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

$A r o p e 5 m l o n g i s f a s t e n e d t o t w o h o o k s$ $4.0 m a p a r t o n a h o r i z o n t a l$ $c e i l i n g . t o t h e r o p e i s a t t a c h e d a 10 k g$ $m a s s s o t h a t t h e s e g m e n t s o f t h e r o p e$ $a r e 3.0 m a n d 2.0 m . c o m p u t e t h e$ $t e n s i o n i n e a c h s e g m e n t$
	Answered by mr W last updated on 12/Oct/20

	Commented by mr W last updated on 12/Oct/20

	$\cos α = \frac{3^{2} + 4^{2} - 2^{2}}{2 \times 3 \times 4} = \frac{7}{8}$ $\cos β = \frac{2^{2} + 4^{2} - 3^{2}}{2 \times 2 \times 4} = \frac{11}{16}$ $\frac{T_{1}}{\sin (\frac{π}{2} - β)} = \frac{T_{2}}{\sin (\frac{π}{2} - α)} = \frac{m g}{\sin (α + β)}$ $\Rightarrow T_{1} = \frac{m g \cos β}{\sin (α + β)} = \frac{10 \times 10 \times \frac{11}{16}}{\frac{\sqrt{15}}{8} \times \frac{11}{16} + \frac{7}{8} \times \frac{\sqrt{135}}{16}}$ $= \frac{55 \sqrt{15}}{3} = 71.00 N$ $\Rightarrow T_{2} = \frac{m g \cos α}{\sin (α + β)} = \frac{10 \times 10 \times \frac{7}{8}}{\frac{\sqrt{15}}{8} \times \frac{11}{16} + \frac{7}{8} \times \frac{\sqrt{135}}{16}}$ $= \frac{70 \sqrt{15}}{3} = 90.37 N$
	Commented by Tawa11 last updated on 06/Sep/21

	$great sir$
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