Let's solve the given problem step by step.

Problem Recap:

A heavy object of mass 80 kg is suspended from two cables as shown in Figure 1. Assuming the system is in equilibrium, find the tension in each cable (ignoring the mass of the cables). Express these as vectors $T_1$ in the direction $\overrightarrow{OA}$ and $T_2$ in the direction $\overrightarrow{OB}$. You can use $g = 9.8 \, \text{m/s}^2$.

Step-by-Step Solution:

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

2. Equilibrium Conditions: Since the object is in equilibrium, the sum of forces in both the x and y directions must be zero.

3. Decompose the Tensions into Components: Let's denote the angles between the cables and the vertical as $\theta_1$ and $\theta_2$ for cables $T_1$ and $T_2$ respectively.

   We can write the tension vectors as: $T_1 = T_{1x} \hat{i} + T_{1y} \hat{j}$ $T_2 = T_{2x} \hat{i} + T_{2y} \hat{j}$

   where: $T_{1x} = T_1 \sin \theta_1 \quad \text{and} \quad T_{1y} = T_1 \cos \theta_1$ $T_{2x} = T_2 \sin \theta_2 \quad \text{and} \quad T_{2y} = T_2 \cos \theta_2$

4. Equilibrium in the x-direction: $T_1 \sin \theta_1 = T_2 \sin \theta_2$

5. Equilibrium in the y-direction: $T_1 \cos \theta_1 + T_2 \cos \theta_2 = W$

Solution Strategy:

• Use the equilibrium conditions to solve for $T_1$ and $T_2$.
• Since the angles $\theta_1$ and $\theta_2$ are not provided, we assume symmetry or calculate based on geometry if the lengths are provided or can be inferred.

Assuming symmetry for simplicity (which means $\theta_1 = \theta_2 = \theta$):

Symmetric Case:

If the system is symmetric, $\theta_1 = \theta_2 = \theta$, then:

$T_1 \sin \theta = T_2 \sin \theta \implies T_1 = T_2$

And for the y-component:

$2 T_1 \cos \theta = W \implies T_1 \cos \theta = \frac{W}{2} = \frac{784}{2} = 392 \, \text{N}$

Therefore:

$T_1 = \frac{392}{\cos \theta}$

$T_2 = T_1 = \frac{392}{\cos \theta}$

So, the tensions in the cables $T_1$ and $T_2$ can be expressed as:

$T_1 = T_2 = \frac{392}{\cos \theta}$

Final Expression in Vector Form:

• $T_1 = \left(\frac{392 \sin \theta}{\cos \theta}\right) \hat{i} + 392 \hat{j}$
• $T_2 = \left(\frac{392 \sin \theta}{\cos \theta}\right) \hat{i} + 392 \hat{j}$

Conclusion:

The tensions in the cables $T_1$ and $T_2$ are given by $\frac{392}{\cos \theta} \, \text{N}$ in magnitude. For exact numerical values, the angles $\theta$ must be known or calculated.

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 happens to the tensions if the angles are unequal?
2. How would you calculate the angles if the lengths of the cables were provided?
3. What are the horizontal components of the tensions in the cables?
4. How do you express the vector forms of the tensions if the angles are known?
5. How would the problem change if the mass of the cables were not negligible?
6. Can this method be applied to more than two cables supporting the object?
7. How does the tension change if the mass of the object increases?
8. What role does the gravitational constant $g$ play in the tension calculations?

Tip: Always consider the symmetry in problems involving equilibrium, as it can greatly simplify the calculations.