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

[intdx]

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. :tongue:

Thank you!

 

[azizlwl]

From data given, maximum frictional force=mg<5N(applied force)
The direction of acceleration should be positive.

 

[Simon Bridge]

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.

 

[PhanthomJay]

Site is pretty bad. I'm surprised it is not first proofread and checked by a teacher before posting.

 

[Nickie]

It is important to note that the coefficient of friction is dependent on the materials in contact and the surface conditions, so it is not a constant value. However, using the given information, we can calculate an approximate value for the coefficient of friction between the book and the floor.

First, let's correct the mistake in the given question. The applied force should be -5N, as the book is sliding in the opposite direction of the applied force. The acceleration should also be positive, as the book is decelerating due to friction. This will give us a more accurate calculation.

Using the equation F=ma, we can find the force of friction acting on the book:
F_f = m*a = 0.4 kg * 1.5 m/s^2 = 0.6 N

Next, we can use the equation F_f = μF_N to find the coefficient of friction. We know that F_N = mg, so we can substitute that into the equation:
0.6 N = μ * 0.4 kg * 9.8 m/s^2
μ = 0.6 N / (0.4 kg * 9.8 m/s^2) = 0.153

So, the coefficient of friction between the book and the floor is approximately 0.153. However, as mentioned before, this value may vary depending on the materials and surface conditions. It is always important to double check your calculations and make sure all the given information is correct.