Une masse de 10 kg est suspendue à deux câbles de même longueur. Si l’angle entre les câbles est de 45°, quelle est la force de tension dans chacun des câbles?

1. Break Down the Problem

We have a mass $m = 10 \, \text{kg}$ suspended by two cables forming an angle of $\theta = 45^\circ$. We need to find the tension $T$ in each cable.

2. Relevant Concepts

We can use the following concepts:

• The weight of the mass is given by $W = m \cdot g$, where $g \approx 9.81 \, \text{m/s}^2$.
• The system is in equilibrium, meaning that the vertical components of the tension must balance the weight of the mass.

3. Analysis and Detail

1. Calculate the weight of the mass: $W = m \cdot g = 10 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 98.1 \, \text{N}$

2. Since there are two cables forming an angle $\theta$ with respect to the vertical, the vertical component of the tension in each cable can be expressed as: $T_y = T \cdot \cos\left(\frac{\theta}{2}\right) = T \cdot \cos\left(22.5^\circ\right)$

3. The total vertical force from both cables is: $2 T_y = W \Rightarrow 2 T \cdot \cos\left(22.5^\circ\right) = 98.1 \, \text{N}$

4. Solving for $T$: $T = \frac{98.1}{2 \cdot \cos(22.5^\circ)}$

5. Calculate $\cos(22.5^\circ)$: $\cos(22.5^\circ) \approx 0.9239$

6. Plugging this value into the equation: $T = \frac{98.1}{2 \cdot 0.9239} \approx \frac{98.1}{1.8478} \approx 53.2 \, \text{N}$

4. Verify and Summarize

• The weight calculated is $98.1 \, \text{N}$.
• The force of tension in each cable calculated is approximately $53.2 \, \text{N}$, and this checks out with the equilibrium conditions.

Final Answer

The force of tension in each cable is approximately $53.2 \, \text{N}$.