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Question Number 224790 by Tawa11 last updated on 04/Oct/25
Commented by Tawa11 last updated on 04/Oct/25
Commented by Tawa11 last updated on 04/Oct/25
Am not sure sir
$$\mathrm{Am}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 05/Oct/25
your solution or other people′s solution?
$${your}\:{solution}\:{or}\:{other}\:{people}'{s} \\ $$$${solution}? \\ $$
Commented by Tawa11 last updated on 05/Oct/25
Someone else
$$\mathrm{Someone}\:\mathrm{else} \\ $$
Answered by mr W last updated on 05/Oct/25
Commented by mr W last updated on 05/Oct/25
tan γ=((0.75(√3))/(1.2+0.75))=((5(√3))/(13)) ⇒sin γ=((5(√3))/(2(√(61)))), cos γ=((13)/(2(√(61)))) ΣM_0 =0: T×1.2 sin γ=mg×((0.75)/2) ⇒T=((2(√(61))×0.75mg)/(2×1.2×5(√3)))=(((√(183)) mg)/(24))≈99.53 N ΣM_B =0: R_y ×1.2=mg(1.2+((0.75)/2)) ⇒R_y =((21mg)/(16))≈231.76 N ΣF_x =0: R_x =T cos γ=((13(√3) mg)/(48))≈82.83 N ⇒R=(√(R_x ^2 +R_y ^2 ))=(((√(1119)) mg)/(24))≈246.12 N
$$\mathrm{tan}\:\gamma=\frac{\mathrm{0}.\mathrm{75}\sqrt{\mathrm{3}}}{\mathrm{1}.\mathrm{2}+\mathrm{0}.\mathrm{75}}=\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{13}}\: \\ $$$$\Rightarrow\mathrm{sin}\:\gamma=\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{61}}},\:\mathrm{cos}\:\gamma=\frac{\mathrm{13}}{\mathrm{2}\sqrt{\mathrm{61}}} \\ $$$$\Sigma{M}_{\mathrm{0}} =\mathrm{0}: \\ $$$${T}×\mathrm{1}.\mathrm{2}\:\mathrm{sin}\:\gamma={mg}×\frac{\mathrm{0}.\mathrm{75}}{\mathrm{2}} \\ $$$$\Rightarrow{T}=\frac{\mathrm{2}\sqrt{\mathrm{61}}×\mathrm{0}.\mathrm{75}{mg}}{\mathrm{2}×\mathrm{1}.\mathrm{2}×\mathrm{5}\sqrt{\mathrm{3}}}=\frac{\sqrt{\mathrm{183}}\:{mg}}{\mathrm{24}}\approx\mathrm{99}.\mathrm{53}\:{N} \\ $$$$\Sigma{M}_{{B}} =\mathrm{0}: \\ $$$${R}_{{y}} ×\mathrm{1}.\mathrm{2}={mg}\left(\mathrm{1}.\mathrm{2}+\frac{\mathrm{0}.\mathrm{75}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{R}_{{y}} =\frac{\mathrm{21}{mg}}{\mathrm{16}}\approx\mathrm{231}.\mathrm{76}\:{N} \\ $$$$\Sigma{F}_{{x}} =\mathrm{0}: \\ $$$${R}_{{x}} ={T}\:\mathrm{cos}\:\gamma=\frac{\mathrm{13}\sqrt{\mathrm{3}}\:{mg}}{\mathrm{48}}\approx\mathrm{82}.\mathrm{83}\:{N} \\ $$$$\Rightarrow{R}=\sqrt{{R}_{{x}} ^{\mathrm{2}} +{R}_{{y}} ^{\mathrm{2}} }=\frac{\sqrt{\mathrm{1119}}\:{mg}}{\mathrm{24}}\approx\mathrm{246}.\mathrm{12}\:{N} \\ $$
Commented by Tawa11 last updated on 05/Oct/25
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by mr W last updated on 05/Oct/25
Commented by mr W last updated on 05/Oct/25
tan γ=((0.75(√3))/(1.2+0.75))=((5(√3))/(13)) ⇒γ=tan^(−1) ((5(√3))/(13))≈33.67° tan α=((0.5×0.75)/((1.2+0.5×0.75) tan γ))=((13(√3))/(63)) ⇒α=tan^(−1) ((13(√3))/(63))≈19.67° β=(π/2)−α−γ ⇒sin β=cos (α+γ) T=((mg sin α)/(sin β))=((mg sin α)/(cos (α+γ)))≈99.53 N R=((mg sin (α+β))/(sin β))=((mg cos γ)/(cos (α+γ)))≈246.12 N
$$\mathrm{tan}\:\gamma=\frac{\mathrm{0}.\mathrm{75}\sqrt{\mathrm{3}}}{\mathrm{1}.\mathrm{2}+\mathrm{0}.\mathrm{75}}=\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{13}}\: \\ $$$$\Rightarrow\gamma=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{13}}\approx\mathrm{33}.\mathrm{67}° \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{0}.\mathrm{5}×\mathrm{0}.\mathrm{75}}{\left(\mathrm{1}.\mathrm{2}+\mathrm{0}.\mathrm{5}×\mathrm{0}.\mathrm{75}\right)\:\mathrm{tan}\:\gamma}=\frac{\mathrm{13}\sqrt{\mathrm{3}}}{\mathrm{63}} \\ $$$$\Rightarrow\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{13}\sqrt{\mathrm{3}}}{\mathrm{63}}\approx\mathrm{19}.\mathrm{67}° \\ $$$$\beta=\frac{\pi}{\mathrm{2}}−\alpha−\gamma\: \\ $$$$\Rightarrow\mathrm{sin}\:\beta=\mathrm{cos}\:\left(\alpha+\gamma\right) \\ $$$${T}=\frac{{mg}\:\mathrm{sin}\:\alpha}{\mathrm{sin}\:\beta}=\frac{{mg}\:\mathrm{sin}\:\alpha}{\mathrm{cos}\:\left(\alpha+\gamma\right)}\approx\mathrm{99}.\mathrm{53}\:{N} \\ $$$${R}=\frac{{mg}\:\mathrm{sin}\:\left(\alpha+\beta\right)}{\mathrm{sin}\:\beta}=\frac{{mg}\:\mathrm{cos}\:\gamma}{\mathrm{cos}\:\left(\alpha+\gamma\right)}\approx\mathrm{246}.\mathrm{12}\:{N} \\ $$
Commented by Tawa11 last updated on 05/Oct/25
I really appreciate sir
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir} \\ $$
Answered by fantastic last updated on 07/Oct/25
T=(((18×9.8×((7.5)/2))/(1.5sin (60^0 −tan^(−1) (((1.3)/(1.95))))))
$${T}=\left(\frac{\mathrm{18}×\mathrm{9}.\mathrm{8}×\frac{\mathrm{7}.\mathrm{5}}{\mathrm{2}}}{\mathrm{1}.\mathrm{5sin}\:\left(\mathrm{60}^{\mathrm{0}} −\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}.\mathrm{3}}{\mathrm{1}.\mathrm{95}}\right)\right.}\right) \\ $$

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