Figure 6-50 shows the cross section of a road cut into the...

Figure $6-50$ shows the cross section of a road cut into the side of a mountain. The solid line $A A^{\prime}$ 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 is $1.8 \times 10^{7} \mathrm{~kg}$, the dip angle $\theta$ of the bedding plane is $24^{\circ}$, and the coefficient of static friction between block and plane is $0.63 .$ (a) Show that the block will not slide. (b) Water seeps into the joint and expands upon freezing, exerting on the block a force $\vec{F}$ parallel to $A A^{\prime} .$ What minimum value of $F$ will trigger a slide?

Key Concepts

Static Friction

Static friction is the resistive force that prevents relative motion between two surfaces in contact until a threshold force is exceeded. In problems involving sliding on an inclined surface, static friction provides the necessary resistance against the component of gravity acting parallel to the plane, and its magnitude is determined by the normal force times the coefficient of static friction.

Forces on an Inclined Plane

Analyzing forces on an inclined plane involves decomposing the gravitational force into two components: one perpendicular to the plane and one parallel to it. The perpendicular component determines the normal force, which influences both frictional resistance and overall stability, while the parallel component drives potential sliding motion.

Equilibrium Conditions

Equilibrium conditions describe the balance of forces acting on an object, meaning that the net force is zero and the object remains at rest. In the context of this problem, establishing equilibrium involves showing that the downslope force due to gravity is counteracted by friction, and determining the additional force needed to overcome this balance when an external force is applied.

Gravitational Force Decomposition

Decomposing gravitational force is a crucial step in problems involving inclined planes. The weight of an object is broken down into two components: one along the plane, which contributes to the tendency to slide, and one perpendicular to the plane, which influences the normal force and thus the maximum static friction available.

Effect of External Forces

External forces, such as the force exerted by freezing water expanding in joints, can alter the force balance acting on a block. Analyzing the effect of such forces involves determining how they add to or counteract the existing components of gravitational and frictional forces along the plane, thereby changing the conditions for initiating motion.

Transcript

00:01 Previously, we've established that the coefficient of static friction on a slope determines the largest angle of that slope before the object will start sliding.

00:14 And this equation, which is worked out in one of the example problems, is given as the static friction needs to be greater than or equal to tangent of that angle.

00:25 If the static friction is not greater than the tangent of angle, then your object will slide down the slope.

00:36 So we just need to plug in our value for the angle here, which is 24 degrees, and see if that plays out.

00:46 Well, tangent of 24 degrees is 0 .44, and the coefficient of static friction that we are told in the problem is 0 .6 .3.

00:58 Three.

00:58 So this proves to ourselves that this mass will not start sliding just yet.

01:09 Part b of this problem asks us to figure out how much additional force given by a freezing body of water in the crack will be required to make this mass begin to move.

01:24 And so we're going to start by figuring out the forces that are acting on this thing.

01:32 In the wide direction, we have a portion of the weight force, mg cosine theta, and that is balanced by the normal force.

01:44 So normal force minus mg cosine theta.

01:51 It's going to be equal to zero, or in other words, normal force is equal to mg cosine theta.

01:59 Then in the parallel direction here, we've got a couple of forces...

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