Given the angles $\theta_1 = 45^\circ$ for $T_1$ and $\theta_2 = 60^\circ$ for $T_2$, we can solve for the tensions in each cable using the equilibrium conditions.

Step-by-Step Solution with Angles:

1. Determine the Weight of the Object: $W = mg = 80 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 784 \, \text{N}$

2. Decompose the Tensions into Components: $T_1 = T_1 \sin 45^\circ \hat{i} + T_1 \cos 45^\circ \hat{j}$ $T_2 = T_2 \sin 60^\circ \hat{i} + T_2 \cos 60^\circ \hat{j}$

   Using $\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos 60^\circ = \frac{1}{2}$, the components become: $T_{1x} = T_1 \frac{\sqrt{2}}{2}, \quad T_{1y} = T_1 \frac{\sqrt{2}}{2}$ $T_{2x} = T_2 \frac{\sqrt{3}}{2}, \quad T_{2y} = T_2 \frac{1}{2}$

3. Equilibrium in the x-direction: $T_1 \frac{\sqrt{2}}{2} = T_2 \frac{\sqrt{3}}{2}$ Solving for $T_1$: $T_1 = T_2 \frac{\sqrt{3}}{\sqrt{2}} = T_2 \frac{\sqrt{6}}{2}$

4. Equilibrium in the y-direction: $T_1 \frac{\sqrt{2}}{2} + T_2 \frac{1}{2} = W$ Substituting $T_1$ from the previous equation: $\left(T_2 \frac{\sqrt{6}}{2}\right) \frac{\sqrt{2}}{2} + T_2 \frac{1}{2} = 784$ Simplifying: $T_2 \frac{\sqrt{12}}{4} + T_2 \frac{1}{2} = 784$ $T_2 \frac{2\sqrt{3}}{4} + T_2 \frac{1}{2} = 784$ $T_2 \left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right) = 784$ $T_2 \left(\frac{\sqrt{3} + 1}{2}\right) = 784$ $T_2 (\sqrt{3} + 1) = 1568$ $T_2 = \frac{1568}{\sqrt{3} + 1}$

   Rationalizing the denominator: $T_2 = \frac{1568}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{1568 (\sqrt{3} - 1)}{3 - 1} = 784 (\sqrt{3} - 1)$ $T_2 = 784 (\sqrt{3} - 1) \approx 784 (1.732 - 1) \approx 784 \times 0.732 \approx 574.69 \, \text{N}$

5. Calculate $T_1$: $T_1 = T_2 \frac{\sqrt{6}}{2} = 574.69 \frac{\sqrt{6}}{2} \approx 574.69 \times 1.2247 \approx 703.68 \, \text{N}$

Tensions in Vector Form:

• $T_1 = 703.68 \left( \frac{\sqrt{2}}{2} \hat{i} + \frac{\sqrt{2}}{2} \hat{j} \right) = 703.68 \times 0.707 \hat{i} + 703.68 \times 0.707 \hat{j} \approx 497.69 \hat{i} + 497.69 \hat{j} \, \text{N}$
• $T_2 = 574.69 \left( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) = 574.69 \times 0.866 \hat{i} + 574.69 \times 0.5 \hat{j} \approx 497.69 \hat{i} + 287.35 \hat{j} \, \text{N}$

Conclusion:

The tensions in the cables are: $T_1 \approx 703.68 \, \text{N}$ $T_2 \approx 574.69 \, \text{N}$

Would you like more details on any part of this solution, or do you have any questions?

Here are 8 related questions to expand your understanding:

1. What if the object was not in equilibrium? How would the solution change?
2. How would you approach this problem if the angles were unknown?
3. What if the mass of the cables could not be ignored?
4. How would the tensions change if the angles were 30° and 60° instead?
5. How can we verify the calculated tensions using trigonometric identities?
6. What happens to the tensions if the mass of the object increases?
7. How do you calculate the horizontal and vertical components of the tensions?
8. What is the importance of breaking down vectors into components in such problems?

Tip: Always verify your final results with equilibrium conditions to ensure accuracy in problems involving forces.