11. an 85 kg object is suspended from a ceiling and attached to a wall. what is the tension in the left…

Explanation:

Step1: Calculate the weight of the object

The weight of the object is given by $F = mg$, where $m = 85\ kg$ and $g=9.8\ m/s^{2}$. So $F=85\times9.8 = 833\ N$.

Step2: Set up equilibrium equations

Let the tension in the left - hand rope be $T_1$ and the tension in the other rope be $T_2$. In the vertical direction, the sum of the vertical components of the tensions equals the weight of the object. The vertical component of $T_1$ is $T_1\sin70^{\circ}$ and the vertical component of $T_2$ is $T_2\sin65^{\circ}$. In the horizontal direction, the horizontal components of the tensions balance each other, so $T_1\cos70^{\circ}=T_2\cos65^{\circ}$, which gives $T_2=\frac{\cos70^{\circ}}{\cos65^{\circ}}T_1$. Substituting into the vertical - equilibrium equation: $T_1\sin70^{\circ}+T_2\sin65^{\circ}=mg$. Replace $T_2$ with $\frac{\cos70^{\circ}}{\cos65^{\circ}}T_1$: [T_1\sin70^{\circ}+\frac{\cos70^{\circ}}{\cos65^{\circ}}T_1\sin65^{\circ}=833] [T_1\left(\sin70^{\circ}+\frac{\cos70^{\circ}\sin65^{\circ}}{\cos65^{\circ}}\right)=833] We know that $\sin70^{\circ}\approx0.9397$, $\cos70^{\circ}\approx0.3420$, $\sin65^{\circ}\approx0.9063$, $\cos65^{\circ}\approx0.4226$. [T_1\left(0.9397+\frac{0.3420\times0.9063}{0.4226}\right)=833] [T_1\left(0.9397 + 0.730\right)=833] [T_1\times1.6697 = 833]

Step3: Solve for $T_1$

[T_1=\frac{833}{1.6697}\approx500\ N]

Answer:

C. 500 N