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Question-198363




Question Number 198363 by BHOOPENDRA last updated on 18/Oct/23
Answered by mahdipoor last updated on 18/Oct/23
F_A =A(cos30j+sin30i)  F_B =B_x i+B_y j  ΣF=0 { ((j⇒((√3)/2)A−((12)/(13))(350)+B_y =0)),((i⇒(1/2)A−(5/(13))(350)+B_x =0)) :}  ΣM_B =0=(5/(13))(350)×2.5+700−(1/2)A×2.5  −((√3)/2)A×12=0  ⇒ { ((A=89.03 lb)),((B_x =90.10 lb     B_y =245.97)) :}
$${F}_{{A}} ={A}\left({cos}\mathrm{30}\boldsymbol{\mathrm{j}}+{sin}\mathrm{30}\boldsymbol{\mathrm{i}}\right) \\ $$$${F}_{{B}} ={B}_{{x}} \boldsymbol{\mathrm{i}}+{B}_{{y}} \boldsymbol{\mathrm{j}} \\ $$$$\Sigma{F}=\mathrm{0\begin{cases}{\boldsymbol{\mathrm{j}}\Rightarrow\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{A}−\frac{\mathrm{12}}{\mathrm{13}}\left(\mathrm{350}\right)+{B}_{{y}} =\mathrm{0}}\\{\boldsymbol{\mathrm{i}}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}{A}−\frac{\mathrm{5}}{\mathrm{13}}\left(\mathrm{350}\right)+{B}_{{x}} =\mathrm{0}}\end{cases}} \\ $$$$\Sigma{M}_{{B}} =\mathrm{0}=\frac{\mathrm{5}}{\mathrm{13}}\left(\mathrm{350}\right)×\mathrm{2}.\mathrm{5}+\mathrm{700}−\frac{\mathrm{1}}{\mathrm{2}}{A}×\mathrm{2}.\mathrm{5} \\ $$$$−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{A}×\mathrm{12}=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{{A}=\mathrm{89}.\mathrm{03}\:{lb}}\\{{B}_{{x}} =\mathrm{90}.\mathrm{10}\:{lb}\:\:\:\:\:{B}_{{y}} =\mathrm{245}.\mathrm{97}}\end{cases} \\ $$
Commented by BHOOPENDRA last updated on 18/Oct/23
Thankyou sir I got the same
$${Thankyou}\:{sir}\:{I}\:{got}\:{the}\:{same} \\ $$
Commented by BHOOPENDRA last updated on 18/Oct/23
Answered by mr W last updated on 19/Oct/23
(B_y −350×((12)/(13)))(((2.5)/( (√3)))+7+5)+700+350×(5/(13))×2.5=0  ⇒B_y =350×((12)/(13))−((700+350×(5/(13))×2.5)/(((2.5)/( (√3)))+7+5))             ≈245.973 ≈ 246 lb
$$\left({B}_{{y}} −\mathrm{350}×\frac{\mathrm{12}}{\mathrm{13}}\right)\left(\frac{\mathrm{2}.\mathrm{5}}{\:\sqrt{\mathrm{3}}}+\mathrm{7}+\mathrm{5}\right)+\mathrm{700}+\mathrm{350}×\frac{\mathrm{5}}{\mathrm{13}}×\mathrm{2}.\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow{B}_{{y}} =\mathrm{350}×\frac{\mathrm{12}}{\mathrm{13}}−\frac{\mathrm{700}+\mathrm{350}×\frac{\mathrm{5}}{\mathrm{13}}×\mathrm{2}.\mathrm{5}}{\frac{\mathrm{2}.\mathrm{5}}{\:\sqrt{\mathrm{3}}}+\mathrm{7}+\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\approx\mathrm{245}.\mathrm{973}\:\approx\:\mathrm{246}\:{lb} \\ $$
Commented by BHOOPENDRA last updated on 19/Oct/23
Thankyou sir
$${Thankyou}\:{sir} \\ $$

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