Problem 49 - College Physics

Question

A physicist's car has a small lead weight suspended from a string attached to the interior ceiling. Starting from rest, after a fraction of a second the car accelerates at a steady rate for about \(10 \mathrm{~s}\). During that time, the string (with the weight on the end of it) makes a backward (opposite the acceleration) angle of \(15.0^{\circ}\) from the vertical. Determine the car's (and the weight's) acceleration during the 10 -s interval.

Step-by-Step Solution

Answer

The car's acceleration is approximately \( 2.55 \text{ m/s}^2 \).

Step 1: Draw a Free-Body Diagram

Visualize the forces acting on the weight. The weight experiences a gravitational force downward and a tension force from the string at an angle of \(15.0^{\circ}\) from the vertical. Decompose these forces into vertical and horizontal components.

Step 2: Set Up Equations of Motion

In equilibrium in the horizontal direction: The horizontal component of the tension provides the net force causing acceleration. Use \(T \sin(15.0^{\circ}) = ma\). In the vertical direction: The vertical component of the tension balances the gravitational force, so \(T \cos(15.0^{\circ}) = mg\).

Step 3: Solve for Tension T

From the vertical equation, \(T = \frac{mg}{\cos(15.0^{\circ})}\). Use this equation to express the tension in terms of known variables \(m\) (mass) and \(g\) (acceleration due to gravity).

Step 4: Substitute Tension into Horizontal Equation

Replace \(T\) in the horizontal motion equation: \(\frac{mg}{\cos(15.0^{\circ})} \cdot \sin(15.0^{\circ}) = ma\).

Step 5: Solve for Acceleration a

Cancel out the mass \(m\) from the equation: \(\frac{g \cdot \tan(15.0^{\circ})}{\cos(15.0^{\circ})} = a\). Now, calculate \(a\) using \(g \approx 9.81 \text{ m/s}^2\).

Step 6: Substitute Numerical Values

Calculate the tangent of the angle and substitute: \(a = 9.81 \cdot \tan(15.0^{\circ})\), to get the final result. The tangent value \(\tan(15.0^{\circ}) \approx 0.2679\).

Key Concepts

Free-Body Diagram

When tackling problems involving force and motion, a free-body diagram (FBD) is a crucial starting point. It visually represents the forces acting on a single object, allowing us to analyze the situation clearly. Here, the weight hanging from the car's ceiling is subject to different forces.

To draw the FBD for this problem:

• Identify all the forces: gravitational force downward and tension in the string.
• Decompose the tension into two components: one along the vertical axis and another along the horizontal axis due to the angle formed with the vertical.

The angle of the string, which is backward relative to the car's motion, tells us that there's a horizontal acceleration. This is why the tension has components both up (against gravity) and backward (against acceleration) at 15.0° from the vertical.

Acceleration Calculation

Once the forces are identified using the free-body diagram, we can proceed to calculate the acceleration. The strategy involves solving the equations derived from the balance of forces and motion.

Here's how the calculation unfolds:

• In the horizontal direction, the net force arises from the tension's horizontal component, which is responsible for the car's acceleration.
• We use the equation: \[ T \sin(15.0^{\circ}) = ma \]
• The vertical forces are balanced as such: \[ T \cos(15.0^{\circ}) = mg \]

By solving these equations, particularly focusing on the horizontal motion equation, we derive the expression for acceleration. Importantly, realize that because these equations incorporate the angles, they describe how the forces translate into motion.

Forces and Motion Analysis

After setting up the motion equations and dissecting the forces, we can dig deeper to understand the interplay of forces that results in acceleration.

Through the analysis process:

• We rearrange the vertical force balance equation to find the tension \( T \) by solving: \[ T = \frac{mg}{\cos(15.0^{\circ})} \]
• By substituting the tension \( T \) back into the horizontal equation, we eliminate \( m \):\[ \frac{g \cdot \tan(15.0^{\circ})}{\cos(15.0^{\circ})} = a \]
• Calculating with \( g = 9.81 \text{ m/s}^2 \), and \( \tan(15.0^{\circ}) \approx 0.2679 \), we find the car's acceleration.

This step of combining the equations highlights how Newton's laws reveal the dynamics at play, allowing us to calculate the precise acceleration the car and weight experience over the 10-second interval.