Optimal Slope Angle for Hilltop House Stability

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SUMMARY

The discussion centers on calculating the optimal slope angle for a house built on a hill with a current slope of 45°. The static coefficient of friction between soil layers is given as 0.67. The correct calculation for the angle to prevent slippage involves using the equation μ = tan(θ), leading to a required angle of approximately 33.822°. However, the user is confused about whether this angle represents the amount to reduce the slope by or the new slope angle itself.

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Homework Statement


A house is built on the top of a hill with a nearby 45° slope (Fig. 6-19). An engineering study indicates that the slope angle should be reduced because the top layers of soil along the slope might slip past the lower layers. If the static coefficient of friction between two such layers is 0.67, what is the least angle through which the present slope should be reduced to prevent slippage?
(I attached an image. Hopefully it shows up - You may have to click the attachment once more after it comes in the pop-up)

Homework Equations


f = μn
f = mg*sin ∅
n = mg*cos ∅
Current value of ∅ = 45°
μ = 0.67

The Attempt at a Solution


I drew a free body diagram, and found that the friction force is equal and opposite that of mg*sin∅, and that the normal force was mg*cos∅.
I then set up this equation:
Net force = f - mg*sin∅
μ*n - mg*sin∅ = 0
μ*n = mgsin∅
μ*mgcos∅ = mgsin∅
(mg's cancel)
μ * cos ∅ = sin ∅
μ = tan ∅
tan-1 μ = ∅
I plugged in 0.67 for mu, and got 33.822°, which I am being told is wrong.

Can somebody please help me? I have one more submission left.
 

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Looks like you solved for the correct angle to me.
Are they asking how much the angle should be reduced by? Or reduced to?
 

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