A house is built on the top of a hill with a nearby slope at angle 45° slope (Figure 6-55). 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 0.42, what is the least angle ϕ through which the present slope should be reduced to prevent slippage?
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To determine the least angle through which the present slope should be reduced to prevent slippage, we can use the concept of limiting equilibrium.
The static coefficient of friction between the two layers of soil is given as 0.42. The angle of the slope is 45°.
Let's assume that the slope angle after reducing is ϕ.
To prevent slippage, the driving force acting downhill on the slope must be balanced by the resisting force provided by the friction between the soil layers. The driving force can be calculated using the angle of the slope and the weight of the soil.
The driving force is given by:
Driving force = Weight of soil * sin(ϕ)
The resisting force provided by the friction between the soil layers is given by:
Resisting force = Weight of soil * cos(ϕ) * coefficient of friction
To prevent slippage, the resisting force must be equal to or greater than the driving force. So we have:
Weight of soil * sin(ϕ) ≤ Weight of soil * cos(ϕ) * coefficient of friction
Canceling the Weight of soil on both sides:
sin(ϕ) ≤ cos(ϕ) * coefficient of friction
Dividing both sides by cos(ϕ):
tan(ϕ) ≤ coefficient of friction
Substituting the given value of coefficient of friction (0.42):
tan(ϕ) ≤ 0.42
Now, we need to find the inverse tangent of 0.42 to determine the angle ϕ.
ϕ = atan(0.42)
Using a calculator, we find that ϕ ≈ 22.48°
Therefore, the least angle ϕ through which the present slope should be reduced to prevent slippage is approximately 22.48°.
To determine the least angle through which the slope should be reduced to prevent slippage, we need to consider the forces acting on the top layers of soil along the slope.
First, let's break down the forces involved:
1. Gravity will act vertically downwards, perpendicular to the slope.
2. The component of the weight of the soil acting parallel to the slope will try to cause the soil to slide downwards.
3. The frictional force between the soil layers will act opposite to the direction of motion.
In order to prevent slippage, the frictional force between the soil layers should be equal to or greater than the component of the weight of the soil trying to cause the soil to slide. This can be expressed mathematically as:
Frictional force (Ff) ≥ Component of weight of soil (Wsinϕ)
where ϕ is the angle through which the slope is reduced.
Given that the static coefficient of friction between the soil layers is 0.42, we can determine the equation for the frictional force:
Ff = μN
where μ is the coefficient of friction and N is the normal force between the soil layers. Since the normal force is perpendicular to the slope, it can be expressed as:
N = Wcosϕ
where W is the weight of the soil.
Combining the above equations, we have:
μN ≥ Wsinϕ
Substituting the values, we get:
0.42(Wcosϕ) ≥ Wsinϕ
Dividing both sides by W, we get:
0.42cosϕ ≥ sinϕ
Now, we can solve for the least value of ϕ that satisfies this inequality.
To do this, we can take the inverse tangent of both sides of the equation. This will give us the angle ϕ for which the tangent is less than or equal to 0.42 (i.e., the inequality is satisfied).
tan^(-1)(0.42) ≥ ϕ
Using a calculator, we find:
ϕ ≤ 22.5°
Therefore, the least angle through which the present slope should be reduced to prevent slippage is 22.5 degrees.