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(B) Let $F$ be the applied force at an angle $	heta = 60^{\circ}$ with the horizontal.The vertical components of forces are the normal force $N$,the

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$\frac{20}{2 + \sqrt{3}} \, N$

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(B) Let $F$ be the applied force at an angle $	heta = 60^{\circ}$ with the horizontal.The vertical components of forces are the normal force $N$,the vertical component of the applied force $F \sin 60^{\circ}$,and the weight $mg$.Balancing vertical forces: $N + F \sin 60^{\circ} = mg \Rightarrow N = mg - F \sin 60^{\circ}$.Given $m = 1 \, kg$,$g = 10 \, m/s^2$,and $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we have $N = 10 - \frac{\sqrt{3}}{2} F$.The block just starts to move when the horizontal component of the applied force equals the limiting friction: $F \cos 60^{\circ} = \mu N$.Given $\mu = 0.5$ and $\cos 60^{\circ} = 0.5$,we substitute the values: $0.5 F = 0.5 (10 - \frac{\sqrt{3}}{2} F)$.Dividing both sides by $0.5$: $F = 10 - \frac{\sqrt{3}}{2} F$.Rearranging terms: $F + \frac{\sqrt{3}}{2} F = 10 \Rightarrow F (1 + \frac{\sqrt{3}}{2}) = 10 \Rightarrow F (\frac{2 + \sqrt{3}}{2}) = 10$.Thus,$F = \frac{20}{2 + \sqrt{3}} \, N$.

$A$ block of mass $1 \, kg$ is at rest on a horizontal table. The coefficient of static friction between the block and the table is $0.5$. The magnitude of the force acting upwards at an angle of $60^{\circ}$ from the horizontal that will just start the block moving is

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