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Homework Help: Simple question-but I have no clue!

  1. Aug 28, 2004 #1
    Simple question--but I have no clue!

    Ok, so I am doing an exercise in analyzing errors in measurement, and I have to measure the three dimensions of my lab room--so I did that and then I have to calculate the mass of air in the room... So I got the volume of the room, with Length times width times height--and then I had to do uncertainties in measurement.. so I have the volume, but I need to find the mass--- what formula do I use? I am given that the density of air is 1.21 kg/m cubed... and then I have to calculate the weight of air in both Newtons and points... are there specific formulas for this?
  2. jcsd
  3. Aug 28, 2004 #2
    The formula is:

    [tex]density = \frac{mass}{volume}[/tex]

    Solving for the mass gives:

    [tex]mass = (density)(volume) = (1.21 \frac{kg}{m^3})(volume)[/tex]

    To calculate the weight in Newtons, use the following formula:

    [tex]weight = (mass)(g)[/tex]

    The acceleration due to gravity is denoted by the symbol [itex]g[/itex], and has a value of [itex]9.81 \frac{meters}{second^2}[/itex].

    Therefore, the weight in Newtons is:

    [tex]weight = (mass)(9.81 \frac{m}{s^2})[/tex]

    Your units will be in [itex]\frac{kg \centerdot m}{s^2}[/itex], and note that [itex]1 Newton = 1 \frac{kg \centerdot m}{s^2}[/itex].

    You say you have to calculate the weight in "points" ... Do you mean pounds?

    Note that:

    [tex]1 pound = 4.448 Newtons[/tex]

    So the conversion formula for the weight in pounds would be:

    [tex]weight_{pounds} = (weight_{Newtons}) \frac{1 lb}{4.448 Newtons}[/tex]

    I hope this helps.
    Last edited: Aug 28, 2004
  4. Aug 29, 2004 #3
    Thank you for all your help--and... I still have questions :) Ok, so I pretty much feel like my answers aren't realistic--and my units aren't really working out right--for instance, for the three dimensions of my room, I got 882 cm +/- 2 cm (my measurements of uncertainty), 780 cm +/- 2 cm, and 292.3 cm +/- 2 cm. So I calculated the Volume and I got 201090708 cm cubed +/- 6 cm. Is this correct so far? Should the 6 cm also be cubed, or not?

    Ok, so after that, I started to calculate the mass--using what you told me-- so I did Mass = 1.21 kg/m cubed (201090708 cm cubed +/- 6 cm) and I got 243319756.7 kg/m cubed--and then I have the cm cubed +/- 6 cm from the Volume calculation--and I don't exactly know what to do with that! Can you help me out, again? LOL.

    Also, I was given that 1 Newton = 0.225 lb -- Not what you gave me of .448 I believe...

    Im sorry-- I have never taken a physics class in my life, and Im going into Physical Therapy and this course is a requirement--so I'm absolutely clueless--I need some basic background help first off--and my professor refuses to do anything of that sort... So anyways--can ya help?
    Last edited: Aug 29, 2004
  5. Aug 30, 2004 #4
    First, you need to convert your units so they match and cancel. Multiplying 2x10^8 cm^3 and 1.21kg/m^3 will give you 2.42x10^11 kg cm^3/m^3, not pretty.
    When you convert your volume to meters cubed then multiply by the density of air, the m^3 will cancel and you will get an answer in kg, which is what you are looking for.

    This is the same as "NoPhysicsGenuis" said,
    1lb=4.448 newtons
    1 newton=0.225lb
  6. Aug 30, 2004 #5
    If I'm correct measurement of uncertainty don't work like noranl measurements. You can't just cube the error in x to get the error in x^3.

    You have to follow the formula for any uncertainty in amvalue that is a function of a measured quantity.
  7. Aug 30, 2004 #6


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    What krusty's suggesting is that you write the dimensions of the room in meters instead of centimeters. That way, the volume is in m-cubed (+/- error ). Now multiplying with the given density will give you the mass in kg, as the m-cubeds will cancel off. Now multiply with 9.8 to find the weight in Newtons.

