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Question Number 64775 by Rio Michael last updated on 21/Jul/19

Commented by Rio Michael last updated on 21/Jul/19

$t h e f i g u r e a b o v e s h o w s a s t r i n g P Q R S, P a n d S a r e a t t a c h e d t o a f i x e d s u p p o r t a n d$ $m a s s, m, a n d 2.5 kg a r e a t t a c h e d a t t h e p o i n t s Q a n d R r e s p e c t i v e l y a n d t h e s y s t e m$ $i s i n e q u i l i b r u i m . C a l c u l a t e$ $a) t h e m a s s, m .$ $b) t h e t e n s i o n T .$ $c) t h e a n g l e θ (r e p r e s e n t e d a s \emptyset o n t h e d i a g r a m)$
	Answered by behi83417@gmail.com last updated on 21/Jul/19
	$30sin30^• +T.sin∅=2.5+m (1) −30cos30^• +T.cos∅=0 (2) 30sin30^• ×QR−m.QR=0 (3) (3)⇒m=30.sin30^• =30×0.5=15^(kg) (2),(1)⇒ { ((T.cos∅=30cos30^• )),((T.sin∅=2.5)) :} ⇒tg∅=((2.5)/(30cos30^• ))=(1/(6(√3)))⇒∅=5.5^• T=(√(2.5^2 +(15(√3))^2 ))=26.101$
	${30 \sin 30}^{∙} + T . \sin \emptyset = 2.5 + m (1)$ $- {30 \cos 30}^{∙} + T . \cos \emptyset = 0 (2)$ ${30 \sin 30}^{∙} \times QR - m . QR = 0 (3)$ $(3) \Rightarrow m = 30 . {\sin 30}^{∙} = 30 \times 0.5 = 15^{kg}$ $(2), (1) \Rightarrow {\begin{cases} T . \cos \emptyset = {30 \cos 30}^{∙} \\ T . \sin \emptyset = 2.5 \end{cases}$ $\Rightarrow tg \emptyset = \frac{2.5}{{30 \cos 30}^{∙}} = \frac{1}{6 \sqrt{3}} \Rightarrow \emptyset = 5 . 5^{∙}$ $T = \sqrt{2 . 5^{2} + {(15 \sqrt{3})}^{2}} = 26.101$
	Commented by Rio Michael last updated on 21/Jul/19

	$e x c u s e m e s i r, {i t}^{'} s 60 °$
	Commented by Rio Michael last updated on 21/Jul/19

	$b u t t h a n k s s i r$
	Commented by Rio Michael last updated on 22/Jul/19

	$30 s i n 60 ° + T s i n θ = 2.5 + m . . . . . . . . . (i)$ $3 o c o s 60 ° + T c o s 60 ° = 0 . . . . . . . . . . . . . . . . (i i)$ $30 s i n 60 ° \times Q R - m Q R = 0 . . . . . . . . . (i i i)$ $m = 30 s i n 60 ° = 15 \sqrt{2} k g$
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