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Simple math problem

  1. Dec 13, 2005 #1
    Sorry, I'm very rusty with this... :confused:

    I have two equations:

    T = za

    and

    T - mg = -ma

    Somehow, they get "combined" and to solve for "a" it is:

    a = (m / (z + m) ) * g

    can someone please explain the process of how it got to this?

    I figured out that we substitue for T in the second one to get:

    za - mg = -ma

    but I get really lost trying to solve for "a".

    Please help!


    b.t.w. sorry, I'm not a math or physics student, I'm trying to learn this stuff myself. :confused:
     
  2. jcsd
  3. Dec 13, 2005 #2

    Hurkyl

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    Why are you lost? Do you know? I suspect it is because there are two a's in the equation and you don't know what to do about it. (Would you agree? Diagnosing why you're lost is often a key step in trying to get back on track)

    So, given the diagnosis, the question you should immediately ask yourself is: "Can I combine these two terms with a in them into a single term?"
     
  4. Dec 13, 2005 #3
    Ok, I think I'm figuring something out here...

    za - mg = -ma

    can be changed to:

    mg = -ma - za

    Then we can get "a" into a single term with:

    mg/a = -m - z

    ok, now I'm stuck here.. Is my math right so far though?
     
  5. Dec 13, 2005 #4

    Hurkyl

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    You have made mistakes. Until you get better, you should never skip any steps. Clearly you're subtracting za from both sides of your equation, so you should write that down first without simplifying, and then simplify one step at a time.


    Your approach to the problem is good, though. So let's now look at:

    mg/a = -m - z

    (Or, instead, we can look at the right result when you get there; they're similar in appearance)

    I offered hints at a general approach to problem solving in my last post, and they still apply: can you figure out exactly what aspect of this equation is giving you trouble?
     
  6. Dec 13, 2005 #5
    so
    mg/a = -m - z
    is incorrect?

    I was dividing both sides by "a" to get this. Having the two terms with the variable I need to solve for is definitely what's causing my problems here, possibly the negatives are throwing me off too. Show I maybe convert the "-m - z" to "-(m + z)" ?
     
  7. Dec 13, 2005 #6

    Hurkyl

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    Yes, but only because

    mg = -ma - za

    is wrong. Your division by a was done correctly. (Though, you did skip some steps! But again, they are common steps to skip)


    That is certainly a reasonable thing to do. It may or may not help.


    In any case, once you get to

    mg/a = -m - z

    (or the correct version), a is no longer appearing twice, so that cannot be the reason you can't figure out the next step. So you need to figure out what it is about this new equation that is stumping you!
     
  8. Dec 13, 2005 #7
    Well if the negatives are throwing you off, it may be easier to look at the equation as being
    za + ma = mg instead (it's the same thing you had, but now the -ves are gone)
    Now you're looking to isolate "a" so you can factor that out on the left hand side to get
    a(z+m) = mg
    Now see what you can do...
     
  9. Dec 13, 2005 #8
    oh ok, I see why
    mg = -ma - za
    is wrong, I forgot the negative on the left side, so it should be:
    -mg = -ma - za

    then, I should have:

    -mg/a = -(m + z)

    so, now another question, since both sides are negative, can I just cancel that out and make it:

    mg/a = m + z

    I think that my real problem is dealing with fractions, would "(mg)/a" be the same as "m * g/a"? I guess it would because it would be "m/1 * g/a", which is equal to mg/a. Alright, assuming my reasoning is right here, I should be able to get:

    m * g/a = m + z

    I don't know exactly what would be best to do here, maybe to divide out the m, but I'm not sure if it's right:

    g/a = z/m

    It doesn't really look right to me, but I'm still confused by the fractions I guess.
     
  10. Dec 13, 2005 #9

    Hurkyl

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    Band-aids are no good! He's making a mistake in manipulating negatives; showing him a way to avoid negatives in this problem does not help him fix his problem!
     
