Find coefficient of kinetic friction on an inclined plane.

[shandsy]

Homework Statement

(there's a diagram that shows a crate on a plane inclined 19.29° above the horizontal)

Find the coefficient of kinetic friction.

θ = 19.29°
m_crate = .205 kg

Homework Equations

F_f = F_N*μ_k
F_g = mg
F_N = cosθ*mg

The Attempt at a Solution

I drew out a free-body diagram for the crate as it slid down the incline including the x and y components of F_g. I came up with the following:

F_g = 2.009 N
F_gy = cos(19.29°)*2.009N = 1.896 N
F_gx = sin(19.29°)*2.009N = 0.664 N

F_N = F<sub>gy = 1.896 N
Ff = Fgx = 0.664 N

However, if I plug these values back in, I get .350 as the value for μk, but I know that the correct answer is supposed to be .93.

Am I right in assuming that Fgx = Ff? I feel like I'm completely missing something...</sub>

 

[Simon Bridge]

Welcome to PF;
Not enough information to work out if you are right - need to know the acceleration of the crate? [1]

Your mistake is to work out the numbers too soon ... doing that hides a ot of the concepts that let you troubleshoot when something goes wrong.

In a slope of angle $\theta$ to horizontal, you write the sum of the forces along the slope = mass times acceleration.

Thus:$$mg\sin(\theta) - \mu mg\cos(\theta) = ma$$... if the acceleration is zero (you found the angle that gave a constant speed) then a=0 and you can solve the equation for $\mu$, simplify it ... then plug your numbers in.

That way, if your answer still disagrees with the model answer, you know you have got it right!

-----------------------------

[1] I can work it out from the model answer - but I'm not showing you because that would give the game away ;)

 

[shandsy]

Sorry if I left the acceleration out-- the problem itself is part of a lab in which we're supposed to be finding the acceleration in a pulley system, but this was the setup my teacher had us use to calculate μ.

Here's the specific question in the lab asking us to set up the system:

"You first need to find the coefficient of kinetic friction between the passenger mass and the cart. Put the Pasco cart on the track and place the passenger mass at the front of the cart's tray. Hold the Pasco cart so that it is not allowed to move and gradually incline the track to the minimum angle that allows the passenger mass to slowly slide down the Pasco cart at a constant speed."

As said previously, we calculated the angle to be 19.29°.

If I'm to assume that acceleration is constant in this set up, I still get .35 for μ, so I still think there's something I'm missing. I know that .93 has to be the real value for μ because it works out with the rest of the lab. I wasn't there when my lab group came up with .93 and I have no way of contacting them at the moment, so I don't know how they came to that value. All I know is that we had the same values for everything else...

Thank you for your help anyway!

 

[Simon Bridge]

Hold the Pasco cart so that it is not allowed to move and gradually incline the track to the minimum angle that allows the passenger mass to slowly slide down the Pasco cart at a constant speed."

As said previously, we calculated the angle to be 19.29°.

?? surely you measured this angle? (How: ratio of two lengths and an inverse trig? arctan(y/x) perhaps?)

If I'm to assume that acceleration is constant in this set up, I still get .35 for μ

So do I :)

so I still think there's something I'm missing. I know that .93 has to be the real value for μ because it works out with the rest of the lab.

Do you mean that using the value of 0.93 gets you the right answers in the rest of the exercizes in the lab book or that the other students in the lab doing the same experiment all get 0.93 out?

Because if they do, and that is how I'm reading what you are telling me, then they are not doing the same lab as you just described.

I wasn't there when my lab group came up with .93 and I have no way of contacting them at the moment, so I don't know how they came to that value. All I know is that we had the same values for everything else...

There is something else going on ...

The relation is simple: $\mu=\tan(\theta)$
(You realize that if the angle was measured by arctan(y/x) then y/x was actually the value you needed?)

Thus: $0.93=\tan(42.93^\circ)$ ... a 23.63° difference.
(Maybe look through the raw data for a systematic error in how the angle was obtained?)

You have a choice: fudge the results for θ=42.93, so you can use the high μ value, or trust the results you have recorded (and hope everyone else did the fudge) and use your value for μ.