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Question Number 27202 by Tinkutara last updated on 03/Jan/18

	Answered by mrW1 last updated on 03/Jan/18

	$N = n o r m a l f o r c e b e t w e e n m a n d M$ $T = t e n s i o n i n t h r e a d$ $N = m g \cos θ$ $T = N \sin θ = m g \sin θ \cos θ = 3 \times 10 \times \frac{1}{2} = 15 N$
	Commented by Tinkutara last updated on 03/Jan/18
	Why we can't resolve tension along the inclined plane and normal to it and then balance the forces?
	Commented by mrW1 last updated on 03/Jan/18

	$b e c a u s e t h e n o r m a l f o r c e b e t w e e n b l o c k$ $M a n d f l o o r a s w e l l a s M g h a v e a l s o c o m p o n e n t$ $a l o n e t h e i n c l i n e d p l a n e a n d n o r m a l$ $t o i t .$
	Commented by Tinkutara last updated on 03/Jan/18
	Thank you very much Sir! I got the answer.
	Commented by mrW1 last updated on 03/Jan/18

	$w e s e e T = \frac{m g \sin 2 θ}{2}$ $s o i f t h e q u e s t i o n i s w h a t i s t h e$ $i n c l i n a t i o n s u c h t h a t t h e t e n s i o n i n$ $t h r e a d i s m a x i m u m ?$ $w e k n o w t h e a n s w e r i s w h e n θ = 45 ° .$
	Commented by Tinkutara last updated on 03/Jan/18
	Yes thanks Sir.
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