Problem 5 - College Physics

Question

A 250-N force is applied at a \(37^{\circ}\) angle to the surface of the end of a square bar. The surface is \(4.00 \mathrm{~cm}\) on a side. What are (a) the compressional stress and (b) the shear stress on the bar?

Step-by-Step Solution

Answer

Compressional stress is approximately 124,778 N/m², and shear stress is approximately 94,500 N/m².

Step 1: Understand the Problem

We need to find both the compressional stress and the shear stress on a square bar when a force is applied at an angle. Compressional stress is due to the component of force perpendicular to the surface, and shear stress is due to the component of force parallel to the surface.

Step 2: Calculate Surface Area

The surface area of the bar is given by the square of its side length. Since each side of the square surface is 4.00 cm, convert to meters for standard units: \[A = (0.04 ext{ m})^2 = 0.0016 ext{ m}^2\]

Step 3: Resolve Force into Components

The force of 250 N needs to be resolved into two components: one perpendicular to the surface (compressional) and one parallel to it (shear).- Compressional (normal) component: \[F_{ ext{normal}} = 250 imes ext{cos}(37^{ heta})\]- Shear component: \[F_{ ext{shear}} = 250 imes ext{sin}(37^{ heta})\]

Step 4: Calculate Compressional Stress

Use the normal component of the force to find the compressional stress, which is force per unit area:\[\sigma_{ ext{compressional}} = \frac{F_{ ext{normal}}}{A} = \frac{250 imes ext{cos}(37^ heta)}{0.0016}\]Calculate the value \[\sigma_{ ext{compressional}} \approx \frac{250 imes 0.7986}{0.0016} \approx 124,777.5 ext{ N/m}^2\]

Step 5: Calculate Shear Stress

Use the shear component of the force to find the shear stress, which is also force per unit area:\[\tau_{ ext{shear}} = \frac{F_{ ext{shear}}}{A} = \frac{250 imes ext{sin}(37^ heta)}{0.0016}\]Calculate the value \[\tau_{ ext{shear}} \approx \frac{250 imes 0.6018}{0.0016} \approx 94,500 ext{ N/m}^2\]

Key Concepts

Compressional Stress

Compressional stress is an important concept in physics that involves the squeeze or contraction experienced by a material under the application of force. When a force is applied perpendicular to the surface area of an object, it leads to compressional stress. This stress is calculated by dividing the normal, or perpendicular, force component by the surface area over which the force is distributed. In formula terms, compressional stress \( \sigma_\text{compressional} \) is given by:

• \( \sigma_\text{compressional} = \frac{F_\text{normal}}{A} \)

The normal force component, \( F_\text{normal} \), can be found using the cosine of the angle of the force with respect to the perpendicular:

• \( F_\text{normal} = F \times \cos(\theta) \)

This formula highlights the role of angles in determining how much of the applied force actually compresses the object. Understanding this can help when calculating forces on beams or building materials.

Shear Stress

Shear stress refers to the force per unit area exerted parallel to the plane of interest, causing deformation in the shape of the object. When discussing shear stress, the focus is on how the forces lead to a sliding motion within the material. It is particularly crucial in determining the behavior of materials under torsion or tangential forces.The formula for shear stress \( \tau_\text{shear} \) involves the parallel component of the force divided by the area:

• \( \tau_\text{shear} = \frac{F_\text{shear}}{A} \)

Here, the shear force component, \( F_\text{shear} \), is found using the sine of the angle between the force vector and the perpendicular:

• \( F_\text{shear} = F \times \sin(\theta) \)

Shear stress is important in engineering because it can predict how materials might fail due to sliding or tearing, which is crucial for designing structures that can withstand tangential loads.

Force Components

Understanding force components is essential for decomposing a force vector into parts that have different physical effects on an object. These components are identified by their directions relative to a reference frame, typically parallel and perpendicular to a surface.The force components are:

• Compressional or normal component: \( F_\text{normal} = F \times \cos(\theta) \)
• Shear component: \( F_\text{shear} = F \times \sin(\theta) \)

Where \( F \) is the total applied force and \( \theta \) is the angle of application.By recognizing the distinct effects of different force components, engineers and physicists can predict and manipulate how materials react to various forces; for example, in ensuring a bridge remains stable under pedestrian traffic or a building stands firm against wind.

Surface Area Calculation

Calculating surface area is a fundamental step in stress analysis as it allows for the accurate distribution of force over the object's surface. Surface area, denoted as \( A \), must often be in standard units such as square meters to align with other units in physics formulas.For a square surface with sides of length \( l \), the area is calculated as:

• \( A = l^2 \)

In this exercise, for a side length of 4.00 cm, it is first converted to meters as 0.04 m. Thus, the calculation becomes:

• \( A = (0.04 \text{ m})^2 = 0.0016 \text{ m}^2 \)

This conversion is crucial for applications ranging from laboratory experiments to designing industrial components, ensuring precision and consistency in calculations and applications.