## marianopolis one year ago A rope is tied to a large crate, which is sitting on a flat surface. The coefficient of static friction between the crate and the ground is 0.9. If a person is to pull on the rope with the minimum force needed such that the crate begins to slide, the angle between the rope and the ground should be A) greater than 0 degrees but less than 90 degrees B) 0 degrees (rope is horizontal) C) 90 degrees I know the answer is A.

1. marianopolis

Now, I was trying to prove it: horizontal pulling: F=ukmg. greater than 0 degrees but less than 90 degrees: Horizontal component: uk(mg-sinthetaF) (Let F be magnitude of force.) Vertical component: costhetaF Total magnitude: (Pythagoras) and (sin^2theta+cos^2theta=1) 0.81mg-1.8sinthetaF+F^2=Magnitude. How do I prove 0.81mg-1.8sinthetaF+F^2 > 0.9mg?

2. theEric

Hi! I want to follow this from the beginning. You are using two different $$F$$'s, right? So let's say $$F_f=\mu_kmg$$ and $$F$$ is the minimal force we're looking for. Then we'll define $$\theta$$ to be the angle between the ground and rope length, like you did I think.. Then the vertical component is $$\mu_kF_N=\mu_k(mg-F\sin\theta)$$ And the horizontal is $$F\cos\theta$$ Besides switching the horizontal and vertical components, we're on the same page. And then you look to identify the total force. $$\sqrt{\left(\mu_k(mg-F\sin\theta)\right)^2+\left(F\cos\theta\right)^2~~}$$ $$=\sqrt{\mu_k^2(mg-F\sin\theta)^2+\left(F\cos\theta\right)^2~~}$$ $$=\sqrt{\mu_k^2(m^2g^2-mgF\sin\theta+F^2\sin^2\theta)+\left(F\cos\theta\right)^2~~}$$ $$=\sqrt{\mu_k^2(m^2g^2-mgF\sin\theta+F^2\sin^2\theta)+F^2\cos^2\theta~~}$$ $$=\sqrt{\mu_k^2m^2g^2-\mu_k^2mgF\sin\theta+\mu_k^2F^2\sin^2\theta+F^2\cos^2\theta~~}$$ And we can't isolate the $$\sin^2\theta+\cos^2\theta$$. Also, I see different powers of $$F$$ in there.

3. theEric

Although I guess we should use $$\mu_s$$ for static friction.

4. theEric

Wait... I did that wrong. I looked at everything all wrong... It's been a long day!

5. marianopolis

Dang!

6. theEric

Haha, I'm about to take another look, sorry.

7. marianopolis

No problem. What you have done, I already consider great ;)

8. theEric

Haha, it's wrong, don't look at it too long! :)

9. theEric

I guess we can say the horizontal component of $$F$$ is $$F\cos\theta$$ And this should be greater than the maximum static friction force, which is given by $$F_f=F_N\mu_s=(mg-F\sin\theta)\mu_s$$ So, $$F\cos\theta>(mg-F\sin\theta)\mu_s=mg\mu_s-F\sin\theta\ \mu_s$$ $$\implies F\cos\theta+F\mu_s\sin\theta>mg\mu_s\\ \implies F(\cos\theta+\mu_s\sin\theta)>mg\mu_s\\ \implies F>\dfrac{mg\mu_s}{\cos\theta+\mu_s\sin\theta}$$ Note for the inequality: when multiplying or dividing the inequality, a negative sign will flip the inequality. We are looking at the range from $$0^\circ$$ to $$90^\circ$$, in which both $$\sin\theta$$ and $$\cos\theta$$ are positive. So, $$\cos\theta+\mu_s\sin\theta$$ is positive, and we don't flip the inequality for those angles. Since $$\cos\theta+\mu_s\sin\theta$$ is in the denominator, it can't be $$0$$. But that's not a worry unless $$\mu_s=0$$ and $$\theta=90^\circ$$, but $$\mu_s\neq0$$, so our denominator is okay. The whole term that $$F$$ must be greater than is a minimum for the angles where the denominator is largest. So... And I would imagine that you're sick of the denominator, but $$\cos\theta+\mu_s\sin\theta$$ must be maximum. Which really depends on $$\mu_s$$. If $$\mu_s$$ is close to $$0$$, then the denominator is about $$\cos\theta+0\dot\ \sin\theta=\cos\theta$$ which maxes at $$0^\circ$$. If $$\mu_s$$ is huge, like $$9999$$, then the denominator is $$\cos\theta+9999\dot\ \sin\theta$$ So, it makes sense that $$9999\dot\ \sin\theta$$ will be the most deciding term. So, if you want the denominator big, put your stock in the sine function. For the sine function to be greater, you want $$\theta$$ closer to $$90^\circ$$. If $$\mu_s$$ something lower, then you'll want $$\theta$$ to be somewhere in-between.

10. marianopolis

Okay, I am half-way through.

11. marianopolis

Eric, when we say "The whole term that F must be greater than is a minimum for the angles where the denominator is largest.", what de we mean? Why does that have to be true? Now, I seem to be missing the central point here ([ashamed]). How did we prove that F directed at a certain angle is the minimal force?

12. marianopolis

do we mean*

13. theEric

I actually haven't solved for the best angle sorry.. Just thinking... And that is from $$F>\dfrac{mg\mu_s}{\cos\theta+\mu_s\sin\theta}$$ So, $$F$$ can be smaller if the denominator is smaller. I think I tried to say too much in one sentence, sorry! That's what math is for anyway. But, if we have $$\frac1x$$, it's smallest when the denominator is largest. So our denominator should be largest. So we pick the $$\theta$$ to make the denominator largest. That is the optimal angle (that I can't find).

