Let's break down the problem step by step: ### Given Data: - \(\alpha = 15^\circ\) - \(\theta = 5^\circ\) - \(\gamma_d = 110~\text{lb/ft}^3\) - \(\phi' = 39^\circ\) - \(\delta = \frac{2}{3}\phi' = \frac{2}{3} \times 39^\circ = 26^\circ\) ### Coulomb's Active Earth Pressure Coefficient (\(K_A\)) Formula: \[ K_A = \frac{\cos^2(\phi' - \theta)}{\cos^2\theta \cos(\delta + \theta) \left[1 + \sqrt{\frac{\sin(\phi' + \delta) \sin(\phi' - \alpha)}{\cos(\delta + \theta) \cos(\theta - \alpha)}} \right]^2} \] Now, let's plug in all the values: - \(\phi' = 39^\circ\) - \(\theta = 5^\circ\) - \(\alpha = 15^\circ\) - \(\delta = 26^\circ\) #### Step 1: Calculate Numerator \[ \cos^2(\phi' - \theta) = \cos^2(39^\circ - 5^\circ) = \cos^2(34^\circ) \] #### Step 2: Calculate Denominator - \(\cos^2\theta = \cos^2(5^\circ)\) - \(\cos(\delta + \theta) = \cos(26^\circ + 5^\circ) = \cos(31^\circ)\) Now, inside the brackets: - \(\sin(\phi' + \delta) = \sin(39^\circ + 26^\circ) = \sin(65^\circ)\) - \(\sin(\phi' - \alpha) = \sin(39^\circ - 15^\circ) = \sin(24^\circ)\) - \(\cos(\delta + \theta) = \cos(31^\circ)\) - \(\cos(\theta - \alpha) = \cos(5^\circ - 15^\circ) = \cos(-10^\circ) = \cos(10^\circ)\) So, \[ \left[1 + \sqrt{\frac{\sin(65^\circ) \sin(24^\circ)}{\cos(31^\circ) \cos(10^\circ)}} \right]^2 \] #### Step 3: Plug in and Calculate Each Term - \(\cos(34^\circ) \approx 0.8290\) - \(\cos^2(34^\circ) \approx 0.6873\) - \(\cos(5^\circ) \approx 0.9962\) - \(\cos^2(5^\circ) \approx 0.9924\) - \(\cos(31^\circ) \approx 0.8572\) - \(\sin(65^\circ) \approx 0.9063\) - \(\sin(24^\circ) \approx 0.4067\) - \(\cos(10^\circ) \approx 0.9848\) Now, calculate inside the root: \[ \frac{\sin(65^\circ) \sin(24^\circ)}{\cos(31^\circ) \cos(10^\circ)} = \frac{0.9063 \times 0.4067}{0.8572 \times 0.9848} \approx \frac{0.3686}{0.8445} \approx 0.4365 \] \[ \sqrt{0.4365} \approx 0.6606 \] \[ 1 + 0.6606 = 1.6606 \] \[ (1.6606)^2 \approx 2.7586 \] Now the denominator: \[ \cos^2(5^\circ) \cos(31^\circ) (1.6606^2) = 0.9924 \times 0.8572 \times 2.7586 \approx 0.8511 \times 2.7586 \approx 2.3471 \] #### Step 4: Compute \(K_A\) \[ K_A = \frac{0.6873}{2.3471} \approx 0.293 \] --- ### **Final Answer** \[ \boxed{K_A = 0.29} \] (Rounded to two decimal places.)

Let's break down the problem step by step: ### Given Data: - \(\alpha = 15^\circ\) - \(\theta = 5^\circ\) - \(\gamma_d = 110~\text{lb/ft}^3\) - \(\phi' = 39^\circ\) - \(\delta = \frac{2}{3}\phi' = \frac{2}{3} \times 39^\circ = 26^\circ\) ### Coulomb's Active Earth Pressure Coefficient (\(K_A\)) Formula: \[ K_A = \frac{\cos^2(\phi' - \theta)}{\cos^2\theta \cos(\delta + \theta) \left[1 + \sqrt{\frac{\sin(\phi' + \delta) \sin(\phi' - \alpha)}{\cos(\delta + \theta) \cos(\theta - \alpha)}} \right]^2} \] Now, let's plug in all the values: - \(\phi' = 39^\circ\) - \(\theta = 5^\circ\) - \(\alpha = 15^\circ\) - \(\delta = 26^\circ\) #### Step 1: Calculate Numerator \[ \cos^2(\phi' - \theta) = \cos^2(39^\circ - 5^\circ) = \cos^2(34^\circ) \] #### Step 2: Calculate Denominator - \(\cos^2\theta = \cos^2(5^\circ)\) - \(\cos(\delta + \theta) = \cos(26^\circ + 5^\circ) = \cos(31^\circ)\) Now, inside the brackets: - \(\sin(\phi' + \delta) = \sin(39^\circ + 26^\circ) = \sin(65^\circ)\) - \(\sin(\phi' - \alpha) = \sin(39^\circ - 15^\circ) = \sin(24^\circ)\) - \(\cos(\delta + \theta) = \cos(31^\circ)\) - \(\cos(\theta - \alpha) = \cos(5^\circ - 15^\circ) = \cos(-10^\circ) = \cos(10^\circ)\) So, \[ \left[1 + \sqrt{\frac{\sin(65^\circ) \sin(24^\circ)}{\cos(31^\circ) \cos(10^\circ)}} \right]^2 \] #### Step 3: Plug in and Calculate Each Term - \(\cos(34^\circ) \approx 0.8290\) - \(\cos^2(34^\circ) \approx 0.6873\) - \(\cos(5^\circ) \approx 0.9962\) - \(\cos^2(5^\circ) \approx 0.9924\) - \(\cos(31^\circ) \approx 0.8572\) - \(\sin(65^\circ) \approx 0.9063\) - \(\sin(24^\circ) \approx 0.4067\) - \(\cos(10^\circ) \approx 0.9848\) Now, compute inside the root: \[ \frac{\sin(65^\circ) \sin(24^\circ)}{\cos(31^\circ) \cos(10^\circ)} \approx \frac{0.9063 \times 0.4067}{0.8572 \times 0.9848} \approx \frac{0.3686}{0.8445} \approx 0.4365 \] \[ \sqrt{0.4365} \approx 0.6606 \] \[ 1 + 0.6606 = 1.6606 \] \[ (1.6606)^2 \approx 2.7586 \] Now the denominator: \[ 0.9924 \times 0.8572 \times 2.7586 \approx 0.8511 \times 2.7586 \approx 2.3471 \] #### Step 4: Compute \(K_A\) \[ K_A = \frac{0.6873}{2.3471} \approx 0.293 \] --- ### **Final Answer** \[ \boxed{K_A = 0.29} \] (Rounded to two decimal places.)

Let's break down the problem step by step: ### Given Data: - \(\alpha = 15^\circ\) - \(\theta = 5^\circ\) - \(\gamma_d = 110\, \text{lb/ft}^3\) - \(\phi' = 39^\circ\) - \(\delta = \frac{2}{3}\phi' = \frac{2}{3} \times 39^\circ = 26^\circ\) ### Coulomb's Active Earth Pressure Coefficient (\(K_A\)) Formula: \[ K_A = \frac{\cos^2(\phi' - \theta)}{\cos^2 \theta \cos(\delta + \theta) \left[1 + \sqrt{\frac{\sin(\phi' + \delta) \sin(\phi' - \alpha)}{\cos(\delta + \theta) \cos(\theta - \alpha)}} \right]^2} \] ### Step 1: Calculate numerator \[ \cos^2(\phi' - \theta) = \cos^2(39^\circ - 5^\circ) = \cos^2(34^\circ) \] ### Step 2: Calculate denominator components - \(\cos^2 \theta = \cos^2(5^\circ)\) - \(\cos(\delta + \theta) = \cos(26^\circ + 5^\circ) = \cos(31^\circ)\) Inside the brackets: - \(\sin(\phi' + \delta) = \sin(39^\circ + 26^\circ) = \sin(65^\circ)\) - \(\sin(\phi' - \alpha) = \sin(39^\circ - 15^\circ) = \sin(24^\circ)\) - \(\cos(\delta + \theta) = \cos(31^\circ)\) (already noted) - \(\cos(\theta - \alpha) = \cos(5^\circ - 15^\circ) = \cos(-10^\circ) = \cos(10^\circ)\) The expression inside the square root becomes: \[ \frac{\sin(65^\circ) \sin(24^\circ)}{\cos(31^\circ) \cos(10^\circ)} \] ### Step 3: Numerical calculations Using approximate values: | Function | Value | Explanation | | --- | --- | --- | | \(\cos(34^\circ)\) | 0.8290 | from calculator | | \(\cos^2(34^\circ)\) | 0.6873 | square of above | | \(\cos(5^\circ)\) | 0.9962 | from calculator | | \(\cos^2(5^\circ)\) | 0.9924 | square of above | | \(\cos(31^\circ)\) | 0.8572 | from calculator | | \(\sin(65^\circ)\) | 0.9063 | from calculator | | \(\sin(24^\circ)\) | 0.4067 | from calculator | | \(\cos(10^\circ)\) | 0.9848 | from calculator | Calculate the ratio inside the root: \[ \frac{0.9063 \times 0.4067}{0.8572 \times 0.9848} \approx \frac{0.3686}{0.8445} \approx 0.4365 \] Square root: \[ \sqrt{0.4365} \approx 0.6606 \] Add 1: \[ 1 + 0.6606 = 1.6606 \] Square: \[ (1.6606)^2 \approx 2.7586 \] Calculate denominator: \[ 0.9924 \times 0.8572 \times 2.7586 \approx 0.8511 \times 2.7586 \approx 2.3471 \] Calculate numerator: \[ \cos^2(34^\circ) \approx 0.6873 \] Finally, compute \(K_A\): \[ K_A = \frac{0.6873}{2.3471} \approx 0.293 \] ### **Final Answer:** \[ \boxed{K_A \approx 0.29} \] (Rounded to two decimal places.)