}=\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}+\mathrm{sin}\:\theta \\ $$$$\mathrm{tan}\:\varphi=\frac{\mathrm{cos}\:\theta}{\frac{\left({a}+{b}\right)\:\mathrm{cos}\:\left(\beta+\theta\right)}{\left({b}−{a}\right)\:\mathrm{sin}\:\beta}−\mathrm{sin}\:\theta} \\ $$$${let}\:\lambda=\frac{{a}+{b}}{{b}−{a}} \\ $$$$\Rightarrow\mathrm{tan}\:\varphi=\frac{\mathrm{1}}{\frac{\lambda}{\mu}−\left(\mathrm{1}+\lambda\right)\mathrm{tan}\:\theta} \\ $$$$\Rightarrow\varphi=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \left\{\frac{\lambda}{\mu}−\left(\mathrm{1}+\lambda\right)\mathrm{tan}\:\theta\right\} \\ $$$${from}\left({i}\right): \\ $$$$\Rightarrow\frac{{b}}{{h}}\:\mathrm{sin}\:\theta+\frac{{l}}{{h}}\:\mathrm{sin}\:\left[\mathrm{tan}^{−\mathrm{1}} \left\{\frac{\lambda}{\mu}−\left(\mathrm{1}+\lambda\right)\:\mathrm{tan}\:\theta\right\}\right]=\mathrm{1} \\ $$$${with}\:{h}=\mathrm{4},{l}=\sqrt{\mathrm{13}},{a}=\frac{\sqrt{\mathrm{10}}}{\mathrm{6}},{b}=\frac{\sqrt{\mathrm{10}}}{\mathrm{3}},\mu=\frac{\mathrm{1}}{\mathrm{3}}: \\ $$$$\Rightarrow\lambda=\mathrm{3} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{10}}}{\mathrm{12}}\:\mathrm{sin}\:\theta+\frac{\sqrt{\mathrm{13}}}{\mathrm{4}}\:\mathrm{sin}\:\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{9}−\mathrm{4}\:\mathrm{tan}\:\theta\right)\right]=\mathrm{1} \\ $$$$\Rightarrow\theta=\mathrm{23}.\mathrm{9883}°\:{or}\:\mathrm{61}.\mathrm{5076}° \\ $$$$\Rightarrow\varphi=\mathrm{90}°−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{9}−\mathrm{4}\:\mathrm{tan}\:\theta\right) \\ $$$$=\mathrm{7}.\mathrm{8855}°\:{or}\:\mathrm{31}.\mathrm{5199}° \\ $$
Commented by ajfour last updated on 18/Nov/18
$${Thank}\:{you}\:{Sir}.\:{You}\:{got}\:{only} \\ $$$${correct}\:{answers}. \\ $$
Answered by ajfour last updated on 18/Nov/18
$${T}\mathrm{cos}\:\varphi+{N}\:=\:{mg}\:\:\:\:\:\:\:\:…\left({i}\right) \\ $$$${T}\mathrm{sin}\:\varphi\:=\:\mu{N}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\:{T}\:{b}\mathrm{cos}\:\left(\theta−\varphi\right)={mg}\left(\frac{{a}+{b}}{\mathrm{2}}\right)\mathrm{cos}\:\theta\:\:..\left({iii}\right) \\ $$$$\:\:{l}\mathrm{cos}\:\varphi+{b}\mathrm{sin}\:\theta\:=\:{h}\:\:\:\:\:\:\:\:…\left({iv}\right) \\ $$$${from}\:\left({i}\right)\&\left({ii}\right) \\ $$$$\:\:\:\:\:\:\:\:\mu{T}\mathrm{cos}\:\varphi+{T}\mathrm{sin}\:\varphi\:=\:\mu{mg} \\ $$$$\Rightarrow\:\:\:{T}\:=\:\frac{\mu{mg}}{\mu\mathrm{cos}\:\varphi+\mathrm{sin}\:\varphi} \\ $$$$\:{using}\:{in}\:\left({iii}\right) \\ $$$$\:\:\:\frac{\mu{mgb}\mathrm{cos}\:\left(\theta−\varphi\right)}{\mu\mathrm{cos}\:\varphi+\mathrm{sin}\:\varphi}\:=\:{mg}\left(\frac{{a}+{b}}{\mathrm{2}}\right)\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:\frac{\mathrm{cos}\:\theta\mathrm{cos}\:\varphi+\mathrm{sin}\:\theta\mathrm{sin}\:\varphi}{\mu\mathrm{cos}\:\varphi+\mathrm{sin}\:\varphi}\:=\:\left(\frac{{a}/{b}+\mathrm{1}}{\mathrm{2}\mu}\right)\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}+\mathrm{tan}\:\theta\:\mathrm{tan}\:\varphi}{\mu+\mathrm{tan}\:\varphi}\:=\:\lambda\:=\:\frac{{a}+{b}}{\mathrm{2}\mu{b}} \\ $$$$\Rightarrow\:\:\:\:\mathrm{tan}\:\theta\:=\:\frac{\lambda\left(\mu+\mathrm{tan}\:\varphi\right)−\mathrm{1}}{\mathrm{tan}\:\varphi} \\ $$$$\&\:{from}\:\left({iv}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{sin}\:\theta\:=\:\frac{{h}−{l}\mathrm{cos}\:\varphi}{{b}} \\ $$$$\theta\:{is}\:{acute},\:{so} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\Rightarrow\:\:\frac{{h}−{l}\mathrm{cos}\:\varphi}{\:\sqrt{{b}^{\mathrm{2}} −\left({h}−{l}\mathrm{cos}\:\varphi\right)^{\mathrm{2}} }}=\frac{\lambda\left(\mu+\mathrm{tan}\:\varphi\right)}{\mathrm{tan}\:\varphi} \\ $$$$\:\:{where}\:\:\lambda\:=\:\frac{\mathrm{1}}{\mathrm{2}\mu}\left(\mathrm{1}+\frac{{a}}{{b}}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.\: \\ $$
Commented by peter frank last updated on 18/Nov/18
$$\mathrm{pls}\:\mathrm{help}\:\mathrm{Qn}\:\mathrm{47979} \\ $$