Find the percentage uncertainty of a wooden block

  • Thread starter Thread starter agentlee
  • Start date Start date
  • Tags Tags
    Block Uncertainty
AI Thread Summary
To find the percentage uncertainty of a wooden block's volume and density, measurements for length, width, and height must be taken using three different instruments, with inherent uncertainties of ±0.005 cm due to the precision of the measurements. The discussion highlights the need to calculate the average volume and apply standard techniques to estimate errors, suggesting that fractional errors from the measurements can be combined to assess total uncertainty. It emphasizes that the net fractional error should be treated as the sum of independent normally distributed random variables, acknowledging potential cancellation effects. The importance of having a standard volume for comparison is also noted, as it aids in determining the uncertainty accurately. Understanding these principles is crucial for effectively calculating the percentage uncertainty in this context.
agentlee
Messages
18
Reaction score
0

Homework Statement


I have to find the percentage uncertainty of a wooden block for its volume and density. Using 3 different types of measurements for length, width, and height.

Let's say that this is my data (cm):
Instrument 1: L= 1.1 W= 1.1 H= 1.3 V=1.573
Instrument 2: L= 1.13 W= 1.13 H= 1.31 V=1.67
Instrument 3: L= 1.12 W= 1.12 H= 1.29 V=1.62

Average Mass could be 2.56g

Homework Equations



I was given ΔV/V and ΔP/P for the uncertaintiy

The Attempt at a Solution


I am not sure how to do this, without an acceptable given volume for the block, I am a bit lost.
 
Physics news on Phys.org
It looks like each measurement is to the nearest .01, so there's an inherent uncertainty of +/- .005 in those. On top of that, the measurements are not entirely consistent, suggesting a further source of error. You can apply standard techniques to estimate, say, one standard deviation on those. Your problem then is to combine into a volume error. If you have two numbers with known fractional errors, e.g. x1(1+δ1), x2(1+δ2), then for small errors the product is approximately x1x2(1+δ12). I.e. the fractional errors just add up. But since some may be plus and some minus, cancellation can occur. So treat the net fractional error as the sum of independent normally distributed random variables with zero mean. It will help if you have some basis for supposing the three dimensions have the same variance.
 
ΔV = V_measured - V_standard, V = V_standard
etcetera for density

You should have been given a standard, have to look one up, or incorrectly take an average of your results as a standard.

I've never used this method for errors, except in several physics labs, and it always seemed kind of hokey to me...
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top