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(D) Let the tension in the upper part of the rope be $T_1$ and the tension in the lower part be $T_2$. Since the $1 \,kg$ mass is in equilibrium, the

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$10 \,N$

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(D) Let the tension in the upper part of the rope be $T_1$ and the tension in the lower part be $T_2$. Since the $1 \,kg$ mass is in equilibrium, the tension in the lower part of the rope is $T_2 = mg = 1 \,kg 	imes 10 \,m/s^2 = 10 \,N$. Now, consider the equilibrium of the point where the force $F$ is applied. Resolving the forces into horizontal and vertical components: For horizontal equilibrium: $T_1 \sin 45^{\circ} = F$ For vertical equilibrium: $T_1 \cos 45^{\circ} = T_2 = 10 \,N$ Dividing the two equations: $\frac{T_1 \sin 45^{\circ}}{T_1 \cos 45^{\circ}} = \frac{F}{10}$ $	an 45^{\circ} = \frac{F}{10}$ Since $	an 45^{\circ} = 1$, we get $1 = \frac{F}{10}$, which implies $F = 10 \,N$.

$A$ $1 \,kg$ mass is suspended from the ceiling by a rope. $A$ horizontal force $F$ is applied at the midpoint of the rope so that the upper part of the rope makes an angle of $45^{\circ}$ with respect to the vertical axis as shown in the figure. The magnitude of $F$ is:

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