I'm working on a problem involving coffecient of friction and a sloped surface, and I was wondering if the coefficient changed depending on where the person was on the slope. Here's the problem for reference:
A skier traveling 12.0 m/s reaches the foot of a steady upward 18 degrees and glides 12.2 m up along this slope before coming to rest. What was the average coefficient of friction?
Now, does the 12.2 m matter in this problem to determine the coefficient of friction? No matter how I look at it:
u = Ff / Fn = F parallel / F perpendicular = tan (pheta)
I would also like to know if there is a difference between "coefficient of friction" and "average coefficient of friction". Please help~!
2006-08-30 17:01:01 · 6 answers · asked by Moosehead 2 in Science & Mathematics ➔ Physics
I like this problem!
Initial KE = (mv^2)/2 = 72m
Final PE = mgl*sin 18 = 36.946m ............. (l = 12.2)
Fn = mg cos 18
Ff = uFn = umgcos 18 = 9.32m u
Now, the friction energy loss must equal the difference in the kinetic and potential energy:
Ff*l = 9.32m u * 12.2 = 72m - 36.946m
Solving this for u gives u = 0.308
douglas, Lighten UP!
2006-08-30 18:07:23 · answer #1 · answered by Steve 7 · 1⤊ 0⤋
The coefficient of friction depends only upon the two surfaces in contact.
The actual frictional force is the product of the coefficient of friction and the normal reaction.
In a horizontal plane surface the weight of an object acts vertically down ward and hence the normal reaction is vertically upward. The friction acts horizontally and is equal to the product of coefficient of friction and the weight.
When the plane is at an angle, the normal reaction tilts through that angle from the vertical direction. Now the normal reaction is weight of that object x cosine of the angle of inclination. The frictional force is coefficient of friction times the normal reaction.
If the slope is increased, friction decreases, but the coefficient remains the same.
For any plane surface there is an angle of inclination at which the coefficient of friction is the tangent of that angle. That angle is also called angle of friction.
In the given problem using the angle 18 degree and the length 12.2 m we can calculate the coefficient of friction as 0.308.
The angle of friction is 17.11 degree.
2006-08-30 23:48:49 · answer #2 · answered by Pearlsawme 7 · 0⤊ 0⤋
As Steve showed, the distance is doubly significant, as the initial energy of the skier goes into 1) gaining altitude, and 2) frictional braking (= force x distance). It says "average coefficient of friction" because unless you know speed versus position, you can't know whether the friction was uniform.
2006-08-30 19:37:49 · answer #3 · answered by injanier 7 · 0⤊ 0⤋
The coefficent of friction doesn't change with distance. Average coefficient of friction would only differ if you were changing surfaces.
2006-08-30 17:05:55 · answer #4 · answered by Anonymous · 1⤊ 0⤋
What matters is the effect of gravity on the Y force coefficent. The distance and angle determine height.
2006-08-30 17:08:34 · answer #5 · answered by Richard B 4 · 0⤊ 0⤋
It's "theta" you idiot, not pheta.
2006-08-30 17:08:33 · answer #6 · answered by MrZ 6 · 0⤊ 2⤋