ok my 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
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
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)
To help you understand how to arrive at the equation a = (m1 + m2)^-1 (m2g - m1g sin theta - (μ kinetic)m1g cos theta), let's break it down step by step:
Starting from the equation you provided:
m1a = m2g - m2a - m1g sin θ - (μ kinetic)m1g cos θ
Step 1: Group like terms
Combine the terms containing 'a' on the left side of the equation and the terms without 'a' on the right side:
m1a + m2a = m2g - m1g sin θ - (μ kinetic)m1g cos θ
Step 2: Simplify
Combine the terms with 'a' using the distributive property:
(a(m1 + m2)) = m2g - m1g sin θ - (μ kinetic)m1g cos θ
Step 3: Isolate 'a'
Divide both sides of the equation by (m1 + m2) to solve for 'a':
a = (m2g - m1g sin θ - (μ kinetic)m1g cos θ) / (m1 + m2)
This is the final form of the equation provided in your textbook.
It's worth noting that the term (m1 + m2)^-1 is equivalent to 1 / (m1 + m2), which represents the reciprocal of the sum of m1 and m2. So, the equation in your textbook is just another way of expressing the same thing.
I hope this helps clarify how to arrive at the given equation!