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Figure 6-22 shows the cross section of a road cut into the side of a mountain. The solid lineAA'represents a weak bedding plane along which sliding is possible. Block B directly above the highway is separated from uphill rock by a large crack (called a joint), so that only friction between the block and the bedding plane prevents sliding. The mass of the block islocalid="1654084347613" 1.8×107kg, the dip anglelocalid="1654084361257" θof the bedding plane islocalid="1654084374565" 24°, and the coefficient of static friction between block and plane islocalid="1654084400008" 0.63. (a) Show that the block will not slide under these circumstances. (b) Next, water seeps into the joint and expands upon freezing, exerting on the block a forceparallel tolocalid="1654084460188" AA'. What minimum value of force magnitudelocalid="1654084470850" Fwill trigger a slide down the plane?

Short Answer

Expert verified

(a) Hence, the block will not slide

(b) The minimum value of force magnitudeF, that will trigger a slide down the plane is 3.0×107N

Step by step solution

01

Given

Mass,m=1.8×107kg

Coefficient of static friction,μs=0.63

Inclined angle,θ=24°

02

Determining the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Use the Newton's 2nd law of motion along vertical and horizontal direction.

Formula:

FNet=ma

where, F is the net force, m is mass and a is an acceleration.

03

Determining thefree body diagram of block

Free body diagram of Block:

04

(a) Showing that the block will not slide under given circumstances

By using Newton’s 2nd law along vertical direction (along y),

Fy=may

Since block is not moving upward, ay=0

N-Fgcos24°=0N=Fgcos24°

Relation between static frictional force and normal force is ,

fs=μsN=μsFg=(0.63)Fgcos24°=1.02×108N

And,

Fgsin24°=(1.8×107)(9.81)sin24°=7.18×107N

In this case, upward forcefsis very greater than the downward force Fgsin24°

i.e.fs>Fgsin24°.

Therefore, the block will not slide.

05

(b) Determining the minimum value of force magnitudeF that will trigger a slide down the plane

Consider the force applied by the ice is F, then by using the Newton’s 2nd law of motion,

Fx=max

ax=0,Since, block is not moving

fs-Fgsin24°-F=0F=fs-Fgsin24°=1.02×108N-7.18×107N=3.0×107N=3.0×107N

Therefore, the minimum value of force magnitudeF, that will trigger a slide down the plane is3.0×107N

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