/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 76 A house is built on the top of a... [FREE SOLUTION] | 91Ó°ÊÓ

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A house is built on the top of a hill with a nearby slope at angle \(\theta=45^{\circ}\) (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 \(\phi\) through which the present slope should be reduced to prevent slippage?

Short Answer

Expert verified
The slope should be reduced by \( 18.43^{\circ} \).

Step by step solution

01

Understanding the Problem

We need to determine the least angle \( \phi \) through which the current slope angle of \( 45^{\circ} \) must be reduced to prevent slippage between two layers of soil. We know the coefficient of static friction is \( 0.5 \).
02

Calculate the maximum angle for no slippage

The maximum angle \( \alpha \) at which slippage does not occur can be determined using the formula for the maximum angle of repose, where \( \mu_s = \tan(\alpha) \), with \( \mu_s \) being the coefficient of static friction. So, \( \alpha = \tan^{-1}(0.5) \).
03

Determine \( \alpha \)

Calculate \( \alpha \) using the inverse tangent function: \( \alpha = \tan^{-1}(0.5) \approx 26.57^{\circ} \). This is the maximum angle of slope that can exist without slippage given the static friction coefficient.
04

Find the Reduction Angle \( \phi \)

We originally have a slope of \( 45^{\circ} \). The slope must be reduced to \( 26.57^{\circ} \) to prevent slippage. Therefore, the reduction angle \( \phi = 45^{\circ} - 26.57^{\circ} \).
05

Calculate \( \phi \)

Subtract the maximum angle \( \alpha \) from the current angle: \( \phi = 45^{\circ} - 26.57^{\circ} = 18.43^{\circ} \). So, the angle \( \phi \) by which the slope should be reduced is \( 18.43^{\circ} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Static Friction
Static friction plays a critical role in determining whether materials slide over each other. Imagine static friction as the invisible glue that keeps things from slipping. It resists the initial motion of an object and only comes into play when there is a tendency for movement.
The static friction force can be calculated using the coefficient of static friction (\(\mu_s\)) multiplied by the normal force (\(N\)). This formula is generally expressed as \(f_s = \mu_s \cdot N\). More simply, \(\mu_s\) represents a numerical value that defines how 'sticky' the surfaces are without actual movement.
  • If the force trying to move an object doesn't exceed this static friction, it stays put.
  • Once the static friction threshold is crossed, motion might begin, and kinetic friction takes over.
In our exercise, the coefficient of static friction between two layers of soil is given as \(0.5\). This tells us that the friction is moderate and helps us calculate the angle at which slippage begins.
Angle of Repose
The angle of repose is the steepest angle at which a sloping surface formed of a particular othersrticle will maintain its position without sliding. It’s like finding the perfect pitching angle of a stack before it topples. This angle depends heavily on static friction.
In technical terms, the angle of repose (\(\alpha\)) can be calculated using \(\mu_s = \tan(\alpha)\). Given the coefficient of static friction \(\mu_s = 0.5\), the angle of repose can be determined as \(\alpha = \tan^{-1}(0.5)\), which calculated gives \(\approx 26.57^{\circ}\).
Understanding this helps us figure out how steep the slope can be before the soil layers start to move. In our scenario, knowing that the slope is initially \(45^{\circ}\) shows us it's significantly steeper than the angle of repose, indicating potential instability. Given this angle, the slope must be reduced to avoid movement of the soil layers.
Slope Stability
Slope stability is about keeping a hilly area, either natural or man-made, safe from sliding. It's a balance between the downhill force caused by gravity and the uphill resistive force given by friction.
  • An unstable slope might experience landslides or soil slippage.
  • Engineers strive to optimize this balance to prevent catastrophic events.
In our specific problem, determining whether the hill slope is stable involves comparing the current slope angle to that of the angle of repose.
The original angle of \(45^{\circ}\) exceeds the calculated angle of repose \(26.57^{\circ}\), signaling danger. To maintain slope stability, the slope must be reduced by \(18.43^{\circ}\) to stay below this critical angle. By adjusting the slope, we ensure that the soil layers stay in place, making the hillside a safer place for structures like houses.

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Most popular questions from this chapter

A \(110 \mathrm{~g}\) hockey puck sent sliding over ice is stopped in \(15 \mathrm{~m}\) by the frictional force on it from the ice. (a) If its initial speed is \(6.0 \mathrm{~m} / \mathrm{s},\) what is the magnitude of the frictional force? (b) What is the coefficient of friction between the puck and the ice?

A bolt is threaded onto one end of a thin horizontal rod, and the rod is then rotated horizontally about its other end. An engineer monitors the motion by flashing a strobe lamp onto the rod and bolt, adjusting the strobe rate until the bolt appears to be in the same eight places during each full rotation of the rod (Fig. \(6-42\) ). The strobe rate is 2000 flashes per second; the bolt has mass \(30 \mathrm{~g}\) and is at radius \(3.5 \mathrm{~cm} .\) What is the magnitude of the force on the bolt from the rod?

What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is \(29 \mathrm{~km} / \mathrm{h}\) and the \(\mu_{s}\) between tires and track is \(0.32 ?\)

A \(2.5 \mathrm{~kg}\) block is initially at rest on a horizontal surface. A horizontal force \(\vec{F}\) of magnitude \(6.0 \mathrm{~N}\) and a vertical force \(\vec{P}\) are then applied to the block (Fig. \(6-17) .\) The coefficients of friction for the block and surface are \(\mu_{s}=0.40\) and \(\mu_{k}=0.25 .\) Determine the magnitude of the frictional force acting on the block if the magnitude of \(\vec{P}\) is (a) \(8.0 \mathrm{~N},\) (b) \(10 \mathrm{~N}\), and (c) \(12 \mathrm{~N}\).

A circular curve of highway is designed for traffic moving at \(60 \mathrm{~km} / \mathrm{h} .\) Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is \(150 \mathrm{~m},\) what is the correct angle of banking of the road? (b) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at \(60 \mathrm{~km} / \mathrm{h} ?\)

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