2 non-rectangular blocks, friction, only gravity

[trogtothedor]

Homework Statement

An attachment with the problem specifications, HW6 - C, is attached. Here is a link in case that is easier to view.

http://www.flickr.com/photos/54849943@N04/5082480364/

a_gravity = 32.2 ft/s²

Homework Equations

$$\Sigma$$F_yA = m_A*a_A
$$\Sigma$$F_xA = 0
$$\Sigma$$F_yB= 0
$$\Sigma$$F_xB= m_B * a_B

The Attempt at a Solution

F_{A on B} = F_{B on A}

Writing my force equations, I get:
$$\Sigma$$F_yA = m_A*a_A = F_{friction,wall}+ F_{friction,B on A}sin(70) + F_{B on A}*sin(20)

$$\Sigma$$F_xA = 0 = F_wall + F _{friction, B on A}*cos(70) - F _{B on A}*cos(20)

$$\Sigma$$F_yB = 0 = N - 100 -F_{A on B} *sin(20) - F<sub>friction, A on B *sin(70)

$$\Sigma$$FyB = mB * aB = FA on B *cos(20) - F friction, ground - F friction, A on B* cos(70)

Also, to simplify,

Ffriction,wall = Fwall * $$\mu$$k

Ffriction, ground = N* $$\mu$$k

For both cases, Ffriction, A on B = FA on B* $$\mu$$k

Also, since both blocks start from rest, 1 foot = (1/2) * aA * (t²)

Using trigonometry, I also concluded that when block A moves one foot down, block B will have moved 1/tan(70) feet to the right. Thus:

t² = 2/(aA) = 2/(aB * tan(70))

Am I wrong in any of this? Either something here is wrong, or my algebra is going haywire throughout the solving process.

Thank you in advance for any assistance!</sub>

 

[trogtothedor]

I anyone can see any area where I've made a mistake, I'd appreciate it! I feel like I covered everything and have all the angles right, but I'm still coming up with the wrong answer.

 

[trogtothedor]

I got it. In my coorrdinate system, acceleration A is negative, so -aA=aBtan(70)