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Fall 2008 -- Purdue University -- West Lafayette, IN
Blog Post LABELS
- Blog Challenge Questions (9)
- Chapter 2 (2)
- Chapter 3 (19)
- Chapter 5 (11)
- Chapter 6 (23)
- Chapter 7 (7)
- Chapter 8 (9)
- Exam 1 (5)
- Exam 2 (4)
- Exam 3 (17)
- Exams (2)
- Final Exam (7)
- General (6)
- Gravity (2)
- Homework (2)
- in class (1)
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- Quizzes (2)
- Spring 2008 Archive (87)
- Spring-Mass Simulator (1)
- Student-Generated Solutions (1)
- Summer 2008 Archive (72)
10 comments:
I hope not.
Where do we even begin with this problem?
May I know if the whole thing is in vertical or horizontal?
seriously. tried it. can't get a right answer. definitely appears ahead of or unrelated to what we're learning.
I haven't figured it out yet, but I assume it would be in the horizontal plane. If not, the normal force acting on the plunger would depend on which position you choose to draw the FBD.
i guess you're supposed to use polar coordinates, which in that case, i assumed at point G an opposite normal and gravitational force in the Er direction, and a negative friction component in the Etheta direction.
i solved for N after summing the forces in the Er, then plugged that value of N in for friction in the Etheta direction to obtain a force that would appear to cause a moment, making the Etheta kinematics irrelevant, which would make sense because you're not told whether theta(dot) is constant.
if those steps are right (hopefully), i still am unsure where i would take a moment.
You can use either polar or path, but polar is probably easier to understand.
There will be no gravitational force, because it's in the horizontal plane like I said before.
Otherwise, you are on the right path. Don't forget your equation will only solve for the moment caused by one plunger, when there are a total of 4 acting on the wheel.
I also think that the problem must take place in the horizontal plane, as otherwise the problem would become a vertical circle problem where the normal force exerted by the wheel on the plunger would be greater at the bottom than at the top, yet you are not given a specific position at which to calculate the force (depending on the angle of the device, the net/total normal force would vary).
Therefore, you can draw the FBD of a plunger B and see that there is a normal force pointing in the En direction (I used path coordinates, since it is a rotating body with no angular accelration (assumption, but if there were, this problem would not be solvable unless we had this value, as the only way moment can be generated is if the force is perpendicular to the moment arm, meaning the force must be in the Et direction)), and the Normal force due to the "rotating spider" pushing from one side and the friction with the wheel (on the bottom of the plunger, if you consider the bottom plunger) pointing in the opposite (-Et) direction. Knowing that the sum of the forces in the Et direction is zero and that the sum of the forces in En correspond to the mass of the plunger times acceleration in the En direction (v^2/p), you can solve the equilibrium equations that you get from the FBD to find the normal force exerted by the spider, which must be the same as the friction force. Since forces have equal and opposite reactions, this same frictional force acts in the opposite direction on the wheel, generating a moment (note that there are 4 plungers). Therefore, by multiplying the net force generated by the friction by the moment arm (distance from the middle of the spider to the wheel), you should get the net moment generated (this gave me the right answer). In order to do this I believe you have to assume that the point of application between the plunger and the wheel is exactly at the wheel, which is not fully true since there is that small part of the wheel that stands out on which the plunger has contact, but I think this is a valid assumption to make.
moneyyy, i got it. neglecting gravity, realizing that one has to account for 4 plungers, and using the entire radius of the wheel, provided a correct result. thanks panda.
Yep, glad I could help
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