If the first Atwood machine is friction-less, would m1g go away or would friction just not be calculated and the numerator remains just m2g-m1g? For the second Atwood machine I got a negative value when I set N = m1cos(25), and plugged that into the numerator as a subtraction value since friction goes against motion. Not sure I did that correctly...
hey, not sure if you figured it out but if the first machine was friction less, Fnet=m2g. You would not subtract m1g as that is not affecting how the system moves, as the sum of your y forces on m1 is 0. So yes, m1g would just not be apart of the equation. In this case, since we have friction as an external force, that is subtracted from the driving force of the system which is m2g.
If the first Atwood machine is friction-less, would m1g go away or would friction just not be calculated and the numerator remains just m2g-m1g?
For the second Atwood machine I got a negative value when I set N = m1cos(25), and plugged that into the numerator as a subtraction value since friction goes against motion. Not sure I did that correctly...
hey, not sure if you figured it out but if the first machine was friction less, Fnet=m2g. You would not subtract m1g as that is not affecting how the system moves, as the sum of your y forces on m1 is 0. So yes, m1g would just not be apart of the equation. In this case, since we have friction as an external force, that is subtracted from the driving force of the system which is m2g.