Does the acceleration cancel out?

[Medgirl314]

Homework Statement

Two masses and a pulley are arranged as shown. m₁ has a mass of 13 kg. The coefficent of static friction between ,₁ and the table in 0.25. How big will m₂ need to be to get m₁ to begin to slide?

Homework Equations

F_net=ma

The Attempt at a Solution

I am having an extraordinarily difficult time trying to derive an equation for this.

F_net=ma
m₂g-f=m₁+m₂a
I tried rearranging to get m₂=kg+f, but for one this doesn't sound right, and for two it seems impossible to solve because there are two variables. Does the acceleration somehow cancel out in this problem, or is there something else I'm missing?

Thanks so much!

 

[voko]

The problem really wants you to find the minimal mass for $m_2$ that causes $m_1$ to slide. The minimal mass causes minimal acceleration - but what is this minimal acceleration?

 

[lightgrav]

if you write f in terms of m_1 , you'll find something to put on the right-hand side.

 

[Medgirl314]

Wow, so sorry for the errors. The mass of m1 is actually 12 kg, and there's at least one grammar error.

voko, that's one of the equations I'm stuck on. I tried deriving an equation for acceleration, but it didn't come together, either. lightgrav, thanks! I've tried a bunch of different ways and I'm still stuck. Could we walk through the steps?

The total mass is m_{}1+m_{}2, correct?

f=m_{}1+m_{}2 * a

Is that correct so far?

Thanks again!

 

[BvU]

Yes. On two counts. Don't forget the brackets...
However, you haven't answered voko and you haven 't written out f in terms of m₁, so lightgrav is also deeply unhappy...
And I think you can easily make them happier!

 

[Medgirl314]

Okay, thanks! Do you have tips on how to do brackets on LaTex? I tried before but it messed up my entire equation.

 

[lightgrav]

what two Forces (|| the rope) cause (m1 + m2)a ?
... do you know a formula for friction?

 

[Medgirl314]

Yes. f=M_{}N

 

[Medgirl314]

(With "M" being mu and "N" being Newtons)

 

[lightgrav]

friction Force F ≤ μ N
N is the "Normal" component of the Force from the table surface
(pushing straight thru the contact Area into the block).
Why does the table need to push up?

Is there tension in the rope? that requires contact Force at _both_ ends!

 

[BvU]

$\TeX$ help: ( or [ or | brackets can simply be typed :
[ itex ] |[(a^2_\epsilon)]|^4 [ /itex ] gives $|[(a^2_\epsilon)]|^4$
If you want nested stuff to be extra clear you can use \bigl ( and \bigr ), \Bigl (, \biggl, \Biggl :
[ itex ] \Biggl ( \biggl( \Bigl( \bigl( () \bigr ) \Bigr) \biggr) \Biggr)^4 [ /itex ] gives $\Biggl ( \biggl( \Bigl( \bigl( () \bigr ) \Bigr) \biggr) \Biggr)^4$

{ has a special meaning as grouping symbol in TeX, but then you can use \{ or \lbrace, again with the extra sizing stuff if so desired.

 

[Medgirl314]

friction Force F ≤ μ N
N is the "Normal" component of the Force from the table surface
(pushing straight thru the contact Area into the block).
Why does the table need to push up?

Is there tension in the rope? that requires contact Force at _both_ ends!

Ah, right. I was mixing up two formulas, thanks. The table doesn't need to push up, it's a stationary object. But it's exerting forces on the masses, since the masses are exerting forces on it.

There's tension in the pulley. Where are we going with this? Do we need to calculate the tension?

Thanks!

 

[Medgirl314]

$\TeX$ help: ( or [ or | brackets can simply be typed :
[ itex ] |[(a^2_\epsilon)]|^4 [ /itex ] gives $|[(a^2_\epsilon)]|^4$
If you want nested stuff to be extra clear you can use \bigl ( and \bigr ), \Bigl (, \biggl, \Biggl :
[ itex ] \Biggl ( \biggl( \Bigl( \bigl( () \bigr ) \Bigr) \biggr) \Biggr)^4 [ /itex ] gives $\Biggl ( \biggl( \Bigl( \bigl( () \bigr ) \Bigr) \biggr) \Biggr)^4$

{ has a special meaning as grouping symbol in TeX, but then you can use \{ or \lbrace, again with the extra sizing stuff if so desired.

Thank you!

 

[lightgrav]

(yes, sliding friction does have an equal sign)

If the table did not push up, the block would sink through it.
in fact, as the block does sink into it, the contact Pressure increases until the sinking stops.

ignore the pulley for now ... it does push on the rope, but its push is perp to the rope.
and we know that the acceleration will be parallel the rope.

always look at components that are || and _|_ the acceleration (if feasible).
There IS a Force applied to the right end of the rope.
We want that one: it contains our desired unknown!

 

[Medgirl314]

Yes. On two counts. Don't forget the brackets...
However, you haven't answered voko and you haven 't written out f in terms of m₁, so lightgrav is also deeply unhappy...
And I think you can easily make them happier!

