Finding Kinetic Coefficient of Friction

[dzimme2]

Homework Statement

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?

Homework Equations

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

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!

 

[tiny-tim]

welcome to pf!

hi dzimme2! welcome to pf!

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

(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?

 

[dzimme2]

hi dzimme2! welcome to pf!

Why thank you, I've frequented the site lately as a guest but finally decided to join. My thanks to you all!

(erm … the whole point of latex brackets is that you put things inside them! )

I'm not exactly sure as to what you are referring, but if it's regarding the 'Fnet' term, I was attempting to subscript "net" with the subscript "x".

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

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?

 

[tiny-tim]

hi dzimme2!

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?

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!)

 

[dzimme2]

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!)

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!

 

[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 .

?

 

[tiny-tim]

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

?

i can't get 0.55

 

[dzimme2]

i can't get 0.55

Do you get 0.1257?

 

[tiny-tim]

Do you get 0.1257?

i make it 0.146