- #1

trogtothedor

- 26

- 0

## 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/s

^{2}

## Homework Equations

[tex]\Sigma[/tex]F

_{yA}= m

_{A}*a

_{A}

[tex]\Sigma[/tex]F

_{xA}= 0

[tex]\Sigma[/tex]F

_{yB}= 0

[tex]\Sigma[/tex]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:

[tex]\Sigma[/tex]F

_{yA}= m

_{A}*a

_{A}= F

_{friction,wall}+ F

_{friction,B on A}sin(70) + F

_{B on A}*sin(20)

[tex]\Sigma[/tex]F

_{xA}= 0 = F

_{wall}+ F

_{ friction, B on A}*cos(70) - F

_{ B on A}*cos(20)

[tex]\Sigma[/tex]F

_{yB}= 0 = N - 100 -F

_{A on B}*sin(20) - F

_{friction, A on B *sin(70) [tex]\Sigma[/tex]FyB = mB * aB = FA on B *cos(20) - F friction, ground - F friction, A on B* cos(70) Also, to simplify, Ffriction,wall = Fwall * [tex]\mu[/tex]k Ffriction, ground = N* [tex]\mu[/tex]k For both cases, Ffriction, A on B = FA on B* [tex]\mu[/tex]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!}