The problem asks to calculate the tension in the cable supporting a lift with a mass of 1200 kg under various conditions. Let's analyze and solve the problem step by step for each scenario.

Key Formula:

The tension in the cable can be calculated using Newton's Second Law: $T = m(g + a)$ Where:

• $T$ is the tension in the cable.
• $m = 1200 \, \text{kg}$ (mass of the lift).
• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity).
• $a$ is the acceleration of the lift. This can be positive (upward) or negative (downward) depending on the motion.

Case (a): Ascending at a constant velocity

At constant velocity, the net acceleration is zero ($a = 0$).

$T = m(g + 0) = 1200 \cdot 9.8 = 11760 \, \text{N}$

Answer for (a): $T = 11,760 \, \text{N}$.

Case (b): Ascending with an upward acceleration of $2.0 \, \text{m/s}^2$

Here, the acceleration is $a = +2.0 \, \text{m/s}^2$.

$T = m(g + a) = 1200(9.8 + 2.0) = 1200 \cdot 11.8 = 14160 \, \text{N}$

Answer for (b): $T = 14,160 \, \text{N}$.

Case (c): Descending with a downward acceleration of $3.0 \, \text{m/s}^2$

Here, the acceleration is $a = -3.0 \, \text{m/s}^2$ (negative because it's downward).

$T = m(g + a) = 1200(9.8 - 3.0) = 1200 \cdot 6.8 = 8160 \, \text{N}$

Answer for (c): $T = 8,160 \, \text{N}$.

Case (d): Descending with an upward acceleration of $4.0 \, \text{m/s}^2$

This might sound counterintuitive, but if the lift is descending with upward acceleration, it means the upward force exceeds the downward pull of gravity. Thus, $a = +4.0 \, \text{m/s}^2$.

$T = m(g + a) = 1200(9.8 + 4.0) = 1200 \cdot 13.8 = 16560 \, \text{N}$

Answer for (d): $T = 16,560 \, \text{N}$.

Summary of Results:

• (a): $T = 11,760 \, \text{N}$
• (b): $T = 14,160 \, \text{N}$
• (c): $T = 8,160 \, \text{N}$
• (d): $T = 16,560 \, \text{N}$

Let me know if you'd like more detailed explanations for any of these cases or further assistance!

Related Questions:

1. What happens to the tension if the lift's mass increases?
2. How is tension affected if the lift is in free fall ($a = -g$)?
3. How would you calculate the force exerted by the cable if friction is included?
4. What is the net force on the lift in case (c)?
5. How does the direction of acceleration impact the tension formula?

Tip:

Always identify the direction of motion and acceleration to determine whether to add or subtract the acceleration term ($a$) from gravity ($g$).