A block of mass sqrt{2} ,kg is placed on a ro | vedclass

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(C) The block is on a horizontal surface. The forces acting on the block are: $1$. Weight $mg$ acting downwards. $2$. Normal reaction $N$ acting upwar

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) The block is on a horizontal surface. The forces acting on the block are: $1$. Weight $mg$ acting downwards. $2$. Normal reaction $N$ acting upwards. $3$. Applied force $F$ at $45^{\circ}$ to the horizontal. Resolving $F$ into components: $F_x = F \cos 45^{\circ}$ and $F_y = F \sin 45^{\circ}$. For vertical equilibrium: $N + F \sin 45^{\circ} = mg$. So, $N = mg - F \sin 45^{\circ} = \sqrt{2} 	imes 10 - F 	imes \frac{1}{\sqrt{2}} = 10\sqrt{2} - \frac{F}{\sqrt{2}}$. The block starts to move when the horizontal component of the force equals the limiting friction: $F \cos 45^{\circ} = \mu N$. Substituting the values: $F 	imes \frac{1}{\sqrt{2}} = 0.25 	imes (10\sqrt{2} - \frac{F}{\sqrt{2}})$. Multiply by $\sqrt{2}$: $F = 0.25 	imes (10 	imes 2 - F) = 0.25 	imes (20 - F)$. $F = 5 - 0.25F$. $1.25F = 5$. $F = \frac{5}{1.25} = 4 \,N$.

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