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A-rope-5m-long-is-fastened-to-two-hooks-4-0m-apart-on-a-horizontal-ceiling-to-the-rope-is-attached-a-10kg-mass-so-that-the-segments-of-the-rope-are-3-0m-and-2-0m-compute-the-tensionin-each-segment-

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Question Number 117594 by mathdave last updated on 12/Oct/20
A rope 5m long is fastened to two hooks 4.0m apart on a horizontal ceiling.to the rope is attached a 10kg mass so that the segments of the rope are 3.0m and 2.0m.compute the tensionin each segment
$${A}\:{rope}\:\mathrm{5}{m}\:{long}\:{is}\:{fastened}\:{to}\:{two}\:{hooks}\: \\ $$$$\mathrm{4}.\mathrm{0}{m}\:{apart}\:{on}\:{a}\:{horizontal} \\ $$$${ceiling}.{to}\:{the}\:{rope}\:{is}\:{attached}\:{a}\:\mathrm{10}{kg}\: \\ $$$${mass}\:{so}\:{that}\:{the}\:{segments}\:{of}\:{the}\:{rope} \\ $$$${are}\:\mathrm{3}.\mathrm{0}{m}\:{and}\:\mathrm{2}.\mathrm{0}{m}.{compute}\:{the} \\ $$$${tensionin}\:{each}\:{segment} \\ $$
Answered by mr W last updated on 12/Oct/20
Commented by mr W last updated on 12/Oct/20
cos α=((3^2 +4^2 −2^2 )/(2×3×4))=(7/8) cos β=((2^2 +4^2 −3^2 )/(2×2×4))=((11)/(16)) (T_1 /(sin ((π/2)−β)))=(T_2 /(sin ((π/2)−α)))=((mg)/(sin (α+β))) ⇒T_1 =((mg cos β)/(sin (α+β)))=((10×10×((11)/(16)))/(((√(15))/8)×((11)/(16))+(7/8)×((√(135))/(16)))) =((55(√(15)))/3)=71.00 N ⇒T_2 =((mg cos α)/(sin (α+β)))=((10×10×(7/8))/(((√(15))/8)×((11)/(16))+(7/8)×((√(135))/(16)))) =((70(√(15)))/3)=90.37 N
$$\mathrm{cos}\:\alpha=\frac{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }{\mathrm{2}×\mathrm{3}×\mathrm{4}}=\frac{\mathrm{7}}{\mathrm{8}} \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{2}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}×\mathrm{4}}=\frac{\mathrm{11}}{\mathrm{16}} \\ $$$$\frac{{T}_{\mathrm{1}} }{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\beta\right)}=\frac{{T}_{\mathrm{2}} }{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\alpha\right)}=\frac{{mg}}{\mathrm{sin}\:\left(\alpha+\beta\right)} \\ $$$$\Rightarrow{T}_{\mathrm{1}} =\frac{{mg}\:\mathrm{cos}\:\beta}{\mathrm{sin}\:\left(\alpha+\beta\right)}=\frac{\mathrm{10}×\mathrm{10}×\frac{\mathrm{11}}{\mathrm{16}}}{\frac{\sqrt{\mathrm{15}}}{\mathrm{8}}×\frac{\mathrm{11}}{\mathrm{16}}+\frac{\mathrm{7}}{\mathrm{8}}×\frac{\sqrt{\mathrm{135}}}{\mathrm{16}}} \\ $$$$=\frac{\mathrm{55}\sqrt{\mathrm{15}}}{\mathrm{3}}=\mathrm{71}.\mathrm{00}\:{N} \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\frac{{mg}\:\mathrm{cos}\:\alpha}{\mathrm{sin}\:\left(\alpha+\beta\right)}=\frac{\mathrm{10}×\mathrm{10}×\frac{\mathrm{7}}{\mathrm{8}}}{\frac{\sqrt{\mathrm{15}}}{\mathrm{8}}×\frac{\mathrm{11}}{\mathrm{16}}+\frac{\mathrm{7}}{\mathrm{8}}×\frac{\sqrt{\mathrm{135}}}{\mathrm{16}}} \\ $$$$=\frac{\mathrm{70}\sqrt{\mathrm{15}}}{\mathrm{3}}=\mathrm{90}.\mathrm{37}\:{N} \\ $$
Commented by Tawa11 last updated on 06/Sep/21
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$

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