    Also, as mentioned above, the conversion you have from Newtons to pounds is just the inverse of the conversion from pounds to Newtons, so there's nothing wrong with either of them.
  8. Aug 30, 2004 #7


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    To find the errors, you must first find the percentage errors (=100*error/measurement) of the lengths. Thrice that is the percentage error of the volume, which is the same as the error in mass and weight (since we are assuming there is no additional error introduced from multiplying with the density, or the value of g).

    And when your error is +/- 2 cm, there is no point in specifying a dimension like 292.3 cm to that extra degree, as it is smaller than the eroor, and hence meaningless. So, just call that 292 cm (+/- 2 cm) or 2.92 m (+/- 0.02m).
  9. Aug 30, 2004 #8


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    There is a rule of thumb that says:
    When adding measurements, the errors add
    When multiplying measurements, the relative errors add.

    That is: if C= x+ y then dC= dx+ dy: the error in measuring x is dx, the error in measuring y is dy: the error in measureing dC is dx+ dy.
    If C= xy, then dC= xdy+ ydx. Dividing both sides by C= xy, we have dC/C= dy/y+ dx/x. dx/x is the relative error in measuring x, dy/y is the relative error in measuring y and dC/C is the relative error in measuring C.

    That last is not exact- it is an approximation. If you want an exact calculation, you need to note that the largest possible value of x is x+ dx while the smallest value is dx-x. Similarly, the largest possible value of y is y+dy while the smallest value is y-dy.
    The largest possible value of C is, then (x+dx)(y+dy)= xy+ ydx+ xdy+ dxdy and the smallest value is (x-dx)(y-dy)= xy- xdy- ydx+ dxdy. The range (the largest possible error) is the difference of those: 2xdy+ 2ydy. Since xy is (approximately) in the middle of that, we have error on each side of xy of dC= xdy+ ydx so dC/C= dx/x+dy/y.
  10. Aug 30, 2004 #9
    Thank you guys soo much, I think I have the basics now--- for the volume, I changed my answer into meters-- so it cancelled out for the mass...so heres my next question-- I have the volume I have the mass, I have the weight in Newtons, now I need the weight in Pounds... so I took weight in newtons and multiplied it by 1 N/ 0.225 lb and I got 5.36 lbs x 10^6 +/- 1 %.... but my question is.. my professor gave me the formulas : 1 KG = 2.2 lbs and 1 N = 0.225 lb... and I never used the first formula, so did I do something wrong?

    For reference, I got 23845336.16 N or 2.38 N x 10^6 for the weight in Newtons...
    Last edited: Aug 30, 2004
  11. Aug 30, 2004 #10


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    No, both formulas are equivalent, so you can use either. Strictly speaking, the pound is a unit of mass, not force. A 'pound of force' refers to the force (weight) exerted by a one-pound (or 0.455 kg) object, which is hence 0.455*9.8 = 4.448 N.

    In other words, it only takes a factor of g=9.8 to convert from your first number (2.2) to your second number (0.225).

    Now, if (and this is unlikely) the question asks for the force in poundals (this is almost archaic), you need to use g = 32 ft/s^2 when you multiply with the mass in lbs.
  12. Aug 30, 2004 #11
    What you are doing is the following:

    [tex]weight_{pounds} = weight_{newtons} \frac{1 N}{0.225 lb}[/tex]

    Do you see the problem here?

    Your units will be [itex]\frac{N^2}{lb}[/itex], not [itex]lb[/itex].

    The trick is to get the units of newtons to cancel, yielding pounds. To do that, multiply your weight in newtons by a factor such that pounds is in the numerator and newtons is in the denominator. That way, the units of newtons will cancel, leaving you with the weight in pounds.

    Here's the correct formula:

    [tex]weight_{pounds} = weight_{newtons} \frac{0.225 lb}{1 N}[/tex]

    Note that you could also use the original formula that I gave you:

    [tex]weight_{pounds} = weight_{newtons} \frac{1 lb}{4.448 N}[/tex]

    Try both formulas to verify that your answer is the same.
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