Problem 64 - Physics

Question

A square plate is \(1.0 \times 10^{-2} \mathrm{~m}\) thick, measures \(3.0 \times 10^{-2} \mathrm{~m}\) on a side, and has a mass of \(7.2 \times 10^{-2} \mathrm{~kg}\). The shear modulus of the material is \(2.0 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\). One of the square faces rests on a flat horizontal surface, and the coefficient of static friction between the plate and the surface is \(0.90 .\) A force is applied to the top of the plate, as in Figure \(10-32 a\). Determine (a) the maximum possible amount of shear stress, (b) the maximum possible amount of shear strain, and (c) the maximum possible amount of shear deformation \(\Delta X\) (see Figure \(10-32 b\) ) that can be created by the applied force just before the plate begins to move.

Step-by-Step Solution

Answer

(a) Maximum shear stress: 705.56 N/m². (b) Maximum shear strain: 3.53 x 10⁻⁸. (c) Shear deformation \( \Delta X \): 3.53 x 10⁻¹⁰ m.

Step 1: Identify Shear Stress Formula

The shear stress \( \tau \) is calculated as the force \( F \) divided by the area \( A \). The area \( A \) of the square face on a side is \((3.0 \times 10^{-2} \text{ m})^2\). The maximum force that can be applied without sliding is impacted by static friction: \( F_{\text{max}} = \mu \cdot \text{normal force (mass \( \times \) gravity)} \). Substitute to find \( \tau = \frac{F_{\text{max}}}{A} \).

Step 2: Calculate Area

The area of the square face is: \[A = (3.0 \times 10^{-2} \text{ m})^2 = 9.0 \times 10^{-4} \text{ m}^2.\]

Step 3: Calculate Maximum Force due to Friction

The maximum force from static friction \( F_{\text{max}} \) is: \[F_{\text{max}} = \mu \cdot m g = 0.90 \cdot 7.2 \times 10^{-2} \text{ kg} \times 9.8 \text{ m/s}^2.\]Calculating gives \( F_{\text{max}} \approx 0.635 \text{ N} \).

Step 4: Compute Maximum Shear Stress

Substitute the values into the shear stress formula:\[\tau = \frac{F_{\text{max}}}{A} = \frac{0.635 \text{ N}}{9.0 \times 10^{-4} \text{ m}^2} = 705.56 \text{ N/m}^2.\]Thus, the maximum shear stress is \( 705.56 \text{ N/m}^2 \).

Step 5: Determine Shear Strain using Shear Modulus

Shear strain \( \gamma \) is related to shear stress \( \tau \) and shear modulus \( G \) by:\[\gamma = \frac{\tau}{G} = \frac{705.56 \text{ N/m}^2}{2.0 \times 10^{10} \text{ N/m}^2}.\]Calculating, \( \gamma \approx 3.53 \times 10^{-8} \).

Step 6: Calculate Shear Deformation \( \Delta X \)

Shear deformation \( \Delta X \) can be found using:\[\Delta X = \gamma \times \text{thickness} = 3.53 \times 10^{-8} \times 1.0 \times 10^{-2} \text{ m } = 3.53 \times 10^{-10} \text{ m} \].

Key Concepts

Shear Strain

Shear strain, denoted as \( \gamma \), is a measure of how much a material deforms under shear stress. It describes the relative displacement between parallel planes within a material. Think of shear strain as observing how much a square distorts into a parallelogram when a force is applied parallel to its surfaces.

In the given exercise, we calculate it using the formula \( \gamma = \frac{\tau}{G} \). Here, \( \tau \) represents the shear stress and \( G \) is the shear modulus of the material, which quantifies how easily the material deforms under shear stress. For example, by applying a calculated shear stress of \( 705.56 \text{ N/m}^2 \) to our material with a shear modulus of \( 2.0 \times 10^{10} \text{ N/m}^2 \), we calculate the shear strain to be approximately \( 3.53 \times 10^{-8} \).

This tiny amount of strain indicates that the material is quite resistant to shear deformation, showing only minimal bending under the applied force. Understanding shear strain is crucial for materials used in structures like bridges or buildings, ensuring they withstand forces without bending too much.

Shear Deformation

Shear deformation refers to the change in shape of a material due to shear stress. While shear strain measures the angle change, shear deformation provides a physical measurement of distortion. It particularly looks at the horizontal shift between layers within the structure.

The formula \( \Delta X = \gamma \times \text{thickness} \), helps calculate this transformation, where \( \Delta X \) indicates the amount of shear deformation and "thickness" is the initial dimension perpendicular to the force. From the exercise, we found the shear strain to be \( 3.53 \times 10^{-8} \) and when multiplied by the thickness \((1.0 \times 10^{-2} \text{ m})\), we determine the shear deformation to be \( 3.53 \times 10^{-10} \text{ m} \).

Considering its minute measure, this shows how resilient the material is against external forces, underscoring why the choice of material and its properties are key in design and engineering.

Frictional Force

Frictional force is the force resisting the relative motion of solid surfaces. In the context of the exercise, it is crucial for preventing the square plate from sliding. The maximum frictional force, \( F_{\text{max}} \), acts before the plate starts to move and is influenced by the static friction coefficient and the normal force.

We calculated it using the formula: \( F_{\text{max}} = \mu \cdot m \cdot g \), where \( \mu = 0.90 \) is the coefficient of static friction, \( m = 7.2 \times 10^{-2} \text{ kg} \) is the mass of the plate, and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. This calculation provided \( F_{\text{max}} \approx 0.635 \text{ N} \).

Understanding the role of frictional force is vital as it ensures stability and prevents unwanted motion in systems, making it an integral part of safe and reliable design.

Shear Modulus

The shear modulus, represented by \( G \), is a fundamental property that describes how a material responds to shear stress. It links the amount of shear stress a material can withstand to its resultant shear strain.

A high shear modulus signifies that the material requires a lot of force to change shape, making it strong and rigid under shear forces. For the embodiment in our exercise, \( G \) is given as \( 2.0 \times 10^{10} \text{ N/m}^2 \), indicating a robust material that deforms minimally under stress.

Such properties are significant when choosing materials for construction or manufacturing, as the shear modulus helps predict how a material will behave when an external force is applied parallel to its surface. Understanding this property assists engineers in ensuring that materials chosen for specific applications have the necessary strength and resilience.