Thanks for pointing that out, but I responded to voko and lightgrav and I are trying to walk through this step by step. We're good. Thanks, though!

 

[Medgirl314]

(yes, sliding friction does have an equal sign)

If the table did not push up, the block would sink through it.
in fact, as the block does sink into it, the contact Pressure increases until the sinking stops.

ignore the pulley for now ... it does push on the rope, but its push is perp to the rope.
and we know that the acceleration will be parallel the rope.

always look at components that are || and _|_ the acceleration (if feasible).
There IS a Force applied to the right end of the rope.
We want that one: it contains our desired unknown!

I'm not sure what you were responding to when you said sliding friction has an equal sign.

Ah, I interpreted your question too literally. Yes, the table definitely exerts a force.

I think you're calling the object in question a rope and I'm calling it a pulley, correct? Or are we talking about different things?

I'm having a hard time translating your words into a variable we can use. So far we have f=m1+m2 * a
Could you please explain what we're trying to do? Are we finding the friction and putting that on the left side as the force? I think I have an good understanding of the concepts to some extent, but I can't figure out how to translate it into numbers.

Thanks again!

 

[lightgrav]

a pulley is a round wheel fixed in place, that can spin freely around an axle.
rope, cord, cable, string are all nearly the same ... string sounds a lot like sPring, so I try to avoid it
("ideal" textbook string has constant length until it breaks ; ideal spring stretches with any tension)

your post 1: m₂ g - f = (m₁ + m₂) a

all you need to do is plug the formula for friction (my post 3),
and decide how small the acceleration can be for m₁ to start sliding (voko's post 2)

 

[Medgirl314]

I don't see a rope in the diagram, maybe that's where the confusion lies.

Okay, I can't believe I'm overcomplicating physics, for crying out loud.

m2 g - MN = (m1 + m2) a

Like that? (With "M" being mu and "N" being the normal force.)

Thank you!

 

[lightgrav]

(what is that thing that keeps the dangling m₂ from falling quickly to the floor?)

What is the Normal Force, in this situation (why isn't it zero?)
... how small can "a" be , to produce some Δx?

 

[Medgirl314]

(what is that thing that keeps the dangling m₂ from falling quickly to the floor?)

What is the Normal Force, in this situation (why isn't it zero?)
... how small can "a" be , to produce some Δx?

Tension?

I'm not sure, can we break this down a little further?

Finding "a" is one of the main points I was confused on for this situation. I haven't dealt with a situation like this one, and I'm not sure if I have enough information to figure it out.

 

[BvU]

I love this. The rope (wire, thread, ) is the thin linefrom m1 over the pulley to m2. Nobody realizes that somebody might not see a rope. Physicists are so preconditioned!

 

[Medgirl314]

That's what I was calling the pulley. I *saw* it, but I thought we were considering the pulleys and the rope as one system.

Yes, they can be.

 

[lightgrav]

would a = 1 m/s² cause the block to slide?
how about ½ m/s² ?
maybe ¼ m/s² would be too small ... ?

 

[Medgirl314]

Oh! So we're not using a specific formula right now, we're simply estimating?

 

[lightgrav]

(Tension is a Force, not a thing that can transmit Force)

2) do you know the table surface Force' Normal component?

acceleration answers:
yes
yes
no, still not too small.

how small can "a" get, but still result in motion?
there is a limit, just like there's a limit to the ≤ sign in the friction formula.

 

[BvU]

You know these movies where they balance a bus on the railing of a bridge after a crash and then somebody tosses a Mars bar from the back to the front and the whole thing falls into the canyon? What lightg is trying to lure you into is to imagine that once you have the equilibrium case, a fly landing on m2 is enough to set the whole lot in motion. Very unrealistic, but physics! And, no, there is no estimating involved! Nonsense of course, but exact nonsense nevertheless ;-)

 

[Medgirl314]

Thank you both for your explanations! I think I understand the concept pretty well, but I still don't really understand the question. How do I get an equation for "a" right now? That was one of my major problems from the beginning, I really need to walk through that part to understand what we're doing. Would someone please help me to find out how to get a, instead of why we're getting a? Thanks so much!

 

[lightgrav]

any acceleration means that it slides ... what they want you to do, is use the limit of smaller and smaller accelerations , till you get to a=0. use it.

 

[Medgirl314]

Oh! So are we just thinking of a reasonable acceleration, testing it, and repeating until we get to a=0?

 

[lightgrav]

technically, you should have a>0 ; but you don't need to calculate anything with "a" bigger than 0, because for any non-zero a, m₂ is more massive than it needs to be. so change the inequality to the limit, which is the equality. put it so the block is just about to slide, and bump the table; the block will slide.

Once the block starts to move, the friction coefficient becomes kinetic, and μ_kinetic < μ_static ... so a becomes > 0 then.

