ok my question

Question

I'm trying to solve this for a

m1 a = m2 g - m2 a - m1 g sin theta - (Mu kinetic)m1 g cos theta

my text book gives me this

a = (m1 + m2)^-1 (m2 g - m1 g sin theta - (Mu kinetic) m1 g cos theta)

ok I don't see how I get to this point tand don't know what to do

a = m1^-1 (m2 g - m2 a - m1 g sin theta - (Mu kinetic)m1 g cos theta)

thanks
Mu kinetic is the coeficent of kinetic friction i couldn't put the greek letter Mu in

Answer 1

I don't know what to do with the m2 a I know I have to do somehthing with that but what I'm not sure

adding it to both sides and dividing by m2 gives me a + a or more simply 2a but then I would have to divide by 2 which isn't in the answer my text book gives me

Answer 2

m1 a = m2 g - m2 a - m1 g sin theta - (Mu kinetic)m1 g cos theta

add m2a to both sides:

a(m1+m2)=g(m2-m1*sinTheta)
divide both sides by m1+m2

a= g(m2-m1sinTheta)/(m1+m2)