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JoshuaR
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1. Tilting trains such as the Acela, which runs from Washinton to new York to Boston, are designed to travel safely at high speeds on curved sections of track which were built for slower, conventional trains. As it enters a curve, each car is tilted by hydraulic actuators mounted on its trucks. The tilting feature of the cars also increases passenger comfort by eliminating or greatly reducing the side force Fs (parallel to the floor of the car) to which passengers feel subjected. For a train traveling at 125mi/h on a curved section of track banked at an angle theta=8 and with a rated speed of 75 mi/h determine (a) the magnitude of the side force felt by a passenger of weight W in a standard car with no tilt (phi=0) (b) teh required angle of tilt phi if the passenger is to feel no side force.
(There is a diagram in the book of a slight bank (8 degrees) and a train on it with trucks tilting another phi degrees).
2. N=mgcos@ F=ma=mv^2/r = mgcos@sin@ (the horizontal component of the normal)
3. So from rated speed and above equations I got a radius of 2726.59 ft. Theoretically, however, another method for getting the radius would be to use a=v^2/R where a is found from arctan(a/g)=8 degrees rather than gcos@sin@=v^2/r This comes out to a radius of 2673.79 ft. Why do these numbers differ? Am I making a wrong assumption somewhere in one of the two methods? The problem is with the first method I get the answer from the book eventually by... a=v^2/r (using the radius found from the first method) and then calculating (a-gcos@sin@)/(gcos@)=coefficient of friction=Fs side force thingy. So using this method a would be 12.327ft/s*s and the answer matches the back of the book of 0.247W. However, if you use the arctan(a/g) method you get an answer more like .255. Why are these numbers different?
Part (b): The most simple solution is arctan(a/g)=sum of theta and phi, where phi is easily solved for as 13.3 degrees, matching the back of the book. But this is only if you use the second method initially of arctan(a/g) to find the radius! If you go by the other radius, hence acceleration, the numbers are slightly off. Worse, you can't use the first method at all here. I assumed mgcos@sin@=mv^2/r still, changing @ in this case to be the sum of theta and phi. This method, however, yields an angle of 17 degrees. (sin2@=2v^2/(gr))Definitely incorrect. Why does this method fail in this case?
As you can see. I've solved the problem, but I had to use different rationale for each part of it to match book answers. I'm unsure as to why these to rationales didn't match up. Thanks!
Josh
(There is a diagram in the book of a slight bank (8 degrees) and a train on it with trucks tilting another phi degrees).
2. N=mgcos@ F=ma=mv^2/r = mgcos@sin@ (the horizontal component of the normal)
3. So from rated speed and above equations I got a radius of 2726.59 ft. Theoretically, however, another method for getting the radius would be to use a=v^2/R where a is found from arctan(a/g)=8 degrees rather than gcos@sin@=v^2/r This comes out to a radius of 2673.79 ft. Why do these numbers differ? Am I making a wrong assumption somewhere in one of the two methods? The problem is with the first method I get the answer from the book eventually by... a=v^2/r (using the radius found from the first method) and then calculating (a-gcos@sin@)/(gcos@)=coefficient of friction=Fs side force thingy. So using this method a would be 12.327ft/s*s and the answer matches the back of the book of 0.247W. However, if you use the arctan(a/g) method you get an answer more like .255. Why are these numbers different?
Part (b): The most simple solution is arctan(a/g)=sum of theta and phi, where phi is easily solved for as 13.3 degrees, matching the back of the book. But this is only if you use the second method initially of arctan(a/g) to find the radius! If you go by the other radius, hence acceleration, the numbers are slightly off. Worse, you can't use the first method at all here. I assumed mgcos@sin@=mv^2/r still, changing @ in this case to be the sum of theta and phi. This method, however, yields an angle of 17 degrees. (sin2@=2v^2/(gr))Definitely incorrect. Why does this method fail in this case?
As you can see. I've solved the problem, but I had to use different rationale for each part of it to match book answers. I'm unsure as to why these to rationales didn't match up. Thanks!
Josh