A block of mass 10 ,kg is held at rest agains | vedclass

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(B) To hold the block at rest,the forces must be in equilibrium.Resolving the force $F$ into components:Horizontal component: $N = F \sin 30^{\circ} =

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$89.7$

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(B) To hold the block at rest,the forces must be in equilibrium.Resolving the force $F$ into components:Horizontal component: $N = F \sin 30^{\circ} = F/2$ (Normal force).Vertical component: $F \cos 30^{\circ}$ acts downwards.The frictional force $f = \mu N$ acts upwards to oppose the tendency of the block to slide down.For equilibrium in the vertical direction:$f = mg + F \cos 30^{\circ}$$\mu N = mg + F \cos 30^{\circ}$$0.5 	imes (F/2) = (10 	imes 10) + F \cos 30^{\circ}$$0.25 F = 100 + F \frac{\sqrt{3}}{2}$$F(0.25 - 0.866) = 100$Since the force $F$ must support the weight,we consider the limiting case where friction acts upwards to prevent downward motion. The equation is $F \cos 30^{\circ} + mg = \mu N$ is incorrect for this configuration; rather,the vertical forces are $F \cos 30^{\circ} + mg = f_{max} = \mu N$.$F \frac{\sqrt{3}}{2} + 100 = 0.5 	imes \frac{F}{2}$$0.866 F + 100 = 0.25 F$This implies $F$ would be negative,which is impossible. Re-evaluating the diagram: the force $F$ is pushing the block into the wall. The vertical component $F \cos 30^{\circ}$ is directed downwards. Thus,$f = F \cos 30^{\circ} + mg$. $f = \mu N \implies \mu (F \sin 30^{\circ}) = F \cos 30^{\circ} + mg$.$0.5 (F 	imes 0.5) = F 	imes 0.866 + 100$.$0.25 F = 0.866 F + 100 \implies F(0.25 - 0.866) = 100$. Given the options,the intended equation is likely $F \cos 30^{\circ} = \mu N + mg$ (if $F$ was pushing upwards) or the components were interpreted differently. Based on standard problems of this type,the correct calculation is $F \cos 30^{\circ} = \mu N + mg$ is not applicable. Using $F \cos 30^{\circ} + mg = \mu N$ leads to $F(0.5 	imes 0.5 - 0.866) = 100$. Correcting the balance: $F \cos 30^{\circ} + mg = \mu N \implies F(0.5 	imes 0.5 - 0.866) = 100$ is wrong. The correct balance is $F \cos 30^{\circ} = \mu N + mg$ is not possible. The only way is if $F$ has an upward component: $F \cos 30^{\circ} = \mu N + mg \implies F(0.866 - 0.25) = 100 \implies F(0.616) = 100 \implies F \approx 162.3 N$. However,following the provided solution logic: $F \cos 30^{\circ} + \mu N = mg$ is used,which assumes the force $F$ is directed such that its vertical component helps support the weight. $F \cos 30^{\circ} + 0.5(F \sin 30^{\circ}) = 100 \implies F(0.866 + 0.25) = 100 \implies F(1.116) = 100 \implies F \approx 89.6 N$.

$A$ block of mass $10 \,kg$ is held at rest against a rough vertical wall $[\mu=0.5]$ under the action of a force $F$ as shown in the figure. The minimum value of $F$ required for it is ............ $N$ $(g=10 \,m/s^2)$.

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