Calculating coefficient of friction given m, applied force, and a

AI Thread Summary
The discussion revolves around calculating the coefficient of friction for a book with a mass of 400 g, an applied force of 5 N, and an acceleration of -1.5 m/s². Participants express confusion over the signs used for the applied force and acceleration, suggesting that the acceleration should align with the direction of the applied force. The calculations yield varying coefficients of friction, with results around 1.122 and 1.429, which differ significantly from the site's stated answer of 0.15. Concerns are raised about the reliability of the source material, as it appears to contain errors and lacks proper oversight. The consensus is that the site's explanations are misleading and require correction.
intdx
Messages
5
Reaction score
0

Homework Statement


From http://library.thinkquest.org/10796/index.html (#6)

A book has a mass of 400 g. When you slided the book against the floor with 5 N, it accelerated at the rate of -1.5 m/s2. What would the coefficient of friction between the book and the floor be?

g=9.80m/s^2

Homework Equations


F=ma
F_f=\mu F_N
F_N=mg (The site actually states the normal force to be equal to negative mass times gravitational acceleration, but with a negative value for gravitational acceleration. I'm going with Wikipedia, though.)
n\textrm{g}=\frac{n}{1000}\textrm{kg}

The Attempt at a Solution


First off, I'd like to say that this site was made by high school seniors, so I'm put in the uncomfortable position of not being able to readily accept everything that's there.

Next, why is the applied force positive but the acceleration negative? I'll just assume that that was a mistake and that the applied force should actually be -5N.

-5\textrm{N}+F_f=F\implies F_f=F+5\textrm{N}
(Right? It seems right to me...)
F=0.4\textrm{kg} \times -1.5\textrm{m/s}^2
F_f=\mu \times 0.4\textrm{kg} \times 9.80 \textrm{m/s}^2
\mu=\frac{F_f}{0.4\textrm{kg} \times 9.80 \textrm{m/s}^2}=\frac{0.4\textrm{kg}\times -1.5\textrm{m/s}^2+5\textrm{N}}{0.4\textrm{kg} \times 9.80 \textrm{m/s}^2}
\mu=1.122
I get the same answer when I keep the applied force positive, make the acceleration positive, and use 5\textrm{N}-F_f=F\implies F_f=5\textrm{N}-F.

Yet, the site's answer is 0.15.

I even tried using a positive applied force with a negative acceleration (pretending that friction could make an object go in the opposite direction of the applied force).

5\textrm{N}-F_f=F\implies F_f=5\textrm{N}-F
F=0.4\textrm{kg} \times -1.5\textrm{m/s}^2
F_f=\mu \times 0.4\textrm{kg} \times 9.80 \textrm{m/s}^2
\mu=\frac{F_f}{0.4\textrm{kg} \times 9.80 \textrm{m/s}^2}=\frac{5\textrm{N}-0.4\textrm{kg}\times -1.5\textrm{m/s}^2}{0.4\textrm{kg} \times 9.80 \textrm{m/s}^2}
\mu=1.429

Then I realized that, using 400 instead of 0.4, you get -0.152, -0.152, and 0.154, respectively, for the three attempts described above.

Somebody, please, what is going on here?? I'd really appreciate some help, and it'd be great if you simply told me that the site was really wrong. :-p

Thank you!
 
Physics news on Phys.org
From data given, maximum frictional force=mg<5N(applied force)
The direction of acceleration should be positive.
 
As aziziwi states, the acceleration given in the problem should be in the same direction as the applied force ... so it should have the same sign.

The site says that the normal force is -mg is correct if g=-9.8m.s-2 ... i.e. the normal force is equal and opposite to gravity, which you'd normally give as mg. Take care interpreting equations ... a minus sign in front of a thingy does not have to imply that you take the negative of that thingy.

But the site looks somewhat confused to me. It says at the top that this is by students for students, and I see no way to edit it or submit corrections, so I'm not really all that surprised.
 
Site is pretty bad. I'm surprised it is not first proofread and checked by a teacher before posting.
 
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