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mmh A toboggan slides down a hill and has a constant velocity. The angle of the hill is $8.00^{\circ}$ with respect to the horizontal. What is the ceefficient of kinetic friction between the surface of the hill and the toboggan?

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University of Washington

Simon Fraser University

Hope College

University of Winnipeg

in this question, this guy uses lighting down with Toby Logan that makes eight degrees sweep the ground acting u netminder three forces the weight forced the normal force and the frictional force. Given that, how can you calculate what is the kinetic coefficient of friction in the situation? For that, we have to use Newton's second law in a special reference frame, which is these one. These will be what I call the white direction, and you will be what I would call the X direction, then applying Newton's second law on the wider action results in the following to the Net force on that direction is equal to the mass off the guy times acceleration off the guy in the direction note that he's not flying away, more going blow the Toby Logan show. Acceleration in this direction is a question zero, Then the net force on the Y direction Is it close to zero? But the net force in that direction is composed by true forces. One is the normal force, which is pointing to the positive direction off the Y axis, and the other one is the Y component off the way force, which he's these one. So this is the Y component off the weight force and it's pointing the negative direction off the Y axis. So and my nose nowhere. You Why? Easy question. Zero. Finally, the normal force is he goes to the y component off the weight force. Okay. Now applying the same equations of applying Newton's second law to the X axis results in the following the Net force on that access is he goes to the mass off the guy times his acceleration in this direction. His is lighting with a constant velocity. So that separation in his direction is it close to zero then and that forced next interaction is close to zero. But the Net force next direction is composed by true forces, the frictional force which points to the negative direction off the X axis and the axe component off the weight force which is point which points to the positive direction off the X axis. So these things w x then w x my news defection or force is equal to zero. Therefore, the frictional force is the course w x. But now what do we do with that? Remember that different channel force is given by the kinetic frictional coefficient times the normal force shoot the kinetically from play fusions times The normal force is the course to W. X. But we want to calculate what is the frictional coefficient. And the frictional coefficient can be given by the X component off the weight forced, divided by the normal force. But as we had seen before, the normal force is equals to the Y component off the weight force. Then UK is the course to w X divided by the way. Why, then, we have to use these rectangle. Try and go here to finish over. In this question, this triangle looks like this. So we have here the high party news one off the sites on the other side and this is a 90 degree angle. Then this is the full weight force disease, though where you acts on the easiest w white. Now what is this angle and this angle to? We can use this under rectum. Who triangle here, whose fault? It These clients triangle. These is the 90 degree angle before these other angle has 82 degrees, meaning that this angle inside the triangle must have eight degrees. So this is an 80 degree angle, and this other one is an 80 degree angle. Now what do we do with that? Well knows that the tangent off this eight degree angle is given by the opposite side off the triangle W X provided by the address inside off the triangle W Y. Then the kinetic frictional coefficient is given by the time gent off this eight degree angle, and these results in a frictional coefficient off approximately 0.1 for one, and he is the answer to these questions.

Brazilian Center for Research in Physics