To solve for the tension in the cable when the scale reads 420 N, we need to approach this problem systematically.

Step 1: Analyze the forces acting on the student.

• The student's weight is given as $600 \, \text{N}$.
• The scale reads $420 \, \text{N}$, which is the normal force acting on the student from the elevator floor.

The difference between the student's weight and the scale reading indicates that the elevator is accelerating.

Step 2: Calculate the student's mass.

To find the mass of the student, we use the formula: $m_{\text{student}} = \frac{F_{\text{weight}}}{g} = \frac{600 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 61.22 \, \text{kg}$ where $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity.

Step 3: Find the acceleration of the elevator.

The normal force $F_N$ acting on the student from the scale is less than their weight, meaning the elevator is accelerating downwards.

The net force on the student is given by Newton's second law: $F_{\text{net}} = m_{\text{student}} \cdot a$ where $a$ is the acceleration of the elevator.

The net force is the difference between the student's weight and the normal force: $F_{\text{net}} = 600 \, \text{N} - 420 \, \text{N} = 180 \, \text{N}$ Now, solving for the acceleration $a$: $a = \frac{F_{\text{net}}}{m_{\text{student}}} = \frac{180 \, \text{N}}{61.22 \, \text{kg}} \approx 2.94 \, \text{m/s}^2$ The elevator is accelerating downwards with an acceleration of $2.94 \, \text{m/s}^2$.

Step 4: Calculate the tension in the cable.

Now, we focus on the entire system, which includes both the elevator and the student. The total mass of the system is $900 \, \text{kg}$.

The net force acting on the system is: $F_{\text{net, system}} = m_{\text{system}} \cdot a = 900 \, \text{kg} \cdot 2.94 \, \text{m/s}^2 = 2646 \, \text{N}$ The gravitational force acting on the system is: $F_{\text{gravity}} = m_{\text{system}} \cdot g = 900 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 8820 \, \text{N}$

Since the system is accelerating downwards, the tension $T$ in the cable is less than the gravitational force: $T = F_{\text{gravity}} - F_{\text{net, system}} = 8820 \, \text{N} - 2646 \, \text{N} = 6174 \, \text{N}$

Final Answer:

The tension in the cable is approximately: $\boxed{6174 \, \text{N}}$

Would you like more details or have any questions?

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Here are 5 related questions to expand your understanding:

1. How would the tension change if the elevator were accelerating upwards?
2. What would the tension be if the elevator moved at constant velocity?
3. How would the normal force on the student change if the acceleration increased?
4. How does the elevator's total mass affect the tension in the cable?
5. What happens to the tension in the cable when the elevator is in free fall?

Tip: In elevator problems, pay close attention to whether the elevator is accelerating or moving at constant velocity—this significantly affects the forces involved.