# Help with algebra in a physics problem

1. Oct 5, 2005

### 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?

Last edited: Oct 5, 2005
2. Oct 5, 2005

### amcavoy

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

3. Oct 5, 2005

### 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...

4. Oct 5, 2005

### JoshHolloway

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

5. Oct 5, 2005

### amcavoy

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

6. Oct 5, 2005

### JoshHolloway

alright just one moment....

7. Oct 5, 2005

### JoshHolloway

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

8. Oct 5, 2005

### JoshHolloway

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

9. Oct 5, 2005

### JoshHolloway

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

10. Oct 5, 2005

### JoshHolloway

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

11. Oct 5, 2005

### JoshHolloway

You are a godsend! Thanks alot friend.

12. Oct 5, 2005