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Question Number 198363 by BHOOPENDRA last updated on 18/Oct/23

	Answered by mahdipoor last updated on 18/Oct/23
	$F_A =A(cos30j+sin30i) F_B =B_x i+B_y j ΣF=0 { ((j⇒((√3)/2)A−((12)/(13))(350)+B_y =0)),((i⇒(1/2)A−(5/(13))(350)+B_x =0)) :} ΣM_B =0=(5/(13))(350)×2.5+700−(1/2)A×2.5 −((√3)/2)A×12=0 ⇒ { ((A=89.03 lb)),((B_x =90.10 lb B_y =245.97)) :}$
	$F_{A} = A (c o s 30 j + s i n 30 i)$ $F_{B} = B_{x} i + B_{y} j$ $Σ F = 0 {\begin{cases} j \Rightarrow \frac{\sqrt{3}}{2} A - \frac{12}{13} (350) + B_{y} = 0 \\ i \Rightarrow \frac{1}{2} A - \frac{5}{13} (350) + B_{x} = 0 \end{cases}$ $Σ M_{B} = 0 = \frac{5}{13} (350) \times 2.5 + 700 - \frac{1}{2} A \times 2.5$ $- \frac{\sqrt{3}}{2} A \times 12 = 0$ $\Rightarrow {\begin{cases} A = 89.03 l b \\ B_{x} = 90.10 l b B_{y} = 245.97 \end{cases}$
	Commented by BHOOPENDRA last updated on 18/Oct/23

	$T h a n k y o u s i r I g o t t h e s a m e$
	Commented by BHOOPENDRA last updated on 18/Oct/23

	Answered by mr W last updated on 19/Oct/23

	$(B_{y} - 350 \times \frac{12}{13}) (\frac{2.5}{\sqrt{3}} + 7 + 5) + 700 + 350 \times \frac{5}{13} \times 2.5 = 0$ $\Rightarrow B_{y} = 350 \times \frac{12}{13} - \frac{700 + 350 \times \frac{5}{13} \times 2.5}{\frac{2.5}{\sqrt{3}} + 7 + 5}$ $\approx 245.973 \approx 246 l b$
	Commented by BHOOPENDRA last updated on 19/Oct/23

	$T h a n k y o u s i r$
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