Problem 96

Question

A rock with mass \(m=3.00 \mathrm{kg}\) is suspended from the roof of an elevator by a light cord. The rock is totally immersed in a bucket of water that sits on the floor of the elevator, but the rock doesn't touch the bottom or sides of the bucket. (a) When the elevator is at rest, the tension in the cord is 21.0 \(\mathrm{N}\) . Calculate the volume of the rock. (b) Derive an expression for the tension in the cord when the elevator is accelerating upward with an acceleration of magnitude a. Calculate the tension when \(a=2.50 \mathrm{m} / \mathrm{s}^{2}\) upward. (c) Derive an expression for the tension in the cord when the elevator is accelerating downward with an acceleration of magnitude \(a\) . Calculate the tension when \(a=2.50 \mathrm{m} / \mathrm{s}^{2}\) downward. (d) What is the tension when the elevator is in free fall with a downward acceleration equal to \(g\) ?

Step-by-Step Solution

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Answer
(a) Volume: 0.000857 m³; (b) Tension (upward) when a=2.5 m/s²: 28.5 N; (c) Tension (downward) when a=2.5 m/s²: 13.5 N; (d) Tension in free fall: 0 N.
1Step 1: Calculate the Buoyant Force
When the elevator is at rest, the forces acting on the rock include the tension in the cord, the gravitational force, and the buoyant force due to water. The equation for equilibrium when the elevator is at rest is: \[ T = mg - F_b \]Given that the tension \( T = 21.0 \, \text{N} \) and \( m = 3.00 \, \text{kg} \), then:\[ 21.0 = 3.00 \times 9.8 - F_b \] Solving for \( F_b \) gives the buoyant force:\[ F_b = 3.00 \times 9.8 - 21.0 = 29.4 - 21.0 = 8.4 \, \text{N} \]
2Step 2: Determine the Volume of the Rock
The buoyant force \( F_b \) is equal to the weight of the displaced water, which is \( \rho V g \), where \( \rho = 1000 \, \text{kg/m}^3 \) for water, \( V \) is the volume of the object, and \( g = 9.8 \, \text{m/s}^2 \). Setting \( F_b = \rho V g \), we get:\[ 8.4 = 1000 \times V \times 9.8 \]Solving for \( V \):\[ V = \frac{8.4}{9800} = 0.000857 \, \text{m}^3 \]
3Step 3: Derive Expression for Tension When Elevator Accelerates Upward
When the elevator accelerates upward, the effective gravitational force \( g' = g + a \). The tension \( T \) is given by:\[ T = m(g + a) - F_b \]Substitute for \( F_b \) in terms of volume \( V \):\[ T = m(g + a) - \rho V g \]For \( a = 2.50 \, \text{m/s}^2\), calculate:\[ T = 3.00(9.8 + 2.5) - 1000 \times 0.000857 \times 9.8 \]\[ T = 3.00 \times 12.3 - 8.4 = 36.9 - 8.4 = 28.5 \, \text{N} \]
4Step 4: Derive Expression for Tension When Elevator Accelerates Downward
When the elevator accelerates downward, the effective gravitational force \( g' = g - a \). The tension is:\[ T = m(g - a) - F_b \]Substituting \( F_b \):\[ T = m(g - a) - \rho V g \]For \( a = 2.50 \, \text{m/s}^2\), calculate:\[ T = 3.00(9.8 - 2.5) - 1000 \times 0.000857 \times 9.8 \]\[ T = 3.00 \times 7.3 - 8.4 = 21.9 - 8.4 = 13.5 \, \text{N} \]
5Step 5: Tension During Free Fall
In free fall, the elevator's acceleration \( a = g \) downward, hence \( g' = 0 \). The only forces are the weight and buoyant force, but due to free fall, effectively the object feels no weight:\[ T = m(g - g) - F_b = 0 - 8.4 = -8.4 \, \text{N} \]This means the tension is zero when considering inner tension only.

Key Concepts

Buoyant forceTension in physicsElevator physics problems
Buoyant force
When an object is immersed in a fluid, it experiences an upward force known as the buoyant force. This force results from the pressure difference between the top and bottom of the object. In this scenario, it helps keep the rock suspended in the water while it is attached to the cord.

The buoyant force is calculated using the formula:
  • \( F_b = \rho V g \)
  • where \( F_b \) is the buoyant force, \( \rho \) is the density of the fluid (1000 kg/m³ for water), \( V \) is the volume of the fluid displaced, and \( g \) is the acceleration due to gravity (9.8 m/s²).
In our exercise, we observed that the tension in the cord is not only due to the weight of the rock but also affected by this buoyant force. The buoyant force essentially reduces the effective weight of the rock, which is why the tension measured when the elevator is stationary is less than its gravitational weight alone.
Tension in physics
Tension is a force exerted along the length of a cord, rope, or any other object in its potential to undergo stretching. In this problem, tension holds the rock suspended within the elevator while also balancing against gravity and buoyant forces.

When analyzing tension, consider:
  • Tension aligns along a single direction, typically opposing the force of gravity.
  • It varies based on other forces at play such as buoyant forces or acceleration.
  • In our problem, if the elevator is at rest, tension equals the gravitational force reduced by the buoyant force: \( T = mg - F_b \).
  • As the elevator accelerates, the tension adjusts due to additional or reduced effective weight based on the direction of acceleration. This is seen in equations for upward \( T = m(g + a) - F_b \) and downward \( T = m(g - a) - F_b \) accelerations.
Understanding tension is crucial as it ensures that the system remains in equilibrium or provides insight into changes when the system is disturbed, such as by motion.
Elevator physics problems
Elevator physics involves understanding how different forces interact when an object experiences acceleration in a confined system like an elevator. These problems often illustrate how apparent weight changes due to acceleration forces beyond gravity.

Consider these key points:
  • In an elevator at rest, forces are balanced, and tension counteracts the gravitational force minus any buoyant forces.
  • When an elevator accelerates upward, the effective weight increases due to additional upward force, thereby increasing tension. This is reflected in the formula \( T = m(g + a) - F_b \).
  • During downward acceleration, effective weight decreases, which reduces the tension as per \( T = m(g - a) - F_b \).
  • In free fall, where acceleration equals gravitational force \( g \), tension becomes zero because the system is in a state of apparent weightlessness.
Understanding these dynamics allows for predictions of force changes within enclosed systems under various motion conditions, helping solve complex physics problems related to real-world applications.