Calculate the force (F) required to cause the block of mass m_1 = 20 kg just to slide under the block of mass m_2 = 10 kg [coefficient of friction μ = 0.25 for all surfaces]


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Question Number 20409 by Tinkutara last updated on 26/Aug/17

$C a l c u l a t e t h e f o r c e (F) r e q u i r e d t o$ $c a u s e t h e b l o c k o f m a s s m_{1} = 20 k g$ $j u s t t o s l i d e u n d e r t h e b l o c k o f m a s s$ $m_{2} = 10 k g [c o e f f i c i e n t o f f r i c t i o n μ$ $= 0.25 f o r a l l s u r f a c e s]$
Commented by Tinkutara last updated on 26/Aug/17

	Answered by ajfour last updated on 26/Aug/17

	$T \sin 53 ° + N_{2} = m_{2} g$ $T \cos 53 ° = μ N_{2}$ $\Rightarrow \frac{m_{2} g - N_{2}}{μ N_{2}} = \tan 53 ° = \frac{4}{3}$ $\frac{100 - N_{2}}{(N_{2} / 4)} = \frac{4}{3} \Rightarrow 300 - 3 N_{2} = N_{2}$ $N_{2} = 75 u n i t s \Rightarrow f_{2} = μ N_{2} = \frac{75}{4}$ $f_{1} = μ N_{1} = μ (m_{1} g + N_{2})$ $= \frac{1}{4} (200 + 75) = \frac{275}{4}$ $F_{m i n i m u m} = f_{1} + f_{2}$ $= \frac{275 + 75}{4} = \frac{350}{4}$ $F_{m i n i m u m} = 87.5 N$
	Commented by Tinkutara last updated on 26/Aug/17

	$Thank you very much Sir!$
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