- #1

francisco300

- 9

- 0

## Homework Statement

Basically, I need to calculate the mass of the pulley and friction using data that I collected.

## Homework Equations

I was given: (m1 - m2)g = (m1 + m2 + mp)a + f

## The Attempt at a Solution

For the first part, I collected data using a sensor to find the acceleration. I kept the total mass attached to each cup on each side of the machine constant but changed the difference in their masses by 10g each run.

After 6 runs, I plotted my data of Fnet vs Acceleration. Fnet is (m1 - m2)g (I converted g into dynes). When I get my slope, the lab manual says that i should "expect our plot of (m1 - m2)g vs a to have a slope

*larger*than

*(m1 - m2)g*by an amount equal to the effective mass of the pulley

*mp*and a positive intercept equal to the frictional force f"

My equation turned out to be y = 143.28x + 323.24 (i used best fit line).

This means 143.28 = m1 + m2 + mp

In which case mp= 28.5g right?

Then the force of friction should = 323.4 right?

Is the unit in dynes?

For part b, did the same procedure but kept the net force constant instead of the total mass. I gathered my data and plotted

1/a vs

*m1 + m2*

I got y = 0.0002x + 0.0059 as my best fit line

Now I have to use this equation to find the force of friction

1/a = ((m1 + m2) )/((m1-m2)g-f) + mp/((m1-m2)g-f)

But i get 4950 which I believe is in Netwons and that would be too much considering 323 dynes = 0.00323 NewtonsI got y = 0.0002x + 0.0059 as my best fit line

Now I have to use this equation to find the force of friction

1/a = ((m1 + m2) )/((m1-m2)g-f) + mp/((m1-m2)g-f)

But i get 4950 which I believe is in Netwons and that would be too much considering 323 dynes = 0.00323 Newtons