As shown in the figure a block of mass 10,kg | vedclass

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Question

$N = Mg - F Sin 30^{\circ}$

$= mg -\frac{ F }{2}=100-\frac{ F }{2}=\frac{200- F }{2}$

$F Cos 30^{\circ}=\mu N$

$\sqrt{3} \frac{ F }{2}=0.25 \

Vedclass · Accepted Answer

$25.2$

Vedclass · Answer

$N = Mg - F Sin 30^{\circ}$

$= mg -\frac{ F }{2}=100-\frac{ F }{2}=\frac{200- F }{2}$

$F Cos 30^{\circ}=\mu N$

$\sqrt{3} \frac{ F }{2}=0.25 	imes\left(\frac{200- F }{2}ight)$

$4 \sqrt{3} F =200- F$

$F =\frac{200}{4 \sqrt{3}+1}=25.22$

As shown in the figure a block of mass $10\,kg$ lying on a horizontal surface is pulled by a force $F$ acting at an angle $30^{\circ}$, with horizontal. For $\mu_{ s }=0.25$, the block will just start to move for the value of $F..........\,N$ : $\left[\right.$ Given $\left.g =10\,ms ^{-2}\right]$

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