Frictional Forces question

1. Jul 8, 2011

Redcutter

Hello, I'm new here. I've browsed before and saw that this is a very helpful community, so I came here for assistance.

1. The problem statement, all variables and given/known data

Blocks A, B, and C are placed as in the figure and connected by ropes of negligible mass. Both A and B weigh 27.5N each, and the coefficient of kinetic friction between each block and the surface is 0.40. Block C descends with constant velocity.

Image attached:
http://session.masteringphysics.com/problemAsset/1038585/38/YF-05-56.jpg

2. Relevant equations

What is the weight of block C?

3. The attempt at a solution

I've found the first part of this problem, which is the tension holding A and B, which is 11N. Part C has me confused. Here's my attempt:
Block B

∑FY=
NB-mBgsin(36.9)=0
NB=27.5Nsin(36.9) = 16.5115562

fK= µNB = .40(16.5115562)= 6.604622479

∑Fx=
T2-T1-fK- 27.5cos(36.9)= 0

T2-T1-6.604622479- 21.99132811= 0

T1= 11N

T2-11-6.604622479- 21.99132811= 0

T2= 17.59595059N.

Block C

∑Fx= T2-mCg = 0
T2 = mCg = 17.59595059N

Is this correct?

Last edited: Jul 8, 2011
2. Jul 8, 2011

elegysix

Since the system is not accelerating, $\sum$F=0.

F1=.4*mg=.4*27.5N

F2=.4*mg*sin(36.9 degrees) = .4*27.5*sin(36.9) N

therefore

F1 + F2 = mg (m = mass of block 3)

.4*27.5(1+sin(36.9)) = mg (the weight of block 3)

3. Jul 8, 2011

jaumzaum

O M = opposite to the movement

Block a:
F in a = F friction in a = N.u = 27,5u O M

Block b:

F in b = Pb sin36,9° and Pb cos36,9°u O M = 27,5.3/5 and 27,5.4/5.u

F friction = 27,5 . 3,6/5 > 27,5.3/5 so they have same direction

27,5.6,6/5 = C = 5,5.6,6 = 36,3N

4. Jul 10, 2011

Redcutter

Thanks, guys.