- #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/
agravity = 32.2 ft/s2
Homework Equations
[tex]\Sigma[/tex]FyA = mA*aA
[tex]\Sigma[/tex]FxA = 0
[tex]\Sigma[/tex]FyB= 0
[tex]\Sigma[/tex]FxB= mB * aB
The Attempt at a Solution
FA on B = FB on A
Writing my force equations, I get:
[tex]\Sigma[/tex]FyA = mA*aA = Ffriction,wall+ Ffriction,B on Asin(70) + FB on A*sin(20)
[tex]\Sigma[/tex]FxA = 0 = Fwall + F friction, B on A*cos(70) - F B on A*cos(20)
[tex]\Sigma[/tex]FyB = 0 = N - 100 -FA on B *sin(20) - Ffriction, 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!