A mass of 10 ; kg is suspended by a rope of l | vedclass

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(A) Let $T$ be the tension in the upper half of the rope. The lower half of the rope is vertical and supports the mass of $10 \; kg$,so the tension in

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$100$

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(A) Let $T$ be the tension in the upper half of the rope. The lower half of the rope is vertical and supports the mass of $10 \; kg$,so the tension in the lower half is $T_{lower} = mg = 10 	imes 10 = 100 \; N$.At the mid-point where force $F$ is applied,we consider the equilibrium of forces.The horizontal component of the tension $T$ in the upper rope must balance the applied force $F$: $T \sin 45^{\circ} = F$.The vertical component of the tension $T$ in the upper rope must balance the downward tension from the lower rope: $T \cos 45^{\circ} = T_{lower} = 100 \; N$.Dividing the two equations: $\frac{T \sin 45^{\circ}}{T \cos 45^{\circ}} = \frac{F}{100}$.$	an 45^{\circ} = \frac{F}{100}$.Since $	an 45^{\circ} = 1$,we get $1 = \frac{F}{100}$,which implies $F = 100 \; N$.

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