Let's break down the step-by-step solutions for each part as shown in the images: --- ## (a) **Normal Stress Calculation** ### **Step 1: Calculate the Volume of the Bar** Given dimensions: - Length = 7 in - Width = 2 in - Height = 4 in Convert inches to meters (since 1 in = 0.0254 m): \[ V = 7 \text{ in} \times 0.0254 \frac{m}{\text{in}} \times 2 \text{ in} \times 0.0254 \frac{m}{\text{in}} \times 4 \text{ in} \times 0.0254 \frac{m}{\text{in}} \] \[ V = 9.176 \times 10^{-4} \, m^3 \] --- ### **Step 2: Calculate the Normal Force (N)** Density of A36: \( \rho = 7800 \frac{kg}{m^3} \) Normal force (N): \[ N = mg\cos\theta = (\rho V)g\cos\theta \] \[ N = 7800 \frac{kg}{m^3} \times 9.176 \times 10^{-4} m^3 \times 9.81 \frac{m}{s^2} \times \cos 40^\circ \] \[ N = 53.786 \, N \] --- ### **Step 3: Calculate the Normal Stress (\(\sigma\))** \[ \sigma = \frac{N}{A} \] where \(A\) is the cross-sectional area: - \( l_x = 0.1778 \) m - \( l_z = 0.1016 \) m \[ A = l_x \times l_z = 0.1778 \times 0.1016 \] \[ \sigma = \frac{53.786}{0.1778 \times 0.1016} \] \[ \sigma = 2977.445 \frac{N}{m^2} \] Convert to psi (\(1 \text{ psi} = 6894.75 \frac{N}{m^2}\)): \[ \sigma = 2977.445 \frac{N}{m^2} \times \frac{1 \text{ psi}}{6894.75 \frac{N}{m^2}} \] \[ \sigma = 0.431 \text{ psi} \] --- ## (b) **Shear Stress Calculation** ### **Step 1: Calculate the Friction Force** \[ f = \mu N \] where \( \mu = 0.4 \) --- ### **Step 2: Calculate Shear Stress (\(\tau\))** \[ \tau = \frac{f}{A} = \frac{\mu N}{l_x l_z} \] \[ \tau = \frac{0.4 \times 53.786}{0.1778 \times 0.1016} \] \[ \tau = 1190.97 \frac{N}{m^2} \] Convert to psi: \[ \tau = 1190.97 \frac{N}{m^2} \times \frac{1 \text{ psi}}{6894.75 \frac{N}{m^2}} \] \[ \tau = -0.172 \text{ psi} \] (Negative since friction opposes motion.) --- ## (c) **Resultant Normal Reaction Force** - It acts at the **center of gravity** of the bar. - **Location**: 3.5 in from the left corner (half of 7 in), and 0.75 in upward from the lower left corner (half of 1.5 in). --- ## (d) **Velocity Calculation Down the Incline** ### **Step 1: Find Net Force and Acceleration** \[ F = ma \] \[ a = \frac{mg\sin\theta - \text{friction}}{m} \] \[ a = \frac{(7800 \times 9.176 \times 10^{-4} \times 9.81)\sin 40^\circ - (0.4 \times 53.786)}{7800 \times 9.176 \times 10^{-4}} \] \[ a = 3.3 \, m/s^2 \] --- ### **Step 2: Find Final Velocity** Using \(v^2 = u^2 + 2as\): Given: - \(u = 0.792 \, m/s\) - \(a = 3.3 \, m/s^2\) - \(s = 1.036 \, m\) \[ v = \sqrt{(0.792)^2 + 2 \times 3.3 \times 1.036} \] \[ v = 7.36 \, m/s \] --- ### **Final Answers:** - **Normal Stress:** \(\sigma = 0.431 \, \text{psi}\) - **Shear Stress:** \(\tau = -0.172 \, \text{psi}\) - **Normal Force Location:** \(3.5 \text{ in}\) to the right and \(0.75 \text{ in}\) upward from the lower left corner - **Velocity at Bottom:** \(v = 7.36 \, \text{m/s}\) If you need more explanation on any step, let me know!

