The figure below shows an infinite-slope model. If \(\tau\) and SS denote the driving shear stress and the resisting shear strength, respectively, then \(\tau\) along the failure plane is \(\times 10^3\) kg/ms\(^2\) (round off to two decimal places). [Use: \(d = 2\) m; \(\rho_{soil} = 2.0\) g/cm\(^3\); \(g = 9.8\) m/s\(^2\)]
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Use the infinite-slope formula tau = rho times g times d times sin(beta) times cos(beta), with d as the vertical soil thickness from the figure, and convert density to kg/m^3 before multiplying.
Step 1: Picture the sliding block.
Take a strip of soil of vertical depth $d$ sitting above a planar failure surface that is tilted at angle $\beta$ to the horizontal. Gravity pulls straight down on this strip with a force equal to its weight per unit plan area, $W = \rho_{soil} g d$.
Step 2: Split gravity into two directions.
On a slope, it helps to split the vertical weight into a part that pushes the soil down the slope (the driving shear stress $\tau$) and a part that presses the soil into the slope (the normal stress). The standard geometry of an infinite slope gives
\[ \tau = W \sin\beta \cos\beta = \rho_{soil} g d \sin\beta\cos\beta \]
The extra $\cos\beta$ appears because only the along-slope share of the tilted weight does the driving work on the failure plane.
Step 3: Put in the numbers.
Convert density to SI units first: $\rho_{soil} = 2.0\ \text{g/cm}^3 = 2000\ \text{kg/m}^3$. The other values are $d = 2\ \text{m}$, $g = 9.8\ \text{m/s}^2$, and $\beta = 30^{\circ}$, so $\sin\beta = 0.5$ and $\cos\beta = 0.8660$.
Step 5: Write the final value.
So $\tau \approx 16974.1\ \text{kg/ms}^2$, which is $16.97 \times 10^3\ \text{kg/ms}^2$ once rounded to two decimal places, right inside the expected band of 16.80 to 17.10.
\[ \boxed{16.97} \]
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