Coefficient of Static Friction error (Lab)

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The discussion focuses on calculating the error in the coefficient of static friction using the formula µs = tan(theta). The user initially struggles with the calculations and receives guidance on using the fractional error formula d(tan(theta))/tan(theta) and the importance of expressing angles in radians. After adjusting their angle measurement and understanding that dθ represents the uncertainty in the angle, the user successfully calculates the error. Ultimately, they conclude with a coefficient of static friction of 0.500 ± 0.001, indicating a clearer understanding of the error calculation process.
mlostrac
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Homework Statement


Need to calculate the error of coefficient of static friction in a lab where: µs = tan(theta)s They say to use the general theory of errors?


Homework Equations


delta[tan(theta)] = [1/cos(theta)]^2 (delta)(theta)


The Attempt at a Solution



Do I just plug in my angle value of 26 into the theta's to get the answer? I tried that and I got a value of 0.40, which doesn't make sense because my coefficient of static friction is 0.50?
 
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You are on the right track, but not there yet. Your fractional error is d(tanθ)/tanθ. To get the percent error multiply by 100. Be sure to do all calculations and express all angles in radians.
 
kuruman said:
You are on the right track, but not there yet. Your fractional error is d(tanθ)/tanθ. To get the percent error multiply by 100. Be sure to do all calculations and express all angles in radians.

Ok so I adjusted my angle from 26.5 degrees to 0.46 rad, and then plugged that into each theta for the above equation giving me:
d(tan(theta)) = 1 x d(theta)

kinda confused on what I do after this
 
mlostrac said:
Ok so I adjusted my angle from 26.5 degrees to 0.46 rad, and then plugged that into each theta for the above equation giving me:
d(tan(theta)) = 1 x d(theta)

kinda confused on what I do after this
I am not sure what you have done here. The fractional error in tanθ is given by the ratio

\frac{d(tan(\theta))}{tan\theta}=\frac{d \theta}{tan\theta*cos^2(\theta)}
 
Last edited:
kuruman said:
I am not sure what you have done here. The fractional error in tanθ is given by the ratio

\frac{d(tan(\theta))}{tan\theta}=\frac{d \theta}{tan\theta*cos^2(\theta)}

d tan(theta)/tan(theta) = d (0.462)/ [tan(0.462) x cos (0.462)^2)]
= 0.46/0.00806
= 57.32 rad

I think my answer is way off, what am I doing wrong and why does something as simple as error calculation seem so complicated?
 
mlostrac said:
d tan(theta)/tan(theta) = d (0.462)/ [tan(0.462) x cos (0.462)^2)]
= 0.46/0.00806
= 57.32 rad

I think my answer is way off, what am I doing wrong and why does something as simple as error calculation seem so complicated?
You seem not to understand the meaning of the symbols in the expression you are plugging in. Here θ is 0.46 radians, true enough. However, dθ is not also 0.46 rad; it represents the uncertainty in the angle. To what accuracy did you measure that angle in radians? That's your dθ. Only you, who did the experiment, can make an educated guess about the size of dθ.
 
kuruman said:
You seem not to understand the meaning of the symbols in the expression you are plugging in. Here θ is 0.46 radians, true enough. However, dθ is not also 0.46 rad; it represents the uncertainty in the angle. To what accuracy did you measure that angle in radians? That's your dθ. Only you, who did the experiment, can make an educated guess about the size of dθ.

Oops, haha. Ok I think I got it right this time. I put my accuracy at .001, and then my answer ended up being 0.12.

So that kinda makes sense. A coefficient of static friction = 0.500 +/- .001.

Thanks for your help kuruman!
 

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