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The principles used to solve this problem are similar to those in Multiple-Concept Example 17. A 205-kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 30.0 with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.900, and the log has an acceleration of magnitude 0.800 $\mathrm{m} / \mathrm{s}^{2}$ . Find the tension in the rope.

$2730 \mathrm{N}$

Physics 101 Mechanics

Chapter 4

Forces and Newton’s Laws of Motion

Newton's Laws of Motion

Applying Newton's Laws

Cornell University

University of Michigan - Ann Arbor

Simon Fraser University

University of Sheffield

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for the squares to now you choose the following reference frame. Are you choose my Y axis, pointing to the direction there is normal to the surface off the ramp. So do they buy why access and I would choose my X axis to point upwards the ramp in this direction. I'm choosing the X axis like that because they walk is accelerating upwards the rent, then my acceleration is positive. If I choose, my exact seems to be pointing up War of the Ramp had they choose in. My exact is pointing to the opposite direction are you have to include the minus time on the acceleration that is given by the problem. So be warning about that now. To calculate the tension force, I should apply Newton's second law on the law using my X axis because the tension force spiral to these X axis. Then, by using Newton's second law, we get the following the X direction. We have a net force component X, which is equals to the mass off the log times its acceleration. The net force next direction is composed by three forces Detention forced the frictional force on the X component off the weight force. Then the net force in the X direction is reading like tension force which points to the positive X direction minus the frictional force that points to the negative X direction minus the weight force component acts which absolute points the negative X direction. This is equals to the mass times acceleration off the log on the X direction. So the tension force is equal to the mass off the log time stops acceleration in the X direction plus the frictional force plus wait force. And now remember that the frictional force in the situation where the log is already moving is the kinetic frictional force which is given by the connected frictional coefficient times the normal force. So the tension is a cost to the mass times acceleration plus muche eight times the normal force plus the weight force component acts. And now we have to calculate the X component off the weight force. To do that, we have to take a look at this triangle here on the situation. So by extending the weight force, we get a 90 degrees angle here and as a consequence, this angle must me a 60 degrees angle in order for the intern angles off this triangle toe. Add up to 180 degrees. Now note that this angle between W Why on the ramp is a 90 degree angle, meaning that these in turn angle here is 30 degrees. Now we can use this 30 degree angle to debt. Remind What is the X component off the weight force? No, the following the sign off 30 degrees in the situation in this triangle is given by they always excite which is the will you x divided by the high part two news which is not where you then no, will you x Is it close to W times the sign off 30 degrees, then the tension force is the mass times acceleration plus, um u k times the normal force plus the weight forced times The sign off 30 degrees and the wait for us, we know is given by the mass off the log times, acceleration of gravity. Then attention forest. He's even by these expression. So now all we have to do is complete. What is the normal force for that? We use Newtown second room on the Y axis the net force on the y axis is G equals to the mass off The Lord Times its acceleration in the Y direction and notes that the log isn't moving on these direction or these direction. Then its acceleration on the Y direction Is it close to zero? So the net force on the widely reaction is it was 20 But the net force in that direction has two components. The normal force and the weight forest component white. And it's written like normal force, which points to the positive by the reduction minus weight forest component. Why? Which points to the negative by direction? This is a question zero. So the way the normal force is the course to the wait for its components why and for the wind force component why we can return to this try and do. But now we use the coastline because the y component largest sense screwed the 30 degrees angle the co sign off 30 degrees on that context is given by they are just inside provided whether party news then the y component off the weight forces the cost of the weight forced times national sign off 30 degrees. So the normal forces because to the weight forced times the go sign off 30 degrees The normal forces, then m times g times the co sign off 30 degrees. Now let me clear the board so that we can finish solving this question. Okay, so we can change change. Just normal force by these expressions for the normal force. The in detention he's given by m Times e was um UK Times, AM times G Co sign off 30 plus m g fine off 30 Now plaguing the values that were given by this problem, we get the attention forces he costume 205 times 0.8 plus 0.9 times 205 times 9.8 time school sign off 30 plus 205 times 9.8 times this sign off 30 and these news is attention. Force off approximately 2730 neutrons

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