Question

a 2000 - lb wheel load is to be supported by aggregate over soil that can withstand a pressure of 1000 lb/sq ft. what depth of aggregate is needed if \\( \theta = 40^{\circ} \\)?

Explanation:

Step1: Use the formula for pressure

Pressure \(P=\frac{F}{A}\), where \(F\) is the force and \(A\) is the area. Here, the force \(F = 2000\) lb. Let the area \(A=\pi r^{2}\) (assuming a circular contact area for simplicity, and the depth \(h = r\)). The pressure \(P = 1000\) lb/sq.ft.

Step2: Rearrange the formula for area

From \(P=\frac{F}{A}\), we can get \(A=\frac{F}{P}\). Substitute \(F = 2000\) lb and \(P=1000\) lb/sq.ft. So, \(A=\frac{2000}{1000}=2\) sq.ft.

Step3: Solve for radius (depth) if \(A = \pi r^{2}\)

If \(A=\pi r^{2}\), then \(r^{2}=\frac{A}{\pi}\). Substitute \(A = 2\) sq.ft. \(r^{2}=\frac{2}{\pi}\approx\frac{2}{3.14}\approx0.637\). So, \(r=\sqrt{0.637}\approx0.8\) ft.

Answer:

The depth of the aggregate needed is approximately \(0.8\) ft.