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Problem 15

The coefficient of static friction between Teflon…


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Problem 14

Figure $6-22$ 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 under these circumstances. (b) Next, 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 force magnitude $F$ will trigger a slide down the plane?


(a) It won't move
(b) $3 \times 10^{7} \mathrm{N}$



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Video Transcript

so here. A lot of the analysis was obtained in sample problem friction applied force at an angle. So in that example, problem the analysis. The analysis gives an equation for the maximum angle for which static friction applies. So we can say that the maximum angle would be equal to Arc tanne of the coefficient of static friction. And this is equaling Tangent Arc Tangent of 0.63 And this is approximately 32 degrees now 32 degrees. However, this is going to be greater then the angle stated in the problem. And if that's the case, my apologies greater than and if that's the case, the block does not slide. So this would be your answer for party because the maximum angle does not exceed the dip angle in the problem. Therefore, the block does not slide now. For Part B. We need to apply suit a Newton's second law on. We can say that the sum of forces in the X direction would be equal to the mass times acceleration in the extraction. We know that this is going to be zero, and we know that there's gonna be the force applied plus mg sign Athena minus the maximum static frictional force. And we can say that the's some of forces in the UAE direction would be equal to the mass times acceleration in the right direction. This would also be equal to zero because the system has translational equilibrium in both the X and Y directions. Ah, this one equal force normal minus m g co sign of data. And in this case, we can say that if, uh, force normal would be equal to m g co sign. If Ada and we have a new angle of 24 degrees and we have a mass equaling 1.8 times 10 to the seventh kilograms, we can find the force needed in order to, um, keep this block stationary essentially so the force would be equal to m g times the coefficient of static friction. This would be Times Co sign off Ada minus sign of Fada. And once we plug in all of our all of our knowns, we also can plug in that the coefficient of static friction equals 0.63 So once we plug in the mass, the coefficient of static friction, the angle as well as the ah ah the force nor force normal into the ex. The sum of forces in the ex direction we can solve for the force applied. So the force applied would be equal to 1.8 times 10 to the seventh kilograms multiplied by 9.8 meters per second. Squared times 0.63 co sign of 24 degrees minus sign of 24 degrees. And we find that the force is going to be equal to 3.0 times 10 to the seventh Newton's. So this would be our final answer again. We have to round 22 significant figures because our mass is only accurate to two significant figures. Therefore, we must around to two significant figures for that force again. 3.0 times 10 to the seventh. Nunes is your friend Lancer. That is the end of the solution. Thank you for watching

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