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

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(A) For the block to just start moving,the horizontal component of the applied force $F$ must be equal to the limiting friction force.$1$. Resolve the

Vedclass · Accepted Answer

$25.2$

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(A) For the block to just start moving,the horizontal component of the applied force $F$ must be equal to the limiting friction force.$1$. Resolve the force $F$ into components:Horizontal component: $F_{x} = F \cos 30^{\circ}$Vertical component: $F_{y} = F \sin 30^{\circ}$$2$. Determine the normal reaction $N$:The vertical forces acting on the block are the weight $mg$ downwards,the normal reaction $N$ upwards,and the vertical component of the applied force $F \sin 30^{\circ}$ upwards.$N + F \sin 30^{\circ} = mg$$N = mg - F \sin 30^{\circ}$Given $m = 10 \ kg$ and $g = 10 \ ms^{-2}$,so $mg = 100 \ N$.$N = 100 - F \sin 30^{\circ} = 100 - 0.5F$$3$. Apply the condition for motion:The limiting friction is $f_{L} = \mu_{s} N$.The block starts to move when $F \cos 30^{\circ} = \mu_{s} N$.$F \cos 30^{\circ} = 0.25(100 - 0.5F)$$F \frac{\sqrt{3}}{2} = 25 - 0.125F$$F(0.866 + 0.125) = 25$$F(0.991) = 25$$F = \frac{25}{0.991} \approx 25.22 \ N$Thus,the block will just start to move for $F \approx 25.2 \ N$.

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 the horizontal. For $\mu_{s} = 0.25$,the block will just start to move for the value of $F$: [Given $g = 10 \ ms^{-2}$] (in $N$)

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