Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

Question Number 47005 by ajfour last updated on 03/Nov/18

Commented by ajfour last updated on 03/Nov/18

A stick of length L is held balanced  against two mutually perpendicular  frictionless walls, with the help  of two strings AC(length b) and  BC(length a). Find Tensions  and Normal reactions T_1 , T_2 , N_1 ,  and N_2 . (mass of stick is m).

$${A}\:{stick}\:{of}\:{length}\:{L}\:{is}\:{held}\:{balanced} \\ $$$${against}\:{two}\:{mutually}\:{perpendicular} \\ $$$${frictionless}\:{walls},\:{with}\:{the}\:{help} \\ $$$${of}\:{two}\:{strings}\:{AC}\left({length}\:{b}\right)\:{and} \\ $$$${BC}\left({length}\:{a}\right).\:{Find}\:{Tensions} \\ $$$${and}\:{Normal}\:{reactions}\:{T}_{\mathrm{1}} ,\:{T}_{\mathrm{2}} ,\:{N}_{\mathrm{1}} , \\ $$$${and}\:{N}_{\mathrm{2}} .\:\left({mass}\:{of}\:{stick}\:{is}\:{m}\right). \\ $$

Answered by MrW3 last updated on 04/Nov/18

Commented by MrW3 last updated on 04/Nov/18

(a cos α)^2 +(b cos β)^2 +(b sin β−a sin α)^2 =L^2   a^2 +b^2 −2ab sin α sin β=L^2   ⇒sin α sin β=((a^2 +b^2 −L^2 )/(2ab))   ...(i)  T_a  sin α=((mg)/2)  ⇒T_a  =((mg)/(2 sin α))  similarly  ⇒T_b  =((mg)/(2 sin β))  N_1 =T_a cos α=((mg cos α)/(2 sin α))=((mg)/(2 tan α))  N_2 =T_b cos β=((mg cos β)/(2 sin β))=((mg)/(2 tan β))  N_1 ×b cos β=N_2 ×a cos α  ⇒((mg)/(2 tan α))×b cos β=((mg)/(2 tan β))×a cos α  ⇒(b/(sin α)) =(a/(sin β))   ...(ii)  (ii) into (i):  (a/b) sin^2  α=((a^2 +b^2 −L^2 )/(2ab))  sin α=(√((a^2 +b^2 −L^2 )/(2a^2 )))  ⇒α=sin^(−1) (√((a^2 +b^2 −L^2 )/(2a^2 )))  similarly  ⇒β=sin^(−1) (√((a^2 +b^2 −L^2 )/(2b^2 )))  ⇒T_a =((mg)/(2 sin α))=mg(a/(√(2(a^2 +b^2 −L^2 ))))  ⇒T_b =((mg)/(2 sin β))=mg(b/(√(2(a^2 +b^2 −L^2 ))))    ⇒N_1 =((mg)/(2 tan α))=((mg)/2)×(√((1/(sin^2  α))−1))  ⇒N_1 =((mg)/2)(√((a^2 +L^2 −b^2 )/(a^2 +b^2 −L^2 )))  ⇒N_2 =((mg)/(2 tan β))=((mg)/2)×(√((1/(sin^2  β))−1))  ⇒N_2 =((mg)/2)(√((b^2 +L^2 −a^2 )/(a^2 +b^2 −L^2 )))

$$\left({a}\:\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +\left({b}\:\mathrm{cos}\:\beta\right)^{\mathrm{2}} +\left({b}\:\mathrm{sin}\:\beta−{a}\:\mathrm{sin}\:\alpha\right)^{\mathrm{2}} ={L}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta={L}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{L}^{\mathrm{2}} }{\mathrm{2}{ab}}\:\:\:...\left({i}\right) \\ $$$${T}_{{a}} \:\mathrm{sin}\:\alpha=\frac{{mg}}{\mathrm{2}} \\ $$$$\Rightarrow{T}_{{a}} \:=\frac{{mg}}{\mathrm{2}\:\mathrm{sin}\:\alpha} \\ $$$${similarly} \\ $$$$\Rightarrow{T}_{{b}} \:=\frac{{mg}}{\mathrm{2}\:\mathrm{sin}\:\beta} \\ $$$${N}_{\mathrm{1}} ={T}_{{a}} \mathrm{cos}\:\alpha=\frac{{mg}\:\mathrm{cos}\:\alpha}{\mathrm{2}\:\mathrm{sin}\:\alpha}=\frac{{mg}}{\mathrm{2}\:\mathrm{tan}\:\alpha} \\ $$$${N}_{\mathrm{2}} ={T}_{{b}} \mathrm{cos}\:\beta=\frac{{mg}\:\mathrm{cos}\:\beta}{\mathrm{2}\:\mathrm{sin}\:\beta}=\frac{{mg}}{\mathrm{2}\:\mathrm{tan}\:\beta} \\ $$$${N}_{\mathrm{1}} ×{b}\:\mathrm{cos}\:\beta={N}_{\mathrm{2}} ×{a}\:\mathrm{cos}\:\alpha \\ $$$$\Rightarrow\frac{{mg}}{\mathrm{2}\:\mathrm{tan}\:\alpha}×{b}\:\mathrm{cos}\:\beta=\frac{{mg}}{\mathrm{2}\:\mathrm{tan}\:\beta}×{a}\:\mathrm{cos}\:\alpha \\ $$$$\Rightarrow\frac{{b}}{\mathrm{sin}\:\alpha}\:=\frac{{a}}{\mathrm{sin}\:\beta}\:\:\:...\left({ii}\right) \\ $$$$\left({ii}\right)\:{into}\:\left({i}\right): \\ $$$$\frac{{a}}{{b}}\:\mathrm{sin}^{\mathrm{2}} \:\alpha=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{L}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$$\mathrm{sin}\:\alpha=\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{L}^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} }} \\ $$$$\Rightarrow\alpha=\mathrm{sin}^{−\mathrm{1}} \sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{L}^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} }} \\ $$$${similarly} \\ $$$$\Rightarrow\beta=\mathrm{sin}^{−\mathrm{1}} \sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{L}^{\mathrm{2}} }{\mathrm{2}{b}^{\mathrm{2}} }} \\ $$$$\Rightarrow{T}_{{a}} =\frac{{mg}}{\mathrm{2}\:\mathrm{sin}\:\alpha}={mg}\frac{{a}}{\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{L}^{\mathrm{2}} \right)}} \\ $$$$\Rightarrow{T}_{{b}} =\frac{{mg}}{\mathrm{2}\:\mathrm{sin}\:\beta}={mg}\frac{{b}}{\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{L}^{\mathrm{2}} \right)}} \\ $$$$ \\ $$$$\Rightarrow{N}_{\mathrm{1}} =\frac{{mg}}{\mathrm{2}\:\mathrm{tan}\:\alpha}=\frac{{mg}}{\mathrm{2}}×\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\alpha}−\mathrm{1}} \\ $$$$\Rightarrow{N}_{\mathrm{1}} =\frac{{mg}}{\mathrm{2}}\sqrt{\frac{{a}^{\mathrm{2}} +{L}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{L}^{\mathrm{2}} }} \\ $$$$\Rightarrow{N}_{\mathrm{2}} =\frac{{mg}}{\mathrm{2}\:\mathrm{tan}\:\beta}=\frac{{mg}}{\mathrm{2}}×\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\beta}−\mathrm{1}} \\ $$$$\Rightarrow{N}_{\mathrm{2}} =\frac{{mg}}{\mathrm{2}}\sqrt{\frac{{b}^{\mathrm{2}} +{L}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{L}^{\mathrm{2}} }} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com