Solving for acceleration with multiple a terms in an equation

[JoshHolloway]

I don't need help with the physics, it is simply the algebra that I can't figure out in this problem. Here am where I am at:
$$-m_{1} a + \mu ( m_{1} g \cos \theta ) + m_{1} g \sin \theta = \frac{\m_{2} a +m_{2} g \cos \theta }{ \cos \theta }$$
I need to solve for a. How the heck do I do this? How can I factor the a's into just one a?

 

[amcavoy]

I don't need help with the physics, it is simply the algebra that I can't figure out in this problem. Here am where I am at:
$$-m_{1} a + \mu ( m_{1} g \cos \theta ) + m_{1} g \sin \theta = \frac{\m_{2} a + m_{2} g \cos \theta }{ \cos \theta }$$
I need to solve for a. How the heck do I do this? How can I factor the a's into just one a?

My suggestion is to multiply both sides by cosθ, thereby eliminating any fractions. From there, rearrange terms so that any term with an a in it is on one side, and everything else is on another. Now you can factor an a out of this and solve.

Alex

 

[JoshHolloway]

in the fraction it is supposed to say m2, not just a subscript two. I don't know what I did wrong.
And I did that. Here I will show you how far I have gotten past that. Just a few minutes...

 

[JoshHolloway]

$$\cos \theta [-m_{1} a + \mu ( m_{1} g \cos \theta ) + m_{1} g \sin \theta] = m_{2} a +m_{2} g \cos \theta$$

 

[amcavoy]

$$\cos \theta [-m_{1} a + \mu ( m_{1} g \cos \theta ) + m_{1} g \sin \theta] = m_{2} a +m_{2} g \cos \theta$$

That's step 1. Now distribute the cosine and put all of the terms containing an a in them on one side. Tell me what you get.

Alex

 

[JoshHolloway]

alright just one moment...

 

[JoshHolloway]

$$-m_{1} a \cos \theta + \mu ( m_{1} g \cos ^2 \theta ) + m_{1} g \sin \theta \cos \theta = m_{2} a +m_{2} g \cos \theta$$

 

[JoshHolloway]

I distributed, now one moment and I will attempt to do the second step you said. By the way this is REALLY helping.

 

[JoshHolloway]

$$\mu ( m_{1} g \cos ^2 \theta ) + m_{1} g \sin \theta \cos \theta - m_{2} g \cos \theta = m_{1} a \cos \theta + m_{2} a$$

 

[JoshHolloway]

$$\mu ( m_{1} g \cos ^2 \theta ) + m_{1} g \sin \theta \cos \theta - m_{2} g \cos \theta = a (m_{1} \cos \theta + m_{2})$$

 

[JoshHolloway]

You are a godsend! Thanks a lot friend.

 

[amcavoy]

You are a godsend! Thanks a lot friend.

Glad I could help