Solve Pulley Problems: Friction & Inertia | Physics Help

[blackice552]

I have to problems that I'm having trouble with today I just can't seem to get my head around a lot of these physics concept so I'm finally coming here Plz help me. I've tried to do these problems and I just don't know what I'm doing.



1.We have two masses hanging from a pulley of inertia mp and constant frictional force f. One mass is 21.5g and the other is 39.8g. When data is taken, the slope of the 1/a vs. (m1 + m2) graph turns out to be 6.77 1/N. What is the value of the frictional force of the pulley? (2 decimals and UNITS!) (g=9.8m/s2)

We have two masses hanging from a pulley of inertia mp and constant frictional force f. One mass is 55.9g and the other is 41.8g. When data is taken, the intercept of the 1/a vs. (m1 + m2) graph turns out to be 0.04s2/m and the frictional force of the pulley 0.31N. What is the value of the inertia of the pulley? (2 decimals and SI UNITS!) (g=9.8m/s2) (If you get a negative answer type it in even though it doesn't make sense physically)

Homework Equations

These are the equations I have:

for no friction:
1/a = (m1 + m2)/((m1 -m2)g)

for a system with friction:
(m1 - m2)g - f = (m1 + m2 + mp)a

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

The Attempt at a Solution

My solutions were totally outrageous I have no idea what I'm doing frankly I've studied this stuff for hours and can't wrap my head around it

 

[member 553083]

.For the first problem:1/a = (21.5 + 39.8)/((21.5 - 39.8)9.8)f = (21.5 - 39.8)9.8 - (21.5 + 39.8)6.77 f = -3.45 NFor the second problem:(55.9 - 41.8)9.8 - f = (55.9 + 41.8 + mp)0.04mp = ((55.9 - 41.8)9.8 - f)(25.1/0.04) - (55.9 + 41.8)mp = 0.31 N * 25.1/0.04 - 97.7mp = 7.60 kg