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Question Number 26982 by Tinkutara last updated on 31/Dec/17

	Answered by mrW1 last updated on 31/Dec/17

	$Q 1$ $m_{B} a = μ_{S} m_{B} g$ $\Rightarrow a = μ_{S} g = 0.6 \times 10 = 6 m / s^{2}$ $\Rightarrow (4)$ $Q 2$ $f = μ_{K} (m_{A} + m_{B}) g = 0.4 \times (1.5 \times 0.5) \times 10 = 8 N$ $\Rightarrow (3)$ $Q 3$ $m_{C} g - m_{C} a = (m_{A} + m_{B}) a + f = (m_{A} + m_{B}) (a + μ_{K} g)$ $m_{C} = (m_{A} + m_{B}) \frac{a + μ_{K} g}{g - a}$ $m_{C} = (m_{A} + m_{B}) \frac{μ_{S} + μ_{K}}{1 - μ_{S}} = (1.5 + 0.5) \times \frac{0.6 + 0.4}{1 - 0.6} = 5 k g$ $\Rightarrow (3)$
	Commented by Tinkutara last updated on 01/Jan/18
	But in first question there is written that consider limiting friction. So why we take 0.6 then?
	Commented by mrW1 last updated on 01/Jan/18

	$T h e l i m i t i n g c o n d i t i o n t h a t b l o c k A$ $a n d B m o v e t o g e t h e r i s t h a t t h e$ $a c c e l e r a t i o n o f t h e m i s n o t m o r e$ $t h a n t h e m a x . s t a t i c f r i c t i o n c a n$ $r e a c h, t h a t i s a_{m a x} = μ_{s} g . i f t h e$ $a c c e l e r a t i o n i s b i g g e r t h a n t h i s, b l o c k$ $B w i l l s l i p .$
	Commented by mrW1 last updated on 01/Jan/18
	There are following situations: 0< mc ≦0.8 kg: no motion, a=0. 0.8 < mc ≦ 5 kg: block A and B move together, 0<a≦6 m/s2. mc > 5 kg: block B slips over block A, aA>6m/s2,aB=6m/s2.
	Commented by Tinkutara last updated on 03/Jan/18
	Thank you very much Sir! I got the answer.
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