Let's follow the same detailed step-by-step approach as in the original solution, now with the new dimensions: **Dimensions given:** - Length \( L = 8\, \text{in} \) - Width \( W = 2.5\, \text{in} \) - Height \( H = 3\, \text{in} \) --- ### Step 1: Convert dimensions to meters Using \( 1\, \text{in} = 0.0254\, \text{m} \): \[ L = 8 \times 0.0254 = 0.2032\, \text{m} \] \[ W = 2.5 \times 0.0254 = 0.0635\, \text{m} \] \[ H = 3 \times 0.0254 = 0.0762\, \text{m} \] --- ### Step 2: Calculate the volume of the bar \[ V = L \times W \times H = 0.2032 \times 0.0635 \times 0.0762 \] \[ V \approx 0.000981\, \text{m}^3 \] --- ### Step 3: Calculate the normal force \( N \) Given: - Density of A36 steel \( \rho = 7800\, \text{kg/m}^3 \) - Acceleration due to gravity \( g = 9.81\, \text{m/s}^2 \) - Inclination angle \( \theta \) (from original problem, assume same as before, i.e., 40°) Compute the weight: \[ mg = \rho V g = 7800 \times 0.000981 \times 9.81 \] \[ mg \approx 75.219\, \text{N} \] Now, the normal force component: \[ N = mg \cos \theta = 75.219 \times \cos 40^\circ \] \[ N \approx 75.219 \times 0.7660 \approx 57.56\, \text{N} \] --- ### Step 4: Cross-sectional area for stress calculation Assuming the normal stress acts over the cross-sectional area perpendicular to the load, which is the face with dimensions \( W \times H \): \[ A = W \times H = 0.0635 \times 0.0762 \approx 0.00484\, \text{m}^2 \] --- ### Step 5: Calculate normal stress \(\sigma\) \[ \sigma = \frac{N}{A} = \frac{57.56}{0.00484} \approx 11,898\, \text{Pa} \] Convert to MPa: \[ \sigma \approx 11.898\, \text{kPa} \] or in psi (\(1\, \text{psi} = 6894.75\, \text{Pa}\)): \[ \sigma = \frac{11,898}{6894.75} \approx 1.73\, \text{psi} \] --- ### **Summary for part (a):** **Normal stress \(\sigma \approx 1.73\, \text{psi}\).** --- ### **Part (b): Shear stress calculation** Assuming the same coefficient of friction \( \mu = 0.4 \): \[ f = \mu N = 0.4 \times 57.56 \approx 23.02\, \text{N} \] Shear stress: \[ \tau = \frac{f}{A} = \frac{23.02}{0.00484} \approx 4,761\, \text{Pa} \] In psi: \[ \tau = \frac{4,761}{6894.75} \approx 0.69\, \text{psi} \] The negative sign indicates direction opposite to motion, as before. --- ### **Part (c): Location of the resultant normal force** - The center of gravity is at the centroid of a rectangular prism, which is at the midpoint: \[ \text{from left (long side)}: \frac{L}{2} = 4\, \text{in} \] \[ \text{from bottom (height)}: \frac{H}{2} = 1.5\, \text{in} \] \[ \text{from front (width)}: \frac{W}{2} = 1.25\, \text{in} \] - The normal reaction acts at the centroid, which is 4 in from the left edge and 1.25 in from the front face, and 1.5 in upward from the bottom. --- ### **Part (d): Velocity calculation** Assuming similar conditions: - Mass \( m = \rho V \approx 75.219\, \text{N} / 9.81\, \text{m/s}^2 \approx 7.67\, \text{kg} \) - The component of gravity along the incline: \[ mg \sin \theta = 75.219 \times \sin 40^\circ \approx 75.219 \times 0.6428 \approx 48.37\, \text{N} \] - Friction force: \[ f_{fric} = \mu N = 0.4 \times 57.56 \approx 23.02\, \text{N} \] - Net force: \[ F_{net} = mg \sin \theta - f_{fric} = 48.37 - 23.02 = 25.35\, \text{N} \] - Acceleration: \[ a = \frac{F_{net}}{m} = \frac{25.35}{7.67} \approx 3.31\, \text{m/s}^2 \] - Initial velocity \( u \) (from previous or assumed), say zero, or as per initial conditions. - Using \( v^2 = u^2 + 2a s \), with \( s \) as the distance traveled (assuming same as previous, or specify if given). For illustration, assuming \( s = 1\, \text{m} \): \[ v = \sqrt{0 + 2 \times 3.31 \times 1} \approx \sqrt{6.62} \approx 2.58\, \text{m/s} \] --- ## **Final summarized answers:** - **Normal stress:** approximately **1.73 psi** - **Shear stress:** approximately **0.69 psi** - **Normal force acts at:** 4 in from the left and 1.25 in from front face, at the centroid. - **Velocity at bottom (if traveled 1 m):** approximately **2.58 m/s** --- If you need calculations for specific distances or initial velocities, please specify!

