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: Define forces in x - direction

Since the person is motionless, the net force in the x - direction is zero. So, $T_{1x}-T_{2x} = 0$. Using trigonometry, $T_{1}\sin15^{\circ}-T_{2}\cos10^{\circ}=0$, which gives $T_{1}=\frac{T_{2}\cos10^{\circ}}{\sin15^{\circ}}$.

Step2: Define forces in y - direction

The net force in the y - direction is also zero. The weight of the person $W = mg$, where $m = 76.0$ kg and $g=9.8$ m/s². So, $T_{1y}-W = 0$. Using trigonometry, $T_{1}\cos15^{\circ}-mg = 0$.

Step3: Substitute $T_{1}$ from step 1 into step 2

Substitute $T_{1}=\frac{T_{2}\cos10^{\circ}}{\sin15^{\circ}}$ into $T_{1}\cos15^{\circ}-mg = 0$. We get $\frac{T_{2}\cos10^{\circ}\cos15^{\circ}}{\sin15^{\circ}}-mg = 0$. Solving for $T_{2}$:
\[

$$\begin{align*} T_{2}&=\frac{mg\sin15^{\circ}}{\cos10^{\circ}\cos15^{\circ}}\\ &=\frac{76\times9.8\times\sin15^{\circ}}{\cos10^{\circ}\cos15^{\circ}}\\ \end{align*}$$

\]
\[

$$\begin{align*} T_{2}&=\frac{76\times9.8\times0.259}{0.985\times0.966}\\ &=\frac{76\times9.8\times0.259}{0.952}\\ &\approx200.7\text{ N} \end{align*}$$

\]

Step4: Calculate $T_{1}$

Substitute $T_{2}$ into $T_{1}=\frac{T_{2}\cos10^{\circ}}{\sin15^{\circ}}$.
\[

$$\begin{align*} T_{1}&=\frac{200.7\times0.985}{0.259}\\ &\approx760.9\text{ N} \end{align*}$$

\]

Answer:

$T_{1}\approx760.9$ N, $T_{2}\approx200.7$ N