91Ó°ÊÓ

First, we need to convert the given weight of the sand sample into stress. To do this, we will find the area of the sand sample and use the formula for stress. Since the diameter of the sand sample is 6.5 cm, the radius is 3.25 cm: $$ r = \frac{6.5}{2} = 3.25 \mathrm{~cm} $$ The area of the sand sample is: $$ A = \pi r^2 = \pi (3.25)^2 = 33.18 \mathrm{~cm^2} $$ Now, let's convert the weight of 5 kg to force (N) and calculate the stress: $$ Force = 5 \mathrm{~kg} \times 9.81 \frac{\mathrm{m}}{\mathrm{s^2}} = 49.05 \mathrm{~N} $$ $$ Stress = \frac{Force}{A} = \frac{49.05 \mathrm{~N}}{33.18 \mathrm{~cm^2}} = 1.48 \frac{\mathrm{N}}{\mathrm{cm^2}} $$ Note that we need to convert the stress to kPa, as the given cell pressure is in kPa: $$ Stress = 1.48 \frac{\mathrm{N}}{\mathrm{cm^2}} \times 10 \frac{\mathrm{kPa}}{\mathrm{N/cm^2}} = 14.8 \mathrm{kPa} $$

Now we will determine the effective stress at the failure point, where the cell pressure (also known as total stress) is 5.2 kPa: $$ Effective \thinspace stress = Total \thinspace stress - Pore \thinspace water \thinspace pressure $$ Since the sand is dry, the pore water pressure is zero: $$ Effective \thinspace stress = 5.2 \mathrm{kPa} - 0 \mathrm{kPa} = 5.2 \mathrm{kPa} $$

Now, we will apply the Mohr-Coulomb failure criterion, considering that the cohesion (c) is zero: $$ \tau = c + \sigma \tan{\phi} $$ Where \(\tau\) is the shear stress, \(\sigma\) is the effective stress, and \(\phi\) is the friction angle. In this exercise, we are asked to find \(\phi\). To calculate the shear stress, we can use the obtained stress value: $$ \tau = 0.5 \times 14.8 \mathrm{kPa} = 7.4 \mathrm{kPa} $$ Now, we can rearrange the Mohr-Coulomb failure criterion equation to isolate \(\phi\): $$ \phi = \arctan \frac{\tau - c}{\sigma} $$ As the cohesion is given to be zero (c = 0), we can simplify the equation: $$ \phi = \arctan \frac{7.4 \mathrm{kPa}}{5.2 \mathrm{kPa}} $$

Finally, calculate the friction angle using the arctangent function: $$ \phi = \arctan \frac{7.4 \mathrm{kPa}}{5.2 \mathrm{kPa}} = 55.11^{\circ} $$ The friction angle of the dry sand sample is approximately 55.11 degrees.