Question

You are designing a simple elevator system for an old warehouse that is being converted to loft apartments. A 22,500-N elevator is to be accelerated upward by connecting it to a counterweight by means of a light (but strong!) cable passing over a solid uniform disk- shaped pulley. The cable does not slip where it is in contact with the surface of the pulley. There is no appreciable friction at the axle of the pulley, but its mass is 875 kg and it is 1.50 m in diameter. (a) What mass should the counterweight have so that it will accelerate the elevator upward through 6.75 m in the first 3.00 s, starting from rest? (b) What is the tension in the cable on each side of the pulley?

Short answer

(a) Counterweight mass: 1940.1 kg. (b) Tensions: 20523.75 N and 21180.05 N.

Step-by-step solution

Calculate Acceleration of Elevator

First, we calculate the required acceleration to move the elevator 6.75 m in 3.00 s starting from rest. We use the kinematic equation: \[ s = ut + \frac{1}{2}at^2 \]where:- \( s = 6.75 \, \text{m} \) is the distance.- \( u = 0 \, \text{m/s} \) is the initial velocity.- \( t = 3.00 \, \text{s} \) is the time.Solving for acceleration \( a \):\[ 6.75 = 0 + \frac{1}{2}a(3.00)^2 \] \[ a = \frac{6.75}{0.5 \times 9} = 1.5 \, \text{m/s}^2 \]

Calculate Mass of Counterweight

Next, we use Newton's Second Law to find the required mass of the counterweight to provide this acceleration:Let \( m_1 = \frac{22,500}{9.81} \) be the mass of the elevator and \( m_2 \) be the mass of the counterweight. The tension \( T \) in the cable and the forces acting can be given by the equations:\[ m_1g - T = m_1a \]\[ T - m_2g = m_2a \]Adding these two equations, \[ m_1g - m_2g = (m_1 + m_2)a \]Solving for \( m_2 \):\[ m_2(g+a) = m_1(g-a) \]\[ m_2 = \frac{m_1(g-a)}{g+a} = \frac{(22,500/9.81)(9.81-1.5)}{9.81+1.5} \approx 1940.1 \, \text{kg} \]

Calculate Pulley Characteristics

The pulley being a solid uniform disk has a moment of inertia \( I = \frac{1}{2}mr^2 \), where \( m = 875 \, \text{kg} \) and \( r = \frac{1.5}{2} = 0.75 \, \text{m} \). Thus, \[ I = \frac{1}{2} \times 875 \times (0.75)^2 = 246.1 \, \text{kg}\cdot\text{m}^2 \]

Calculate Tension in the Cable

To find the tension on both sides of the pulley, we still use Newton’s and rotational dynamics. Using the equation:\[ \tau = I\alpha = T_2r - T_1r \]\( a = \alpha r \), hence \( \alpha = \frac{a}{r} \). Substituting:\[ (T_2 - T_1) \cdot 0.75 = 246.1 \times \frac{1.5}{0.75} \]\[ T_2 - T_1 = \frac{246.1 \times 1.5}{0.75^2} \approx 656.3 \, \text{N} \]Now, from Newton’s laws:- \( T_1 = m_1(g - a) = \frac{22,500}{9.81}(9.81 - 1.5) \approx 20523.75 \, \text{N} \)- \( T_2 = T_1 + 656.3 \approx 21180.05 \, \text{N} \)

Key concepts

Kinematics

Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause this motion. In this elevator physics problem, kinematics helps us determine the acceleration needed for the elevator to travel 6.75 meters in 3 seconds, starting from rest. We use the kinematic equation:

• \( s = ut + \frac{1}{2}at^2 \)

Here, \( s \) is the distance traveled, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time taken. Since the elevator starts from rest, \( u = 0 \), simplifying our equation to:

• \( s = \frac{1}{2}at^2 \)

By plugging in \( s = 6.75 \, \text{m} \) and \( t = 3.00 \, \text{s} \), we find that the elevator needs an acceleration of \( 1.5 \, \text{m/s}^2 \). This kinematic analysis allows us to set the stage for using dynamics to solve the problem.

Newton's Second Law

Newton's Second Law of motion states that the force acting on an object is equal to the mass of that object multiplied by its acceleration (\( F = ma \)). This principle is crucial in solving our elevator problem, as it is used to calculate the mass of the counterweight needed to accelerate the elevator. For the elevator:

• The gravitational force downwards is \( m_1g \), and the net force acting up is \( m_1a \) since it is accelerating upwards.

For the counterweight:

• The gravitational force downwards is \( m_2g \), and the net force acting up is negative \( m_2a \).

When we combine the equations of motion for both the elevator and counterweight, and solve for the counterweight's mass, we can ensure the system provides the correct acceleration. This allows us to find a counterweight mass of approximately 1940.1 kg, balancing the forces correctly.

Rotational Dynamics

When dealing with systems involving rotation, such as this problem with a pulley, we must consider rotational dynamics. The pulley in this elevator system is a solid disk, and as the cable moves over the pulley, it rotates. The rotational motion is characterized by torque (\( \tau \)) and angular acceleration (\( \alpha \)). The relationship between torque and angular acceleration is given by:

• \( \tau = I \alpha \)

Here, \( I \) is the moment of inertia of the pulley. The torque in this case arises due to the tension difference on either side of the pulley, noted as \( T_2 \) and \( T_1 \). The angular acceleration \( \alpha \) is directly related to the linear acceleration (\( a = \alpha r \)), enabling us to calculate the differences in tension as the pulley spins.

Moment of Inertia

The moment of inertia (\( I \)) is a property of a rotating object that determines how much torque is needed for a desired angular acceleration. It's similar to mass in linear motion and depends on the shape and mass distribution of the object. For a solid uniform disk, like our pulley, the moment of inertia is calculated using the formula:

• \( I = \frac{1}{2}mr^2 \)

where \( m \) is the mass of the disk, and \( r \) is the radius. In our example, with a mass of 875 kg and a radius of 0.75 m (half the diameter of 1.5 m), the moment of inertia comes out to be 246.1 kg\cdot m^2. This value is vital when determining the torque required to produce the needed angular acceleration, subsequently affecting calculations for tension in the cable across the pulley.