A 90cm uniform lever has a load of 30N suspended at 15cm from one of its ends.If the fulcrum is at the centre of gravity what is the force that must be applied to its other end to keep it in horizontal equilibrium?


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Question Number 52340 by Necxx last updated on 06/Jan/19

$A 90 c m u n i f o r m l e v e r h a s a l o a d$ $o f 30 N s u s p e n d e d a t 15 c m f r o m$ $o n e o f i t s e n d s . I f t h e f u l c r u m i s a t$ $t h e c e n t r e o f g r a v i t y w h a t i s t h e$ $f o r c e t h a t m u s t b e a p p l i e d t o i t s$ $o t h e r e n d t o k e e p i t i n h o r i z o n t a l$ $e q u i l i b r i u m ?$
	Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jan/19

	$F \times 45 = 30 \times 30$ $F = 20 N m o m e n t t a k e n a b o u t f u l c r u m$
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