Homework Help: Friction problem- possible error in Halliday/Resnick

1. Oct 27, 2012

Syrus

1. The problem statement, all variables and given/known data

See attachment. I am only concerned with part b). Part a) is solved.

2. Relevant equations

3. The attempt at a solution

See attachment. The solution is quoted to be tan-1s) = θ0, but, as can be seen by my solution, this is impossible based on the derivation. Is this their error, or mine?

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Last edited: Oct 27, 2012
2. Oct 27, 2012

frogjg2003

What did you do for part a?
Not sure if it will work, but try taking your answer for part a, and having v->0.

3. Oct 27, 2012

Syrus

I solved for F in part a). I display the results (verified correct) at the top of the page. I'd like to see how the value for theta naught is explicity obtained.

4. Oct 27, 2012

frogjg2003

You said the solution is tan(μs) = θ0. Do you mean tan-1s) = θ0?

5. Oct 27, 2012

Syrus

Yes, apologies- it's edited now.

6. Oct 27, 2012

frogjg2003

Well, just take the limit when force goes to infinity. When the angle is less than θ0, no amount of force will be able to move the mop.

7. Oct 27, 2012

Syrus

I considered that, but I wondered if there were any better ways of showing this. Plus, this example is from an elementary (Halliday/Resnick) text, and such un-straightforward proofs seem unwarranted in problem-solving. Just some thoughts. Is an explicit derivation out of the question?

8. Oct 27, 2012

frogjg2003

This seems like one of the more "challenging" problems. As such, more out-of-the-box thinking is required. This method is pretty good as it stands. What would you consider a "better" way?
Why is this an "un-straightforward proof" or not "an explicit derivation"?

9. Oct 27, 2012

Syrus

Fair enough. The only reason I said what I did above was becuase other problems and results derived throughout the book are a bit more explicit in using Newton's laws to demonstrate the results. I can accept this method, just inquiring about other manners.

10. Oct 27, 2012

Syrus

The value for F should be multiplied by μk.

11. Oct 28, 2012

ehild

You overcomplicate the problem a bit.
You got $$F=\frac{mg\mu_k}{\sin(\theta)-\mu_k \cos(\theta)}$$
The mop is pushed, so it can not be negative. What does it mean for theta?

ehild