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AH

Carnegie Mellon University

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Problem 11

A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined $15^{\circ}$ above the horizontal.

(a) If the coefficient of static friction is $0.50,$ what minimum force magnitude is required from the rope to start the crate moving?

(b) If $\mu_{k}=0.35,$ what is the magnitude of the initial acceleration of the crate?

Answer

(a) 304 $\mathrm{N}$

(b) 1.3 $\mathrm{m} / \mathrm{s}^{2}$

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## Discussion

## Video Transcript

free body diagram for the mass going straight up would, of course, be forced. Normal. Going straight down would be mg gravitational force at an angle here defined Fada, we have the tension from the rope and then opposite to the direction of motion the frictional force. After drawing the free body diagram, we can then apply the Newton's second law in the X and my directions. So t the tension co sign of data minus the frictional force would be equaling zero. This is the system at a constant velocity. Or of course, that rest. And we can say that two sign of data plus the force normal minus the gravitational force MG. This will equal zero and we know that data in this case is equaling 15 degrees. We can say that here, the first equation gives that the frictional force would be going to tea co sign of data and the second gives that the forced normal is gonna be equaling to the gravitational force. Mg minus t sign If Ada and we know that if they're create is to remain at rest. This means that the frictional force must be less than the coefficient of static friction multiplied by the force normal. Or we can say that T co sign of data must be less than the coefficient, static friction and then substituting it for the force Normal mg minus t Sign of data we can solve for we can say when the tension forces sufficient to just start the creek moving. So just in order to start the create removing just exceeding that maximum static frictional force, we can say that t co sign of data will be equaling two. The coefficient of static friction mg minus t sign of data and four party we can solve for the tension. Attention. T would then be equaling the coefficient of static friction times the gravitational force divided by coastline of Florida, plus the coefficient of static friction multiplied by sine of data we can solve. This would be 0.50 multiplied by 68 kilograms multiplied by 9.8 meters per second squared. This is gonna be divided by co sign of 15 degrees plus 0.50 multiplied by sine of 15 degrees. And we find that the tension forced t is gonna be equaling two, approximately 304 Newton's so I'd be your answer for part A For part B, we can say that. Rewriting Newton's second law for the moving create. So now applying Set. Applying Newton's second law in the Ex direction Chico sign of data minus the frictional force will equal the mass times acceleration of the system. And then we have t sign of data plus the force normal, minus mg equaling again. Zero. So, in the UAE direction, of course we're way have translational equilibrium. However, in the extraction of course we have. The crate is moving at a certain acceleration. Now we have that the friction is equaling the kinetic. The coefficient of kinetic friction multiplied by the force normal and the force normal is now equaling mg minus t sign of data. And we can then say that the frictional force would be equaling to the coefficient of kinetic friction multiplied by M G minus t sign of data so we can substitute this expression into the first equation here. And we can say that then she co sign of data minus the coefficient of kinetic friction multiplied by M G minus T sign of data. This is gonna be equaling to the mass times acceleration of the crate. And so the acceleration of the crate for Part B would be going t multiplied by co sign of data, plus the coefficient of kinetic friction multiplied by sine. If ADA, this would be divided by the mass minus tickle efficient, kinetic friction multiplied by the acceleration due to gravity, weaken, substitute and solve. 304 Newtons multiplied by co sign of 15 degrees plus 0.35 multiplied by sine of 15 degrees, divided by the mass of 68 kilograms, minus 0.35 multiplied by 9.8 meters per second squared. And we find that the acceleration A isn't going to be equaling to approximately 1.3 meters per second squared. This would be our final answer for part B. That is the end of the solution. Thank you for watching

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