Problem 23 - College Physics

Question

At a construction site, a \(22.0 \mathrm{~kg}\) bucket of concrete is connected over a very light frictionless pulley to a \(375 \mathrm{~N}\) box on the roof of a building. (See Figure \(5.55 .)\) There is no appreciable friction on the box, since it is on roller bearings. The box starts from rest. (a) Make free- body diagrams of the bucket and the box. (b) Find the acceleration of the bucket. (c) How fast is the bucket moving after it has fallen \(1.50 \mathrm{~m}\) (assuming that the box has not yet reached the edge of the roof)?

Step-by-Step Solution

Answer

The bucket's acceleration is approximately \(1.64 \text{ m/s}^2\) and its speed after falling \(1.5 \text{ m}\) is \(2.2 \text{ m/s}\).

Step 1: Draw Free Body Diagrams

Create two free body diagrams: one for the bucket and one for the box. For the bucket, include gravitational force \( F_{g,b} = m_b imes g \) acting downward, where \( m_b = 22.0 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \). For the box, include gravitational force \( F_{g,box} = 375 \text{ N} \) acting down the side, tension force \( T \) acting upward on the bucket, and tension force \( T \) acting horizontally on the box.

Step 2: Calculate the Forces

Calculate the gravitational force on the bucket: \( F_{g,b} = 22.0 \times 9.8 = 215.6 \text{ N} \). For the box, the gravitational force is already given as \( 375 \text{ N} \). Use these calculations to understand the forces at play.

Step 3: Write Equations for Motion

Write Newton's second law for both the bucket and the box. For the bucket: \( m_b imes a = T - F_{g,b} \). For the box, since it experiences gravitational force only in the direction of motion: \( m_{box} \times a = F_{g,box} - T \), where \( m_{box} = 375 / 9.8 \).

Step 4: Solve for Tension and Acceleration

Combine the equations from the previous step. From the bucket: \( 22.0a = T - 215.6 \). From the box: \((375 / 9.8) \times a = 375 - T\). Add the two equations to eliminate \( T \) and solve for \( a \). After solving, \( a \approx 1.64 \text{ m/s}^2 \).

Step 5: Calculate Final Speed

Use the kinematic equation \( v^2 = u^2 + 2as \) to find the final speed \( v \). Here, \( u = 0 \text{ m/s} \) (since the bucket starts from rest), \( a = 1.64 \text{ m/s}^2 \), and \( s = 1.50 \text{ m} \). Plug the values into the equation to solve: \( v^2 = 0 + 2 \times 1.64 \times 1.50 \), hence \( v \approx 2.2 \text{ m/s} \).

Key Concepts

Free-body diagrams

A free-body diagram is a crucial tool for analyzing the forces acting on an object in a system. It's a simplified representation of the object with vectors showing all the forces exerted on it. For our exercise, we have a bucket and a box connected over a frictionless pulley.

• For the bucket: The gravitational force, \( F_{g,b} = m_b \times g \), acts downwards. Given \( m_b = 22.0 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \), this force becomes \( 215.6 \text{ N} \).
• For the box: It has a gravitational force of \( 375 \text{ N} \), and it experiences a tension force \( T \) from the rope, acting in the direction opposite of the gravitational pull.

These diagrams simplify the complex interaction of forces, allowing us to apply Newton's laws effectively. By isolating the forces acting on each object, we can accurately capture the effects of tension and gravity in the system.

Kinematics

Kinematics involves the motion of objects, described by parameters such as velocity, acceleration, and displacement without considering the forces causing the motion. In this exercise, kinematics helps us calculate the final speed of the bucket after falling a certain distance.

• The bucket starts at rest, so the initial velocity \( u = 0 \text{ m/s} \).
• Using the acceleration \( a \approx 1.64 \text{ m/s}^2 \) found through Newton's laws, we apply the kinematic equation: \( v^2 = u^2 + 2as \).
• Substitute \( a = 1.64 \text{ m/s}^2 \) and \( s = 1.50 \text{ m} \), we find that the final velocity \( v \approx 2.2 \text{ m/s} \).

Understanding kinematics allows us to predict future motion, providing insights into how a system's properties change over time.

Gravitational force

Gravitational force is the force of attraction between two masses. It's vital in this exercise as it directly affects both the bucket and the box. This force is calculated using \( F = m \times g \) where \( m \) is the mass and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity.

• For the bucket of \( 22.0 \text{ kg} \), the gravitational force is \( F_{g,b} = 215.6 \text{ N} \).
• The box, with a pre-calculated gravitational force of \( 375 \text{ N} \), indicates its weight in the typical force of gravity.

Gravitational force is crucial in determining how objects move and the forces that need to counteract or balance for stability. In this case, it sets the bucket in motion when paired with tension, influencing the acceleration and dynamics of the system.