Let's systematically replicate the detailed step-by-step solution as per the original image, now using the new dimensions: ### Given Data: - Length \( L = 8\, \text{in} \) - Width \( W = 2.5\, \text{in} \) - Height \( H = 3\, \text{in} \) - Inclination angle \( \theta = 30^\circ \) --- ### Step 1: Convert dimensions to meters Using \( 1\, \text{in} = 0.0254\, \text{m} \): \[ L = 8 \times 0.0254 = 0.2032\, \text{m} \] \[ W = 2.5 \times 0.0254 = 0.0635\, \text{m} \] \[ H = 3 \times 0.0254 = 0.0762\, \text{m} \] --- ### Step 2: Calculate the volume of the bar \[ V = L \times W \times H = 0.2032 \times 0.0635 \times 0.0762 \] \[ V \approx 0.000981\, \text{m}^3 \] --- ### Step 3: Calculate the weight (force due to gravity) Using density \( \rho = 7800\, \text{kg/m}^3 \): \[ mg = \rho V g = 7800 \times 0.000981 \times 9.81 \] \[ mg \approx 75.219\, \text{N} \] --- ### Step 4: Calculate the normal force \( N \) - The normal component of the weight on the inclined plane: \[ N = mg \cos \theta = 75.219 \times \cos 30^\circ \] \[ \cos 30^\circ = 0.8660 \] \[ N \approx 75.219 \times 0.8660 \approx 65.14\, \text{N} \] --- ### Step 5: Determine the cross-sectional area for stress Assuming the normal stress acts over the face with dimensions \( W \times H \): \[ A = W \times H = 0.0635 \times 0.0762 \approx 0.00484\, \text{m}^2 \] --- ### Step 6: Calculate normal stress \(\sigma\) \[ \sigma = \frac{N}{A} = \frac{65.14}{0.00484} \approx 13,456\, \text{Pa} \] Convert to psi: \[ \sigma = \frac{13,456}{6894.75} \approx 1.95\, \text{psi} \] --- ### **Part (a): Final normal stress** **\(\boxed{\sigma \approx 1.95\, \text{psi}}\)** --- ### Part (b): Shear stress calculation - Friction coefficient \( \mu = 0.4 \) - Friction force: \[ f = \mu N = 0.4 \times 65.14 \approx 26.06\, \text{N} \] - Shear stress: \[ \tau = \frac{f}{A} = \frac{26.06}{0.00484} \approx 5,382\, \text{Pa} \] - Convert to psi: \[ \tau = \frac{5,382}{6894.75} \approx 0.78\, \text{psi} \] Considering the direction, shear stress is negative if opposing motion. **\(\boxed{\tau \approx -0.78\, \text{psi}}\)** --- ### Part (c): Location of the resultant normal reaction force - The center of gravity for a rectangular prism is at the centroid: \[ \text{from left (length)}: \frac{L}{2} = 4\, \text{in} \] \[ \text{from bottom (height)}: \frac{H}{2} = 1.5\, \text{in} \] \[ \text{from front (width)}: \frac{W}{2} = 1.25\, \text{in} \] - The normal force acts at this centroid, located 4 in along the length, 1.25 in from the width, and 1.5 in from the bottom. --- ### Part (d): Velocity calculation along the incline - **Step 1:** Calculate the component of gravitational force along the incline: \[ F_{g,\text{parallel}} = mg \sin \theta = 75.219 \times \sin 30^\circ \] \[ \sin 30^\circ= 0.5 \] \[ F_{g,\text{parallel}} \approx 75.219 \times 0.5 = 37.61\, \text{N} \] - **Step 2:** Net force considering friction: \[ F_{net} = F_{g,\text{parallel}} - f = 37.61 - 26.06 = 11.55\, \text{N} \] - **Step 3:** Calculate acceleration: \[ a = \frac{F_{net}}{m} = \frac{11.55}{75.219/9.81} \approx \frac{11.55}{7.67} \approx 1.51\, \text{m/s}^2 \] - **Step 4:** Find velocity after traveling a distance \( s \). Assuming initial velocity \( u = 0 \), and \( s = 1\, \text{m} \): \[ v = \sqrt{u^2 + 2 a s} = \sqrt{0 + 2 \times 1.51 \times 1} \approx \sqrt{3.02} \approx 1.74\, \text{m/s} \] --- ### **Final Results:** | Part | Result | |---|---| | **(a) Normal stress \(\sigma\)** | **\(\boxed{1.95\, \text{psi}}\)** | | **(b) Shear stress \(\tau\)** | **\(\boxed{-0.78\, \text{psi}}\)** | | **(c) Location of normal force** | 4 in from the left, 1.25 in from front, at the centroid | | **(d) Velocity after 1 m** | **\(\boxed{1.74\, \text{m/s}}\)** | --- If further specific conditions or distances are provided, these calculations can be refined accordingly.

