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Chapter 6: Q76P (page 146)

A house is built on the top of a hill with a nearby slope at angleθ=45°(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 ϕthrough which the present slope should be reduced to prevent slippage?

Short Answer

Expert verified

ϕ=20°

Step by step solution

01

Given data

  • The initial angle of slope,θ=45°
  • Coefficient of static friction,μs=0.50 .
02

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:

θ=tan-1μs

03

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,

θ=tan-1μs

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

θ=tan-10.50θ=27°

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

ϕ=45°-27°≈20°

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

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In Fig. 6-60, a block weighing22 Nis held at rest against a vertical wall by a horizontal force of magnitude 60 N .The coefficient of static friction between the wall and the block is 0.55 , and the coefficient of kinetic friction between them is 0.38 . In six experiments, a second force is applied to the block and directed parallel to the wall with these magnitudes and directions: (a) 34 N , up, (b) 12 N , up, (c) 48 N , up, (d) 62 N, up, (e) 10 N , down, and (f) 18 N, down. In each experiment, what is the magnitude of the frictional force on the block? In which does the block move (g) up the wall and (h) down the wall? (i) In which is the frictional force directed down the wall?

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