Does the acceleration cancel out?

[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 ?

 

[Medgirl314]

m2 g - Mm1 g = 0

I think I see it now. If we divide both sides by g, they cancel out. m2 - Mm1 = 0
We already have the equation set to zero. Do I just fill in the numbers and use the quadratic formula?

 

[BvU]

A little bit yes and a hefty no. You really made three chaps happy by typing m2 - Mm1 = 0. You can make them even happier by typing m2 = Mm1 (because then you "have solved for m2"). No need for a quadratic formula ! Or "the"quadratic formula. Shudder.

There is some latest news flash from post #4 that m2 is not 13 kg but 12 kg.

And M = 0.25 as steadfast as ever.

(<Alt>"230" µ is, I am sorry to say, not TeX but a regular character in the font, just a bit hard to remember and not found on keyboards -- except on greek ones, of course.
${\#}{\#}$ \mu ${\#}{\#}$, $\mu$ ìs TeX and easier to remember, together with a lot more of useful brothers and sisters.
Even more sorry to say that µ looks better when embedded in the regular font...)​

Anyway, the physicists are happiest when they see m₂ = µ m₁. Schoolteachers might prefer m₂ = 3 kg because that's what's in their solutions manual.

 

[Medgirl314]

Haha, not everyone had gotten far enough in physics to know that the quadratic formula taught in algebra is not the only quadratic equation. No need to have a panic attack. Oh, I just didn't move the numbers around enough. If you add m2 to both sides you do get m2=$\mu$Mm1 . So now we just put the numbers in? Did I really overcomplicate that much??

Thanks for the LaTex directions, I couldn't find $\mu$ with the Greek symbols and,unfortunately, I don't have time to learn the entire code.

 

[Medgirl314]

Hmm, I think I added an extra letter here. m2=μMm1

Better? m2=μm1

 

[Medgirl314]

Using this equation:m2=μm1
m2=0.25*12
m2=3

Hmm. I can't decide if this sounds reasonable. At first I thought m2 would have to be bigger than m1, but gravity is probably helping a little. Could someone check the equation/arithmetic? It almost just looks to simple.

Thanks again!

 

[Medgirl314]

Bump.

 

[haruspex]

Using this equation:m2=μm1
m2=0.25*12
m2=3

Hmm. I can't decide if this sounds reasonable. At first I thought m2 would have to be bigger than m1, but gravity is probably helping a little. Could someone check the equation/arithmetic? It almost just looks to simple.

That's the right answer. It has nothing to do with gravity "helping". As your equation shows, gravity canceled out, so as long as it is nonzero you will get the same result.
The point is that m₂ is not lifting m₁.

 

[Medgirl314]

Thank you! That makes sense. I was actually comparing it to lifting, not sliding. Oops. I couldn't lift something 4 times my bodyweight, but I could definitely make it slide on a frictionless surface.