- #61
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Not yet. Do you recognize the ratio in the argument of the log for what it is? Specifically, can you simplify the numerator some?LockeZz said:
Motion with frictional force but without driving force
- Thread starter LockeZz
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- Force Frictional force Motion
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Homework Help Overview
The problem involves an engine moving on horizontal rails under the influence of a frictional force defined as f(v) = α + βv², with an initial velocity v₀. The questions focus on determining the time it takes for the engine to come to rest and the distance covered during this time.
Discussion Character
- Exploratory, Assumption checking, Problem interpretation
Approaches and Questions Raised
- Participants discuss the application of Newton's Second Law and the need to express acceleration in terms of velocity. There are attempts to derive a general expression for velocity as a function of time, and questions arise about the nature of acceleration and whether it remains constant.
Discussion Status
The discussion is ongoing, with participants exploring various interpretations of the problem. Some have provided guidance on the need for correct integration and the implications of changing forces on acceleration. There is no explicit consensus yet, as participants continue to clarify their understanding and correct previous errors.
Contextual Notes
Participants are working under the constraints of homework guidelines, which require them to show their work and reasoning without receiving direct answers. There is an emphasis on understanding the relationship between velocity, force, and acceleration in the context of the problem.
- #62
LockeZz
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i think i can convert to tangent:
<br /> x=-\frac{m}{\beta}(\ln(\tan(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})))<br />
then simplify again:
<br /> x=-\frac{m}{\beta}(\ln(v_0\sqrt{\frac{\beta}{\alpha}}))<br />
<br /> x=-\frac{m}{\beta}(\ln(\tan(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})))<br />
then simplify again:
<br /> x=-\frac{m}{\beta}(\ln(v_0\sqrt{\frac{\beta}{\alpha}}))<br />
- #63
- 15,937
- 9,126
Note that if you let v0 become very large, you get an infinity. I hadn't appreciated that in the beginning. One thing that bothers me is the negative sign up front. Certainly x must be positive and the argument of the log is greater than 1 especially if v0 becomes very large. Check your integral tables and don't forget that you are integrating
<br /> \int Tan(\delta - t)dt<br />
There is a negative sign in front of t.
Check that and you are done.
<br /> \int Tan(\delta - t)dt<br />
There is a negative sign in front of t.
Check that and you are done.
- #64
LockeZz
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hm.. does that mean that the negative sign represent a mistake in the expression?
- #65
- 15,937
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All I am saying is that
<br /> <br /> \int Tan(\delta - t)dt=Log[Cos(t-\delta)]<br /> <br />
<br /> <br /> \int Tan(\delta - t)dt=Log[Cos(t-\delta)]<br /> <br />
- #66
LockeZz
- 34
- 0
I shall check back from the beginning the see if any mistake again.. thanks for the guidance by the way.. i really appreciate on your help.. Thank you very much :)
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