To find the tensions in ropes $b$ and $c$ in a system at equilibrium, let's analyze each section step-by-step, assuming $g = 9.8 \, \text{m/s}^2$ for gravitational acceleration.

Step 1: Identify the Forces

• The 10 kg mass has a force due to gravity $F_{10} = 10 \times 9.8 = 98 \, \text{N}$.
• The 20 kg mass has a force due to gravity $F_{20} = 20 \times 9.8 = 196 \, \text{N}$.
• The 30 kg mass has a force due to gravity $F_{30} = 30 \times 9.8 = 294 \, \text{N}$.

Step 2: Tension in Rope "a"

Rope "a" supports only the 30 kg mass. Therefore, the tension in rope "a" is equal to the weight of the 30 kg mass: $T_a = F_{30} = 294 \, \text{N}$

Step 3: Tension in Rope "b"

Rope "b" supports both the 20 kg and the 30 kg masses. Thus, the tension in rope "b" is the combined weight of these two masses: $T_b = F_{20} + T_a = 196 + 294 = 490 \, \text{N}$

Step 4: Tension in Rope "c"

Rope "c" supports the 10 kg, 20 kg, and 30 kg masses. Thus, the tension in rope "c" is the combined weight of all three masses: $T_c = F_{10} + T_b = 98 + 490 = 588 \, \text{N}$

Final Answers

• Tension in rope "b": $490 \, \text{N}$
• Tension in rope "c": $588 \, \text{N}$

If you need further explanation or have additional questions, please let me know!

1. How would the tensions change if the 30 kg mass was removed?
2. What if the gravitational acceleration was different (e.g., on another planet)?
3. How can we calculate the force if the masses are moving up or down instead of being in equilibrium?
4. Why is it essential to consider equilibrium conditions when analyzing tension?
5. How would we analyze this problem if there were an additional external force acting on one of the masses?

Tip: When solving for tension in a system with multiple masses, start by analyzing from the bottom up to simplify force calculations.