Question

A house is built on the top of a hill with a nearby slope at angle$\mathit{\theta }\mathbf{=}\mathbf{45}\mathbf{°}$(Fig. 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 coefficient of static friction between two such layers is 0.5, what is the least angle $\mathit{\varphi }$through which the present slope should be reduced to prevent slippage?

Short answer

$\varphi =20°$

Step-by-step solution

Given data

• The initial angle of slope,$\theta =45°$
• Coefficient of static friction,${\mu }_s=0.50$ .

Understanding the concept

The problem deals with Newton’s laws of motion which describe the relations between the forces acting on a body and the motion of the body.Using the application of Newton’s laws, the given problem can be solved.

Formula:

$\theta ={\tan }^{-1}\left({\mu }_s\right)$

Calculate the least angle ϕ through which the present slope should be reduced to prevent slippage

The maximum angle for which static friction applies is given by,

$\theta ={\tan }^{-1}\left({\mu }_s\right)$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \theta ={\tan }^{-1}\left(0.50\right) \\ \theta =27° \end{gathered}$$

This implies that the angle through which the slope should be reduced is,

$$\begin{gathered} \varphi =45°-27° \\ \approx 20° \end{gathered}$$

Thus, is the least angle through which the present slope should be reduced to prevent slippage is 20 degrees.