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In figure, two blocks are separated by a uniform strut attached to each block with frictionless pins. Block $A$ weighs $400\,N$, block $B$ weighs $300\,N$, and the strut $A B$ weigh $200\,N$. If $\mu=0.25$ under $B$, determine the minimum coefficient of friction under $A$ to prevent motion.
$0.4$
$0.2$
$0.8$
$0.1$
Solution
(a)
Consider $F BD$ of structure.
Applying equilibrium equations,
$Av + Bv =200 N$
$A _{ H }= B _{ H}{ \ldots \text { (ii) }}$
From F BD of block B,
$B _{ H }+ F _{ B } \cos 60^{\circ}- N _{ B } \sin 60^{\circ}=0$
$N_B \cos 60^{\circ}-B_V-300+F_V \sin 60^{\circ}=0$
$F _{ B }=0.25 N _{ B }$
$B _{ H }-0.74 N _{ B }=0 \ldots \text {.(iii) }$
$- B _{ V }+0.71 N _{ B }=300$
$F _{ A }- A _{ H }=0$
$N _{ A }- A _{ V }=400$
$F _{ A }=\mu_{ A } N _{ A }$
$\therefore \mu_{ A } N _{ A }- A _{ H }=0$.
On solving above equations, we get
$N _{ A }=650 N , F _{ A }=260 N , F _{ A }=\mu_{ A } N _{ A }$
$\therefore \mu_{ A }=\frac{260}{250}=0.4$
