Question

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.

Answer 1

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?

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

Dang!

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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.

Answer 10

Okay, I am half-way through.

Answer 11

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?

Answer 12

do we mean*

Answer 13

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).

Answer 14

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.

Answer 15

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

Answer 16

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.
\(\sqrt{(F\sin\theta)^2+(mg\mu_s-F\sin\theta\ \mu_s)^2~~}\)
\(=\sqrt{F^2\sin^2\theta+m^2g^2\mu_s^2-mg\mu_s^2F\sin\theta+F^2\sin^2\theta\ \mu_s^2~~}\)

Answer 17

OHHH

Answer 18

ok, I reread the whooole post.

Answer 19

I think I am starting to understand.

Answer 20

(I reread thrice.)

Answer 21

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

Answer 22

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?

Answer 23

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

Answer 24

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

Answer 25

That sounds good!

Optimal angle... Time for the thinking cap.

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

I've had lots of time to practice :)

Answer 30

ohhh ok. ;) that's why

Answer 31

Haha, yep!

Answer 32

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

Answer 33

Oh of course! It finds the minimal value!

Answer 34

All those cal courses now come back to me!

Answer 35

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
\(\dfrac{{\rm change\ in\ }f(x)}{{\rm change\ in\ }x}\)

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 \(0\).