Does the acceleration cancel out?

[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 !

 

[Medgirl314]

Haruspex, could I have your input here? I think I'm just not understanding the link between you asking what "N" equals and why BvU told me not to.

BvU, sorry for the typo, I just noticed it. I actually did mean m2 g/M=N.

 

[BvU]

N is the normal force the table exercises on block m1. That force is equal and opposite to the force the block exercises on the table, which is m1 g.

 

[BvU]

Basically we want the tension in the wire to be no bigger than that it compensates the maximum friction force.

Perhaps this exercise becomes a bit clearer if we think of block m1 and a wire only. Forget the pulley and m2 (feels great, right?)

How hard can you pull on the wire without the block m1 starts to move ?

By the definition of $\mu$ which we call M for some reason: $\ F_{max} = \mu N = \mu \ m_1\ g$

 

[Medgirl314]

Okay, BvU, for some reason I'm still confused here. Are we saying the N=m1g, so by finding N, I know the equation for m1g, so if I divide by g I can easily solve for m1?

Thanks!

 

[BvU]

You have a funny way of putting things. You don't have to "solve" for m1. There really, really is no need. m1 is a given. Its value is 13 kg.

 

[Medgirl314]

Sorry, I'm just still confused on why haruspex said to solve for N and then you said I didn't need to. I forgot m1 is given.

 

[BvU]

So now you have N. Don't calculate it, just keep the m1 g . Fill it in in the m2 g - MN = 0 equation. Solve for m2.

Solve for m2 means: write it so that it has the form m2 = blabla.

For example, if you get to solve $5x - 3y = 0$ for x, with $y = \zeta p$, you add $3y$ left and right and you divide left and right by 5, to get: $x = {3\over 5} \zeta p$. Done.

 

[Medgirl314]

Okay, so I'm this far with solving for m2: m2 g - MN = 0

But don't I need to isolate m2 completely? Shouldn't I do something to get the g out of the way?

 

[BvU]

Sorry, I'm just still confused on why haruspex said to solve for N and then you said I didn't need to. I forgot m1 is given.

Can't even find it any more. All I find is haruspex, lightgrav and me begging you to fill in N = m1 g

 

[BvU]

Ah, we type so fast things start to cross. How difficult is it to type m2 g - M m1 g = 0 and solve for m2 ?