# motion with frictional force but without driving force

by LockeZz
Tags: driving, force, frictional, motion
 HW Helper PF Gold P: 3,444 Can you make the expression more compact? What is the difference between two logarithms? Also set t0=0 and see what you get.
 P: 34 ok.. the logarithm still can be simplify : $$x-x_0=-\frac{m}{\beta}(\ln(\frac{\cos(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1)}{\cos(\arctan(v_0\sqrt{\frac{\beta}{\a lpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0)})$$ so.. for t0=0.. i will get : $$x-x_0=-\frac{m}{\beta}(\ln(\frac{\cos(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1)}{\cos(\arctan(v_0\sqrt{\frac{\beta}{\a lpha}}))})$$
 HW Helper PF Gold P: 3,444 Good. I can see the light at the end of the tunnel. Can you? What do you think you should do next?
 P: 34 get rid of the x0 ad substitute the t1?
HW Helper
PF Gold
P: 3,444
 Quote by LockeZz get rid of the x0 ad substitute the t1?
Yes, you may assume that x0=0. Substitute t1 and also the time has come to let v0 become very large.
 P: 34 after the substitution, i get : $$x=-\frac{m}{\beta}(\ln(\frac{\cos(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\pi}{2})}{\cos(\arctan(v_0 \sqrt{\frac{\beta}{\alpha}}))})$$ since $$t_1=m(\frac{\pi}{2})\sqrt{\frac{1}{\alpha\beta}}$$ should i bother about v0?
HW Helper
PF Gold
P: 3,444
 Quote by LockeZz should i bother about v0?
Not yet. Do you recognize the ratio in the argument of the log for what it is? Specifically, can you simplify the numerator some?
 P: 34 i think i can convert to tangent: $$x=-\frac{m}{\beta}(\ln(\tan(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})))$$ then simplify again: $$x=-\frac{m}{\beta}(\ln(v_0\sqrt{\frac{\beta}{\alpha}}))$$
 HW Helper PF Gold P: 3,444 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 $$\int Tan(\delta - t)dt$$ There is a negative sign in front of t. Check that and you are done.
 P: 34 hm.. does that mean that the negative sign represent a mistake in the expression?
 HW Helper PF Gold P: 3,444 All I am saying is that $$\int Tan(\delta - t)dt=Log[Cos(t-\delta)]$$
 P: 34 I shall check back from the beginning the see if any mistake again.. thx for the guidance by the way.. i really appreciate on your help.. Thank you very much :)

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