In the 3rd line you have taken "Ln" separately - that was illegal operation You could take the "Ln" of the whole Right-hand-side - but that don't take you closer to solution. Instead you should do following:Hey all,
Have I gone wrong somewhere in this transposition/final equation, seems way overkill to get there.
Plus the final result came out negative when I thought it should be a positive value? (Friction coefficient)
Any advice would be great.
In the 3rd line you have taken "Ln" separately - that was illegal operation You could take the "Ln" of the whole Right-hand-side - but that take you closer to solution. Instead you should do following:
\(\displaystyle \frac{P}{VT_1} \ = \ 1 - \ e^{-\frac{\mu \theta }{sin( \alpha )}}\)
\(\displaystyle 1 - \ \frac{P}{VT_1} \ = \ e^{-\frac{\mu \theta }{sin( \alpha )}}\)
\(\displaystyle ln \left[ 1 - \ \frac{P}{VT_1} \right] \ = \ ln \left[ e^{-\frac{\mu \theta }{sin( \alpha )}} \right] \)
continue......
Show your work according to your understanding and we will discuss further.Thanks for your reply, I will have another go with that correction. Am I wrong in thinking that line 5 onwards the right side will not change a great deal? A little bit confused.
Show your work according to your understanding and we will discuss further.
Looks good to me.Here's what I came up with.
Looks good to me.
I think I have made a mistake still as my answer is still coming out negative. Coefficient of friction should always be positive right?
Now for a final answer I'm getting -0.2709
Could it be an error where I should have used (-sinx)
You dropped a negative sign at the next to last line.Here's what I came up with.
You dropped a negative sign at the next to last line.
Yes I was pretty sure that was the problem. What's confusing me is I'm almost certain the final solution is positive 2.857×10^-3 but I cant seem to get there?
You were told that you had dropped a "negative" sign. Did you fix that?Yes I was pretty sure that was the problem. What's confusing me is I'm almost certain the final solution is positive 2.857×10^-3 but I cant seem to get there?
You were told that you had dropped a "negative" sign. Did you fix that?
I did have a go. Whether it's right or not is another matter.
I see this time that you took [MATH]P = 5.25\times 10^3[/MATH], and I see that you did that in your original work, too. I suppose that must be right; but you never told us how the variables in the original equation are defined, which is why I assumed it should just be 5.25. Evidently P is in watts, and P is 5.25 kilowatts?
In that case, I, too, get [MATH]\mu = 0.27[/MATH]. Did you try checking it in the original equation?
What you said was that P = 5.25 kW, without saying explicitly that P is defined as in watts, not kW. But it was implied by your work.The values/units for each variable are in post #1 but sorry about that I should have included in the rest.
I would not have moved the negative inside the sine, though it's not wrong (since the sine is an odd function). I would have just put a "-" to the left of the whole expression, since you multiplied by -1 and by [MATH]\sin(\alpha)[/MATH].I was unsure if how I moved the negative sign was legal. e.g flipping the exponent to get sin(-x) on the left side.