14. theEric

Note that $$\mu_s$$ won't ever be $$9,999$$. Unless the crate is sticky, or something. It will probably be a little below $$1$$, like the $$0.9$$ in the problem.

15. marianopolis

Hmm. I am still lost. Where the statement that Fmin is necessarily smaller than Fhorizontal? Sorry, I am really bad at physics.

16. theEric

Don't worry about it! I'm not the best, either. Math is somewhat of a weakpoint, also. But, let's go back to $$F>\dfrac{mg\mu_s}{\cos\theta+\mu_s\sin\theta}$$ and think about $$F=\dfrac{mg\mu_s}{\cos\theta+\mu_s\sin\theta}$$ since it's easier. What would make $$F$$ really small? I mean, we can't $$\rm choose$$ $$m$$, $$g$$, or $$\mu_s$$. All we can adjust is $$\theta$$. To make $$F$$ as small as possible, all we can do is make the denominator as BIG as possible, by changing $$\theta$$. I guess I should go back to the magnitude, for you. Don't read it if you don't want to! Since I did something weird before... First I break up the pulling force again. Horizontal is at least $$F_N\mu_s=(mg-F\sin\theta)\mu_s=mg\mu_s-F\sin\theta\ \mu_s$$ Vertical is just $$F\sin\theta$$. I guess the magnitude would be this, but I don't know how it helps. $$\sqrt{(F\sin\theta)^2+(mg\mu_s-F\sin\theta\ \mu_s)^2~~}$$ $$=\sqrt{F^2\sin^2\theta+m^2g^2\mu_s^2-mg\mu_s^2F\sin\theta+F^2\sin^2\theta\ \mu_s^2~~}$$

17. marianopolis

OHHH

18. marianopolis

ok, I reread the whooole post.

19. marianopolis

I think I am starting to understand.

20. marianopolis

21. marianopolis

So, let me sum up to see if I got this clear.

22. theEric

All that I did is show that it would be best for the angle to be somewhere between $$0$$ and $$90$$ degrees. I didn't find any best angle. Haha, you're trying! I am, too :P So you feel more comfortable with this now?

23. marianopolis

We ended with that relationship. We pluged in 0 and 90 and a random other number. The random number gave a better result.

24. marianopolis

Out of curiosity, how would I find the optimal angle?

25. theEric

That sounds good! Optimal angle... Time for the thinking cap.

26. theEric

So, a formula for the optimal angle as a function of $$\mu_s$$...

27. marianopolis

By the way, if you dont mind me asking, what is your background? You're good at this.

28. theEric

Thanks! I have not shined in this post, but thanks! :) Undergraduate physics and computer science majors at a state university in Pennsylvania.

29. theEric

I've had lots of time to practice :)

30. marianopolis

ohhh ok. ;) that's why

31. theEric

Haha, yep!

32. marianopolis

Could taking the derivative of both sides help? I do not conceptually understand how it would, but the textbook sometimes does that.

33. marianopolis

Oh of course! It finds the minimal value!

34. marianopolis

All those cal courses now come back to me!

35. theEric

I didn't know you knew about derivatives and stuff! Well, derivatives will give you rates of change. It's like a slope. So, you take the derivative of $$f(x)$$ and you get $$\dfrac{{\rm change\ in\ }f(x)}{{\rm change\ in\ }x}$$ |dw:1397356730399:dw|The function has to turn around at maximums and minimums. Then the function's value doesn't change for an instant as the independent variable goes on. Just like when you throw a ball up in the air, it has to stop before it comes down. And at the turn-around point, there is no change, and the derivative is $$0$$. So, if we have a function to describe the force, $$F(\theta)$$, and we differentiate it (find its change) with respect to $$\theta$$... Well, we can look at when it equals $$0$$.

36. theEric

Haha, glad they came back to you!

37. theEric

I don't know how to make it help at the moment, though!

38. theEric

Finding the angle....

39. theEric

Hmm...

40. marianopolis

Well, if we set the derivative to 0, the moment where there will be a min/max value, we'll be able to find the angle.

41. marianopolis

I might be wrong though!

42. theEric

Right! And I didn't think it would work for a certain reason that doesn't matter because it was wrong! But we'd still have to be able to solve for $$\theta$$. If we focus on finding the max of the denominator... $$\dfrac d{dt}(\cos\theta+\mu_s\sin\theta)=-\sin\theta+\mu_s\cos\theta$$

43. theEric

$$-\sin\theta+\mu_s\cos\theta=0$$ $$\implies\mu_s\cos\theta=\sin\theta$$ $$\implies\mu_s=\dfrac{\sin\theta}{\cos\theta}=\tan\theta$$ $$\implies\arctan(\mu_s)=\theta$$ So the best angle is $$\arctan(\mu_s)$$! Cool! :)

44. theEric

The arctan function gives you the angle if you give it the "slope" on the unit circle. So a positive slope is in the first or first or third quadrants|dw:1397360433999:dw| And $$\arctan(x)$$ is always between $$-90^\circ$$ and $$90^\circ$$. So we are limited to the first quadrant, right where we want to be. Long story short, it's good. And there's probably a more algebraic way to go about this. But oh well!

45. theEric

It's weird that we didn't have to find whether it was a minimum or maximum, though.

46. marianopolis

I think it does not matter, we could simply plug the angle back in. This would find F and we could compare with another random value, telling us whether we get a minimum or a max.