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Question Number 70159 by ajfour last updated on 01/Oct/19

Commented by ajfour last updated on 01/Oct/19

$$Underbalancedconditions,find$$$$\theta .Allsurfacesfrictionless.$$

Answered by mr W last updated on 01/Oct/19

Commented by mr W last updated on 01/Oct/19

$$letL=lengthofrods$$$$(N_1-\frac{Mg}{2})L\sin \frac{\theta }{2}=amg...\left(i\right)$$$$(N_2-\frac{Mg}{2})L\sin \frac{\theta }{2}=2bmg\sin \alpha ...\left(ii\right)$$$$\left(ii\right)/\left(i\right):$$$$\frac{N_2-\frac{Mg}{2}}{N_1-\frac{Mg}{2}}=\frac{2b\sin \alpha }{a}=2$$$$2N_2-Mg=2(2N_1-Mg)$$$$\Rightarrow 2N_2=4N_1-Mg$$$$N_1=Mg+\frac{L-b}{2L}mg$$$$N_2=Mg+\frac{L+b}{2L}mg$$$$2Mg+\frac{2(L+b)}{2L}mg=4Mg+\frac{4(L-b)}{2L}mg-Mg$$$$\Rightarrow \frac{b}{L}=\frac{1}{3}(\frac{M}{m}+2)$$$$from\left(i\right):$$$$(Mg+\frac{L-b}{2L}mg-\frac{Mg}{2})L\sin \frac{\theta }{2}=mgb\cos \theta$$$$(\frac{M}{m}+1-\frac{b}{L})\sin \frac{\theta }{2}=2\cos \theta \frac{b}{L}$$$$\frac{2M+m}{2(M+2m)}\sin \frac{\theta }{2}=1-2{\sin }^2\frac{\theta }{2}$$$$2{\sin }^2\frac{\theta }{2}+\mu \sin \frac{\theta }{2}-1=0$$$$\Rightarrow \sin \frac{\theta }{2}=\frac{\sqrt{{\mu }^2+8}-1}{4}$$$$\Rightarrow \theta =2{\sin }^{-1}\frac{\sqrt{{\mu }^2+8}-1}{4}$$$$with\mu =\frac{2M+m}{2(M+2m)}$$

Commented by ajfour last updated on 01/Oct/19

$$looksrightsir,letmecheck$$$$indetailinsomewhile,thanks!$$

Commented by ajfour last updated on 02/Oct/19

$$N_1=Mg+\frac{L-b}{2L}mg$$$$N_2=Mg+\frac{L+b}{2L}mg$$$$Sirpleaseexplainabovetwoeqs.$$

Commented by mr W last updated on 02/Oct/19

$$theweightofthetworodsisequally$$$$distributedbetween{N}_{1}and{N}_2,{that}^{\prime }s$$$$clear.$$$$theweightmgwillbedistributed$$$$unequallyinthisway:$$$$N_1\times 2L\sin \frac{\theta }{2}=mg\times (L-b)\sin \frac{\theta }{2}$$$$\Rightarrow N_1=\frac{L-b}{2L}mg$$$$N_2\times 2L\sin \frac{\theta }{2}=mg\times (L+b)\sin \frac{\theta }{2}$$$$\Rightarrow N_2=\frac{L+b}{2L}mg$$$$therefore:$$$$\Rightarrow N_1=Mg+\frac{L-b}{2L}mg$$$$\Rightarrow N_2=Mg+\frac{L+b}{2L}mg$$

Commented by mr W last updated on 02/Oct/19

$$sir,canyoucheckthequestionagain:$$$$ithinkthesystemcannotbe$$$$inequilibrium.$$$$tokeeptheequilibriumitmustbe$$$$\theta =2{\sin }^{-1}\frac{\sqrt{{\mu }^2+8}-1}{4}$$$$butontheothersideitmustbe$$$$\alpha =90°-\theta =\frac{\theta }{2}\Rightarrow \theta =60°$$

Commented by mr W last updated on 03/Oct/19

Commented by ajfour last updated on 04/Oct/19

$$2Mg+\frac{2(L+b)}{2L}mg=4Mg+\frac{4(L-b)}{2L}mg-Mg$$$$Sirexplainhowyouformthis$$$$eq.anditmightnotleadto$$$$\frac{b}{L}=\frac{1}{3}(\frac{M}{m}+2)$$$$....$$

Commented by mr W last updated on 04/Oct/19

$$(N_1-\frac{Mg}{2})L\sin \frac{\theta }{2}=mga$$$$(N_2-\frac{Mg}{2})L\sin \frac{\theta }{2}=2mgb\sin \alpha =2mga$$$$\Rightarrow N_2-\frac{Mg}{2}=2(N_1-\frac{Mg}{2})$$$$\Rightarrow N_2=2N_1-\frac{Mg}{2}$$$$N_1=Mg+\frac{L-b}{2L}mg$$$$N_2=Mg+\frac{L+b}{2L}mg$$$$\Rightarrow Mg+\frac{L+b}{2L}mg=2Mg+\frac{2(L-b)}{2L}mg-\frac{Mg}{2}$$$$\Rightarrow \frac{3b-L}{L}m=M$$$$\Rightarrow \frac{b}{L}=\frac{1}{3}(\frac{M}{m}+1)$$$$(N_1-\frac{Mg}{2})L\sin \frac{\theta }{2}=mgb\cos \theta$$$$(\frac{L-b}{2L}+\frac{M}{2m})\sin \frac{\theta }{2}=\frac{b}{L}\cos \theta$$$$(1-\frac{b}{L}+\frac{M}{m})\sin \frac{\theta }{2}=\frac{2}{3}(\frac{M}{m}+1)(1-2{\sin }^2\frac{\theta }{2})$$$$(1-\frac{1}{3}(\frac{M}{m}+1)+\frac{M}{m})\sin \frac{\theta }{2}=\frac{2}{3}(\frac{M}{m}+1)(1-2{\sin }^2\frac{\theta }{2})$$$$(1+\frac{M}{m})\sin \frac{\theta }{2}=(\frac{M}{m}+1)(1-2{\sin }^2\frac{\theta }{2})$$$$\sin \frac{\theta }{2}=1-2{\sin }^2\frac{\theta }{2}$$$$2{\sin }^2\frac{\theta }{2}+\sin \frac{\theta }{2}-1=0$$$$\sin \frac{\theta }{2}=\frac{1}{2}$$$$\Rightarrow \theta =60°$$$$wecangetthisdirectlyfrom$$$$\alpha =90-\theta =\frac{\theta }{2}$$$$\Rightarrow \theta =60°$$

Commented by mr W last updated on 04/Oct/19

$$sorry,ihadatypoin$$$$\Rightarrow \frac{b}{L}=\frac{1}{3}(\frac{M}{m}+2)$$$$whichledtowrongconclusion.$$

Commented by ajfour last updated on 04/Oct/19

$$Toomuchfuss,aboutnothing!$$$$haha..$$

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