Problem 32 - University Physics with Modern Physics

Question

A pickup truck is carrying a toolbox, but the rear gate of the truck is missing, so the box will slide out if it is set moving. The coefficients of kinetic and static friction between the box and the bed of the truck are 0.355 and 0.650 , respectively. Starting from rest, what is the shortest time this truck could accelerate uniformly to 30.0 \(\mathrm{m} / \mathrm{s}\) without causing the box to slide? Include a free-body diagram of the toolbox as part of your solution.

Step-by-Step Solution

Answer

The shortest time is approximately 4.71 seconds.

Step 1: Analyze the Forces on the Toolbox

In this step, we identify the forces acting on the toolbox. The forces include the gravitational force pulling it downward, the normal force from the bed of the truck pushing upward, and the frictional force resisting the motion parallel to the truck bed. The free-body diagram should show these forces: the weight of the toolbox downward, the normal force upward, and the maximum static friction force acting in the opposite direction of the truck's acceleration.

Step 2: Calculate the Maximum Static Friction Force

The maximum static frictional force is what keeps the toolbox from sliding. It can be calculated using the formula: \( f_s = \mu_s \cdot N \), where \( \mu_s \) is the coefficient of static friction (0.650) and \( N \), the normal force, equals the weight of the toolbox (\( mg \)). Thus, \( f_s = 0.650 \cdot mg \).

Step 3: Determine the Maximum Allowable Acceleration

The maximum static frictional force also represents the maximum force that can be exerted without the box sliding. Using Newton's second law, \( f_{net} = ma \), where \( f_{net} \) is the force exerted by static friction, we can find the maximum acceleration \( a \). So, \( \mu_s \cdot mg = ma \). Simplifying gives \( a = \mu_s \cdot g = 0.650 \cdot 9.8 \, \text{m/s}^2 \).

Step 4: Calculate Time to Reach Final Velocity

Now, use the kinematic equation \( v = u + at \), where \( v \) is the final velocity (30.0 m/s), \( u \) is the initial velocity (0 m/s), and \( a \) is the maximum allowable acceleration (6.37 m/s²). Solve for time \( t \) as follows: \( t = \frac{v}{a} = \frac{30.0}{6.37} \approx 4.71 \, \text{seconds} \).

Key Concepts

Friction

Friction plays a crucial role in determining whether objects stay put or start sliding. In this scenario, friction is what prevents the toolbox from sliding off the truck. There are two primary types of friction: static and kinetic. Static friction is the force that needs to be overcome to start moving an object at rest, while kinetic friction acts on an object that is already moving.

In the context of this exercise, static friction is what keeps the toolbox from sliding as the truck begins to accelerate. The static frictional force can be calculated using the equation:

• \( f_s = \mu_s \cdot N \)

where \( f_s \) is the static frictional force, \( \mu_s \) is the coefficient of static friction, and \( N \) is the normal force, which is equivalent to the weight of the toolbox in this scenario.

Key Points:

• Higher static friction means more force is required to start moving the object.
• Once the object begins to move, kinetic friction takes over, which is usually less than static friction.

Newton's laws of motion

Newton's laws of motion are fundamental principles in physics explaining the relationship between the motion of an object and the forces acting on it. For this problem, Newton's second law is indispensable. It states:

• \( f_{net} = ma \)

which relates the net force \( f_{net} \) acting on an object to its mass \( m \) and acceleration \( a \). In this exercise, the static frictional force is the net force that prevents the toolbox from sliding by allowing for a maximum acceleration.

By setting the static frictional force equal to \( ma \), we derive that:

• \( \mu_s \cdot mg = ma \)

Solving gives \( a = \mu_s \cdot g \), illustrating how the static friction affects the maximum acceleration possible before the box slides.

Key Points:

• Newton's laws provide a framework for analyzing motion and are essential for understanding mechanics problems.
• Newton's second law helps in calculating forces and acceleration, particularly when friction is involved.

Kinematic equations

Kinematic equations describe the motion of an object under uniform acceleration. In this scenario, we're particularly interested in how long it takes for the truck to reach a certain speed without causing the toolbox to slide. The equation used in the example is:

• \( v = u + at \)

where:

• \( v \) is the final velocity
• \( u \) is the initial velocity
• \( a \) is the acceleration
• \( t \) is the time

For our problem, \( u \) is 0 since the truck starts from rest, \( v \) is given as 30.0 m/s, and \( a \) comes from the maximum allowable acceleration from static friction. Solving the equation gives us the minimum time required to reach the final velocity without sliding.

Key Points:

• Kinematic equations help predict how an object moves over time under constant acceleration.
• Using known values, such as initial velocity, final velocity, and acceleration, it allows for the calculation of unknowns like time.

Free-body diagram

A free-body diagram is a simple, visual way to represent all forces acting on an object. It's an essential step in solving physics problems as it lays out the forces at play. In this problem, the free-body diagram for the toolbox should include:

• The weight of the toolbox acting downward
• The normal force exerted by the truck bed acting upward
• The static frictional force acting in the direction opposite to the truck's acceleration

Creating this diagram is crucial because it visually organizes the force vectors, assisting in the calculation and comprehension of the forces involved.

Key Points:

• A free-body diagram allows for clear visual representation of all forces acting on an object.
• It's a fundamental tool for analyzing forces and solving physics problems, especially those involving friction and motion.