 

[Medgirl314]

Okay. I'm sorry, it still feels like we're going a few steps ahead. Are we? I'm still trying to figure out how to start this. Would I, for example, say 3 m/s and plug it into F=ma?

Thanks again!

 

[lightgrav]

3 m/s² is much too quick . 1/3 m/s² still has m2 with too much mass ;
1/100 m/s² is probably okay, being the size of acceptable roundoff for "g".
1/1000 m/s² is definitely within tolerence even if they require 3 sig.fig's.

Do you see that we are trying to make a small enough to ignore?

 

[Medgirl314]

Somewhat. I've never done guesswork lie this with physics(even exact guesswork), so I'm having a hard time figuring out the process we're using. If you think it would help, would you mind outlining the steps that we'll use for the problem?

Thanks so much!

 

[lightgrav]

What do you do with friction's Force, in a problem that says "friction is negligible"?
I'll bet you have no trouble setting F_friction = 0 ...

Here, you first recognize that you want "a" small enough to ignore -
then you ignore it, by replacing "a" with 0 .

 

[Medgirl314]

Okay, so we don't actually need an equation for "a", we just need to set it to zero?

Thanks!

 

[Medgirl314]

Is anyone still around? How does this look so far?

m2 g - MN = (m1 + m2)0

Would that cancel out m1+m2, leaving m2g-MN?

 

[haruspex]

Is anyone still around? How does this look so far?

m2 g - MN = (m1 + m2)0

Would that cancel out m1+m2, leaving m2g-MN?

Yes, except that's not an equation. What equation do you think it leaves? And what can you substitute for N?

 

[lightgrav]

good so far; it IS an equation, since it has an "=" sign, but you have not yet isolated the desired unknown.
Now write the formula for the Normal Force (from the table) and cancel the common factor.
{fyi: mu can be typed as <Alt>"230" µ}

 

[Medgirl314]

Lightgrav, haruspex was saying that after m1+m2 canceled out, it wouldn't be an equation, because there isn't an "=" sign in m2g-MN. Did I skip a step? Thanks for the tip on LaTex, that was starting to annoy me. XD

 

[lightgrav]

"0" is my favorite number ... I can add "0" to anything, without using a calculator!
N = ?

 

[haruspex]

Lightgrav, haruspex was saying that after m1+m2 canceled out, it wouldn't be an equation, because there isn't an "=" sign in m2g-MN.

Nearly right. I'm saying you didn't do the cancellation correctly because what you wrote afterwards was not an equation.
You had m2 g - MN = (m1 + m2)0. Just simplify the right-hand-side. The equals sign does not go away.

 

[Medgirl314]

m2 g - MN = 0

Is that all for deriving the equation?


Sorry, still can't figure out the LaTex! I'll play around with it and find a tutorial later.

 

[haruspex]

m2 g - MN = 0

Is that all for deriving the equation?

Yes. And N equals?

 

[Medgirl314]

Do we need to move this equation around the find N, and then plug everything we know into the equation above?

 

[haruspex]

Do we need to move this equation around the find N, and then plug everything we know into the equation above?

N is the normal force from a horizontal table on a block placed on it, right? In terms of the mass of the block, what does that equal?

 

[lightgrav]

why does the table push the block upward?
... or, why does the block not accelerate upward in response to being pushed?

 

[Medgirl314]

Sorry this took me so long! I got immensely busy with standardized tests(oh, the joy) and haven't had enough time to work on this problem.

To find N:
m2 g - MN = 0
m2 g-MN=0
+MN +MN
m2 g=MN
m2 g/M=MN/M
m2 g/N=M

Does that look okay so far?

Thanks!

 

[BvU]

Looks OK, beautiful, fine, etcetera. But it does not help you to find N.
In two ways:
1) If it was to help you find N you would write N = m2 g / M , not m2 g / N = M
2) Whichever way you write it, there still is this little problem that you don't know m2. No wonder, because that was what the original exercise was asking for.

Quoting light grav: m1 is not going up and it is not going down. It is not accelerating in the vertical direction, even though gravity is pulling down with a force m1 g. The acceleration is zero.
The acceleration can only be zero if the table exercises a normal force N that is equal and opposite to m1 g.

In other words:

m1 g + N = 0

The usual choice of coordinates is y+ = up, which means that m1 g is pointing in the negative y direction. Something like m1 kg x -9.81 m/s²:

m1 kg x -9.81 m/s² + N = 0 $\Leftrightarrow$ N = m1 * 9.81 kgm/s².

So by now, we can assume N is really a known thing. Right ?

Back to your thingy and now rewrite it so that it helps you find m2 !

 

[Medgirl314]

N is the normal force from a horizontal table on a block placed on it, right? In terms of the mass of the block, what does that equal?

BVU, this is what I was replying to with that derivation. Did I misunderstand the question?

Thanks!

 

[BvU]

Apparently, because m2 is not lying on the table. So it would really be very strange if m2 appeared in an expression for the Normal force a table exercises on m1 !