Question

I recently did a lab experiment. In the experiment there was in incline at 30.4 degrees, and m2 which was on the incline had a string attached to it and was passed over a pulley and had m1 at the end of it. The idea was to add enough weight so m2 would move with constant velocity.

I'm supposed to calculate the coefficient of kinetic friction, muk.
I'm not sure how to even begin solving for this. I did convert m1 to a force, and I got 0.637N.

Here are my knowns:
M1: 0.065kg
M2: 0.0954kg
Theta: 30.4.

Help would be appreciated!

Yes, I've already drawn my force diagram.

Answer 1

is this what you're describing?

Answer 2

Yes!

Answer 3

i literally did this last night, gimme a sec to type it up

Answer 4

\[M_{2}gsin(\theta) + F_{f} = M_{1}g\]\[F_{f} = \mu_{k}M_{2}gcos(\theta)\]
think you can take it from there?

Answer 5

sorry that's not totally correct\[M_{2}gsin(\theta) +F_{f} = T\]
\[T = \frac{M_{1}M_{2} }{M_{1}+M_{2} }g\]

Answer 6

So am I going to be substituting for T and solving for Ff?

Answer 7

essentially yea, isolate muK

Answer 8

Hm, sorry, for some reason isolating muk in this is appearing difficult.
I'm assuming
(m1+m2)M2gSin+M2gcos/m1m2g is definitely wrong.

Answer 9

\[M_{2}gsin(\theta)+\mu_{k}M_{2}\cos(\theta) = \frac{ M_{1}M_{2} }{M_{1}+M_{2} }g\]you can immediately remove g, then rework

\[\mu_{k} = \frac{ \frac{ M_{1}M_{2} }{M_{1}+M_{2} }-M_{2}\sin(\theta) }{M_{2}\cos(\theta) }\]

Answer 10

in the initial equation there should be a g along with muK M2 cos(theta)

Answer 11

Is muK allowed to be a negative? My answer is -0.01186

Answer 12

take the absolute value