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Question Number 64478 by necx1 last updated on 18/Jul/19

Commented by necx1 last updated on 18/Jul/19

$p l e a s e h e l p m e a n a l y s e t h i s f r e e b o d y$ $d i a g r a m . I^{'} l l b e g r a t e f u l s i r s .$
Commented by necx1 last updated on 18/Jul/19

$p l e a s e f i n d R_{l} a n d R_{r}$
	Answered by behi83417@gmail.com last updated on 18/Jul/19
	$R_1 ×8+R_2 ×4=10×?+10×5 R_1 +R_2 =10+10+5 ⇒ { ((2R_1 +R_2 =2.5×?+12.5)),((R_1 +R_2 =25)) :} R_1 =2.5×?−12.5,R_2 =37.5−2.5×? ?=distance from: left 10^N to 5^N force.(not given)$
	$R_{1} \times 8 + R_{2} \times 4 = 10 \times ? + 10 \times 5$ $R_{1} + R_{2} = 10 + 10 + 5$ $\Rightarrow {\begin{cases} {2 R}_{1} + R_{2} = 2.5 \times ? + 12.5 \\ R_{1} + R_{2} = 25 \end{cases}$ $R_{1} = 2.5 \times ? - 12.5, R_{2} = 37.5 - 2.5 \times ?$ $? = distance from : left 10^{N} to 5^{N} force . (not given)$
	Commented by necx1 last updated on 18/Jul/19

	$y e s . . . . {T h a t}^{'} s w h e r e i a l s o d o n o t$ $u n d e r s t a n d . T h o u g h a f r i e n d g o t 10 N$ $a n d 15 N b u t t h e p r o c e s s h e a p p l i e d i$ $d o n t g e t . H e w a s u s i n g s o m e t h i n g$ $+ v e a n d - v e i n r e l a t i o n t o a f r e e l y$ $s u p p o r t e d a n d c a n t i l e v e r b e e m . I a l s o$ $t h i n k t h e q u e s t i o n i s k i n d a n o t c o m p l e t e$ $i n t e r m s o f g i v e n p a r a m e t e r s .$
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