Connected particles on an inclined plane

[Taylan]

Homework Statement

Mass of m1=90kg
Inclination is 40degrees. (see the attachment)

a) What must be the size of m2 , to hold the system in balance (no movement) ?

b)what must be the size of m2 at least, to make m1 move upwards. The static friction coefficient between m1 and the plane is 0.3.

c) what is the acceleration of the both particles in b) if the static coefficient of friction is overcome and the sliding friction coefficient is 0.15=

Homework Equations

F=ma
F_friction= μR

The Attempt at a Solution

I am clear with a and b. the tension pulling both particles and the acceleration of the particles is the same in this kind of questions. I write equations for T (tension) for both particles separately (ex.. for m2 --> T-mg=ma) and substitute them to find a and T.
in a, mass of m2 found to be 57.85kg and in b it is 78.54kg.

I am confused about c. So once again suming up the data I have for c and trying to solve it:
m1= 90gk

m2= 78.53kg

μ= 0.15 ( ı just used the sliding friction coefficent and didn't involve the static one anywhere since the particles are moving. This is where I am confused about. Should I really leave the static friction coefficient out of my calculations?)
a=?

-------writing equations for m2-------------
m2.g-T=m2.a
T=m2.g-m2.a ...(1)

------writing equations for m1---------
T-m1gsin(40)-μ.m1.g.cos40=m1.a
T=m1.a+m1.g.sin(40)+μ.m1.g.cos40.....(2)T (1) = T (2)

m2.g-m2.a=m1.a+m1.g.sin(40)+μ.m1.g.cos40

solving for a give a=1.81m/s²

 

[TSny]

( ı just used the sliding friction coefficent and didn't involve the static one anywhere since the particles are moving. This is where I am confused about. Should I really leave the static friction coefficient out of my calculations?)

Yes. When the blocks are moving, the only friction in play is kinetic.

solving for a give a=1.81m/s²

I believe that's correct. Your work looks good.