The problem shown in the image asks to calculate the tensions $T_1$ and $T_2$ in a system with a mass of 1 kg hanging from two ropes. One rope is horizontal ($T_2$) and the other rope forms a $45^\circ$ angle from the ceiling ($T_1$).

Given:

• The mass $m = 1 \, \text{kg}$
• Angle $\theta = 45^\circ$
• Gravitational acceleration $g = 9.8 \, \text{m/s}^2$

Approach:

The forces acting on the mass can be resolved using Newton's second law. Since the system is in equilibrium, the sum of forces in both the vertical and horizontal directions must equal zero.

1. Vertical force balance (Y-axis): $T_1 \sin(45^\circ) = mg$ Here, $T_1$ has a vertical component balancing the gravitational force.

2. Horizontal force balance (X-axis): $T_1 \cos(45^\circ) = T_2$ The horizontal component of $T_1$ is balanced by the horizontal tension $T_2$.

Solution:

1. From the vertical balance equation, solve for $T_1$: $T_1 \sin(45^\circ) = 1 \times 9.8 = 9.8 \, \text{N}$ $T_1 = \frac{9.8}{\sin(45^\circ)} = \frac{9.8}{\frac{\sqrt{2}}{2}} = 9.8 \times \frac{2}{\sqrt{2}} = 13.86 \, \text{N}$

2. Using this $T_1$ value, substitute into the horizontal balance equation to solve for $T_2$: $T_2 = T_1 \cos(45^\circ) = 13.86 \times \frac{\sqrt{2}}{2} = 9.8 \, \text{N}$

Final Results:

• $T_1 = 13.86 \, \text{N}$
• $T_2 = 9.8 \, \text{N}$

Let me know if you'd like more details or if you have any questions!

Related questions:

1. How would the tensions change if the angle were $30^\circ$ instead of $45^\circ$?
2. What happens to the tensions if the mass is doubled to 2 kg?
3. Can you explain how to derive the equilibrium conditions for more complex systems?
4. How would the system behave if there were friction at the wall?
5. What if the rope connected to $T_1$ was elastic and stretched?

Tip:

In such problems, always resolve forces along the vertical and horizontal directions, ensuring equilibrium conditions are satisfied.