  11. Dec 13, 2005 #10

    Hurkyl

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    Yep! (I'm hoping that, in the future, if you wrote all the steps, you would generally catch this mistake, because then you would write za - mg - za, simplify to 0 - mg, and then see that you should have -mg left over!)


    Yep!


    Well, dividng by m certainly simplifies the term involving a, so it is probably a good thing to do.

    Though you made a mistake again! Again, I'll advise writing out all the steps. :tongue2: I think your mistake here is in the same spirit as your previous one: you realized that you're "cancelling" something, so you blissfully erased what's being cancelled, rather than doing it the "right" way.


    And my previous hint still applies: if you can figure out exactly what about the problem is making it difficult for you to know what to do next, then you might be able to figure out the right next thing to do! (Or how to avoid the thing that's confusing you, which is the direction that giggle's hint was taking you)
     
  12. Dec 13, 2005 #11
    Ok, I got it with giggles hint. I have a problem that when I want to simplify something in an equation, I want to move it to the other side, I need to remember to do things like changing "am + az" to "a(m + z)".

    From there, it's a lot easier for me...

    mg = a(m + z)

    divide by (m + z) and you get

    a = mg/(m + z)

    I'm not really sure why they write it like:

    a = m/(m + z) * g

    in the book though...

    I really appreciate all of your help. I'm not sure what you mean by doing all the steps though, and I'm not sure what I did wrong in the dividing out m in my previous post. I was removing the m from the LHS term and then on the RHS, I divided it out from both terms. Well, I guess I do see a mistake, maybe it should be:

    g/a = m/m + z/m
    =
    g/a = 1 + z/m

    I'd kinda like to figure it out this way too, just so I get over my problem with figuring out these types of eqations.
     
  13. Dec 13, 2005 #12

    Hurkyl

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    Yep, this is done correctly.

    I hate giving away the thing I was trying to hint towards, but I think you've been lost ever since

    mg/a = m + z

    because a is on the bottom of a fraction, so you need to find some way to fix that. (and as you saw, you could avoid having a on the bottom by doing the problem differently. Sometimes avoiding problems is more fruitful than trying to fix them once they arise! But this one is workable)
     
  14. Dec 13, 2005 #13

    JasonRox

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    a = m/(m + z) * g and a = mg/(m + z)

    ...are the same thing. You prove this simply by solving this...

    m/(m + z) * g = a = mg/(m + z)

    See how both equal a, so now you know for sure that...

    m/(m + z) * g = mg/(m + z)

    Now, solve that and see what you get. I'm letting you solving this so you can have some practice.

    Note: I'd also recommend practicing other algebra problems too. It certainly wouldn't hurt, and sometimes they are quite fun, especially when there is a little trick in it.
     
  15. Dec 13, 2005 #14
    T=za , carry this to second equation: za - mg = -ma ;

    carry 'mg' to rightside , and 'ma' to leftside of the equation: za+ma=mg

    a(z+m) = mg ----------> a = (m/(z+m))*g
     
  16. Dec 13, 2005 #15

    JasonRox

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    Read the thread next time because giving the answer is not being helpful.
     
  17. Dec 22, 2005 #16
    I hope it's not too late, but it looks like you were confused with a in the denominator, as in
    [tex]\frac{mg}{a} = m+z[/tex] this is correct, but hasn't solved for a
    Since we can do anything we want to an equation as long as it's done to both sides, let's take the inverse of both sides :
    [tex]\frac{1}{\frac{mg}{a}} = \frac{1}{m+z}[/tex] this is ugly, but if we apply this rule for fractions with fractions:
    [tex]\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ac}{bd}[/tex] it gets better, because then, noting that b=1 and not mg:
    [tex]\frac{a}{mg} = \frac{1}{m+z}[/tex] let's multiply by mg to leave a on the left
    [tex]a = \frac{1*mg}{m+z}[/tex]
     
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