# Finding Kinetic Coefficient of Friction

1. Mar 29, 2012

### dzimme2

1. The problem statement, all variables and given/known data

A 2 kg. block is pulled up a 22° incline with a force of 10N parallel to the incline. What is μ$_{}k$ if it moves up with constant speed?

2. Relevant equations

F$_{}N$ = Normal Force
f$_{}k$ = μ$_{}k$ F$_{}N$

3. The attempt at a solution

Forces perpendicular to plane: y-vector component of gravity, mgcos$\theta$ (negative) , normal force, F$_{}N$ (positive)

Forces parallel to plane: applied force, F (up plane) , kinetic frictional force, f$_{}k$ (up plane), and x-vector component of gravity, mgsin$\theta$ (down plane)

F$_{}net$$_{}y$ = F$_{}N$ - mgcos$\theta$ = 0
F$_{}N$ = mgcos$\theta$

F$_{}net$$_{}x$ = F - mgsin$\theta$ + f$_{}k$ = 0
f$_{}k$ = mgsin$\theta$ - F

μ$_{}k$mgcos$\theta$ = mgsin$\theta$ - F
μ$_{}k$ = [ mgsin$\theta$ - F ] / [ mgcos$\theta$ ]
μ$_{}k$ = ........

The given answer is μ$_{}k$ = 0.55 but I can not come up with this. Please help, thanks!

Last edited: Mar 29, 2012
2. Mar 29, 2012

### tiny-tim

welcome to pf!

hi dzimme2! welcome to pf!
(erm … the whole point of latex brackets is that you put things inside them! )

no, the friction is downhill, like the gravity, isn't it?

3. Mar 29, 2012

### dzimme2

Re: welcome to pf!

I was under the impression that the kinetic frictional force would oppose the block's ability to slide down the plane due to gravity. While the applied force is pushing the box up the plane and gravity is pulling the block down the plane, wouldn't the frictional force be directed up the slope in opposition of the gravitational component?

4. Mar 29, 2012

### tiny-tim

hi dzimme2!
ah, nope

the direction of kinetic friction is always opposite the actual motion

here, the actual motion is up, so the friction is down

(the direction of friction in the static case is more complicated, but in the kinetic case it's very simple!)

5. Mar 29, 2012

### dzimme2

Ohh, I see. Could prove to be the error with other problems as well. I'll try it out and see what I get. Thanks again!

6. Mar 29, 2012

### dzimme2

After changing the sign of f$_{k}$ to restate : F$_{net}$$_{x}$ = F - mgsin$\theta$ - f$_{k}$ = 0;

f$_{k}$ = F - mgsin$\theta$
μ$_{k}$ F$_{N}$ = F - mgsin$\theta$
μ$_{k}$ mgcos$\theta$ = F - mgsin$\theta$
μ$_{k}$ = $\frac{F - mgsin\theta}{mgcos\theta}$
μ$_{k}$ = $\frac{(10 N) - (2kg)(9.8 m/s^2)sin 22}{ (2kg)(9.8m/s^2)cos 22}$

μ$_{k}$ = 0.1257

However, the answer key has μ$_{k}$ = 0.55 .

???

7. Mar 29, 2012

### tiny-tim

i can't get 0.55

8. Mar 29, 2012

### dzimme2

Do you get 0.1257?

9. Mar 29, 2012

### tiny-tim

i make it 0.146