Question

1. a 76.0 - kg person is being pulled away from a burning building as shown in figure 4.40. calculate the tension in the two ropes if the person is momentarily motionless. include a free - body diagram in your solution.

Explanation:

Step1: Calculate the weight of the person

The weight of the person is given by $W = mg$, where $m = 76.0\ kg$ and $g=9.8\ m/s^{2}$. So $W=76.0\times9.8 = 744.8\ N$.

Step2: Set up equilibrium equations

In the x - direction, $\sum F_x=T_2\cos10^{\circ}-T_1\sin15^{\circ}=0$, so $T_2\cos10^{\circ}=T_1\sin15^{\circ}$, and $T_2=\frac{T_1\sin15^{\circ}}{\cos10^{\circ}}$. In the y - direction, $\sum F_y=T_1\cos15^{\circ}+T_2\sin10^{\circ}-W = 0$.

Step3: Substitute $T_2$ into the y - direction equation

Substitute $T_2=\frac{T_1\sin15^{\circ}}{\cos10^{\circ}}$ into $T_1\cos15^{\circ}+T_2\sin10^{\circ}-W = 0$. We get $T_1\cos15^{\circ}+\frac{T_1\sin15^{\circ}}{\cos10^{\circ}}\sin10^{\circ}-744.8 = 0$. Factor out $T_1$: $T_1(\cos15^{\circ}+\frac{\sin15^{\circ}\sin10^{\circ}}{\cos10^{\circ}})=744.8$.
Calculate $\cos15^{\circ}+\frac{\sin15^{\circ}\sin10^{\circ}}{\cos10^{\circ}}\approx0.9659+\frac{0.2588\times0.1736}{0.9848}\approx0.9659 + 0.0457=1.0116$.
Then $T_1=\frac{744.8}{1.0116}\approx736\ N$.

Step4: Calculate $T_2$

Substitute $T_1\approx736\ N$ into $T_2=\frac{T_1\sin15^{\circ}}{\cos10^{\circ}}$. $T_2=\frac{736\times0.2588}{0.9848}\approx193\ N$.

Answer:

The tension $T_1\approx736\ N$ and $T_2\approx193\ N$