YAPhysicsT: Solving a kinetic friction problem (solved)

[Ricemarine]

Problem:
- A person wants to know the static friction and kinetic friction of a box on a plank
- He lifts one side of the plank until @ 30 degrees, it begins to slip and slides down the 2.5m plank in 4 seconds
- What is the kinetic (Mk) and static friction (Ms)??

So I've been trying to find the kinetic friction...

if Fk = Mk * Fn, then
ma = Mk * mgsin(theta), then
a = Mk * gsin(theta)

Plugging it into the equation x = xo + vot + 1/2at^2, it becomes
2.5 = 0.5(Mk * 9.81 * sin 30 ) * 4^2, which becomes
0.3125 = Mk * 9.81 * sin 30, thus
Mk = 0.0637

The question is... is that the right answer? If you look at the Ms, it is much much higher, in whcih I'm not sure if Mk should be this low.

theta = 30 degrees.

When solving for static friction, I get
X-coords = Ff + F = 0, in which F = mgsin(theta)
Y-coords = Fn - mgcos(theta) [weight]
So therefore
Fn = mgcos(theta)
So plugging it back in,
Ff = MsFn ---> Ms = mgsin(theta)/mgcos(theta) --> tan(theta) = 0.58 = Ms

My question for this one is, if 30 degrees was the point where the box began to slip, does that mean that the static point was < 30 degrees, and therefore I have to calculate at a lower theta?

Thanks

Edit: Solved, thanks to DrPizza.

 

[herm0016]

kenitic friction will always be lower than static friction.

 

[Ricemarine]

Originally posted by: herm0016
kenitic friction will always be lower than static friction.

But is the kinetic friction this low??
Most of the other problems in the book don't go X.X * 10^-2 area..., usually 0.XX, not 0.0X

 

[DrPizza]

GRRRRRRRRRRRRRRRrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr!!!!!!!!!!!!!

I spent a while going through step by step... the answer is .54
An error wiped the entire reply out.

 

[DrPizza]

Crap. I showed you step by step another approach to your physics problem, and "oops, error" wiped it out. Okay, I'll try again!

I approached it this way: break gravity into vector components...
Component of gravity parallel to the incline is 9.81m/s^2 * sin 30 degrees = 4.905m/s^2

Calculate the actual acceleration down the slope.
2.5 meters, 4 seconds (starting from rest) gives me an acceleration of 5/16 m/s^2 = .3125 m/s^2

Then, approach it from a net force point of view...

Net force = mass of box * acceleration = mass * .3125m/s^2
Net force = parallel force due to gravity - force of friction.
force of friction = parallel force due to gravity minus net force
coef of kinetic friction * normal force = parallel force due to gravity minus net force
coef of k.f. * mass *9.81 m/s^2 * cos 30 degrees = mass * 4.905 m/s^2 - mass * .3125 m/s^2
divide both sides by mass.
solve for coef of kinetic friction.
.54

Incidentally, in you stated to find the static friction and kinetic friction. Static friction and kinetic friction are forces, measured in newtons. The coefficient of static frictino and the coefficient of kinetic friction are unitless quantities. Your first formula suggests you're looking for the coefficients of friction, not the friction itself.

 

[Howard]

Draw FBD 🙂

 

[Ricemarine]

Ahh now I understand...
I was getting a bit hasty trying to plug in an equation to solve for a when a was actually solvable...

so pretty much

Fnet - Ff = ma ---> Ff = Fnet - ma
also, Fnet = MkFn

So MkFn = Fnet - ma
Mk * mgcos30 = mgsin30 - 0.3125 m/s^2 * m
Mk = (gsin30 - 0.3125)/gcos30
Mk = 0.54...


Alright!!
Thanks a lot DrPizza for taking time to help me solve it 🙂