Let's carefully replicate the solution step-by-step based on the new dimensions and parameters, following the same approach as in the original image. --- ### Given Data: - Length \( L = 8\, \text{in} \) - Width \( W = 2.5\, \text{in} \) - Height \( H = 3\, \text{in} \) - Inclination angle \( \theta = 30^\circ \) - Coefficient of kinetic friction \( \mu_k = 0.2 \) - Density of A36 steel \( \rho = 7800\, \text{kg/m}^3 \) (assumed same as in previous calculations) --- ### **Step 1: Convert dimensions to meters** \[ L = 8 \times 0.0254 = 0.2032\, \text{m} \] \[ W = 2.5 \times 0.0254 = 0.0635\, \text{m} \] \[ H = 3 \times 0.0254 = 0.0762\, \text{m} \] --- ### **Step 2: Calculate the volume \( V \)** \[ V = L \times W \times H = 0.2032 \times 0.0635 \times 0.0762 \approx 0.000981\, \text{m}^3 \] --- ### **Step 3: Calculate the weight \( mg \)** \[ mg = \rho V g = 7800 \times 0.000981 \times 9.81 \approx 75.219\, \text{N} \] --- ### **Part (a): Normal Stress Calculation** **Step 4: Calculate the normal force \( N \)** \[ N = mg \cos \theta = 75.219 \times \cos 30^\circ \] \[ \cos 30^\circ = 0.8660 \] \[ N \approx 75.219 \times 0.8660 \approx 65.14\, \text{N} \] **Step 5: Cross-sectional area for normal stress** Assuming the normal force acts on the face with dimensions \( W \times H \): \[ A = W \times H = 0.0635 \times 0.0762 \approx 0.00484\, \text{m}^2 \] **Step 6: Normal stress \(\sigma\)** \[ \sigma = \frac{N}{A} = \frac{65.14}{0.00484} \approx 13,456\, \text{Pa} \] Convert to psi: \[ \sigma = \frac{13,456}{6894.75} \approx 1.95\, \text{psi} \] --- ### **Part (b): Shear stress due to kinetic friction** **Step 7: Friction force** \[ f_{fric} = \mu_k N = 0.2 \times 65.14 \approx 13.03\, \text{N} \] **Step 8: Shear stress \(\tau\)** \[ \tau = \frac{f_{fric}}{A} = \frac{13.03}{0.00484} \approx 2,690\, \text{Pa} \] Convert to psi: \[ \tau = \frac{2,690}{6894.75} \approx 0.39\, \text{psi} \] *Note:* Since kinetic friction opposes motion, the shear stress is negative relative to the direction of movement, but magnitude is approximately 0.39 psi. --- ### **Part (c): Location of the normal reaction** - The centroid (center of gravity) of the rectangular block is at the midpoint: \[ \text{from left (length)}: \frac{L}{2} = 4\, \text{in} \] \[ \text{from bottom (height)}: \frac{H}{2} = 1.5\, \text{in} \] \[ \text{from front (width)}: \frac{W}{2} = 1.25\, \text{in} \] - The normal force acts at this centroid: 4 in along the length, 1.25 in from the front face, and 1.5 in above the bottom. --- ### **Part (d): Velocity at bottom after sliding** - **Step 9: Components of forces** The component of weight parallel to incline: \[ F_{g,\text{parallel}} = mg \sin \theta = 75.219 \times 0.5 = 37.61\, \text{N} \] - **Step 10: Net force** \[ F_{net} = F_{g,\text{parallel}} - f_{fric} = 37.61 - 13.03 = 24.58\, \text{N} \] - **Step 11: Mass of the block** \[ m = \frac{mg}{g} = \frac{75.219}{9.81} \approx 7.67\, \text{kg} \] - **Step 12: Acceleration** \[ a = \frac{F_{net}}{m} = \frac{24.58}{7.67} \approx 3.21\, \text{m/s}^2 \] - **Step 13: Final velocity after traveling distance \( s \)** Assuming initial velocity \( u=0 \) and \( s=1\, \text{m} \): \[ v = \sqrt{u^2 + 2 a s} = \sqrt{0 + 2 \times 3.21 \times 1} \approx \sqrt{6.42} \approx 2.53\, \text{m/s} \] --- ### **Final summarized answers:** | Part | Result | |---|---| | **(a) Normal stress \(\sigma\)** | **\(\boxed{1.95\, \text{psi}}\)** | | **(b) Shear stress \(\tau\)** | **\(\boxed{0.39\, \text{psi}}\)** | | **(c) Reaction force location** | 4 in from the left, 1.25 in from front face, at the centroid | | **(d) Velocity after 1 m** | **\(\boxed{2.53\, \text{m/s}}\)** | --- If you need calculations for different distances, initial velocities, or other parameters, please specify!