Acceleration of a Block on pulley

[KillerZ]

Homework Statement

Determine the acceleration of block A when the system is released. The coefficient of kinetic friction and the weight of each block are indicated. Neglect the mass of the pulley and cord.

Homework Equations

\sum F = ma

The Attempt at a Solution

w_{B} = 20 lb

w_{A} = 80 lb

\mu_{k} = 0.2

\theta = 60 deg

Block A

\leftarrow\sum F_{x} = ma_{x}

-F_{f}cos60 + Nsin60 - 2Tcos60 = ma_{x}

\downarrow\sum F_{y} = ma_{y}

-F_{f}sin60 + w - Ncos60 - 2Tsin60 = ma_{y}

Block B

\downarrow\sum F_{y} = ma_{y}

w - T = ma_{y}

Kinematics

2S_{A} + S_{B}

2a_{A} + a_{B}

a_{A} = -\frac{a_{B}}{2}

I am confused do I just say N = ma = (\frac{80lb}{32.2ft/s^{2}})(32.2ft/s^{2}) or do I have to solve N as an unknown?

 

[rl.bhat]

What is the value of Ff and N?

 

[KillerZ]

Wouldn't they be:

N = ma = (\frac{80lb}{32.2ft/s^{2}})(32.2ft/s^{2}) = 80lb

and

F_{f} = N(0.2) = (80lb)(0.2) = 16lb

 

[rl.bhat]

No.
N is the normal reaction of the surface on the sliding body.
You have to resolve the weight of the body ( mg) into two components. One pqarallel to the inclined plane and the other along the inclined plane.
Now redraw the FBD.
You have written 2SA = SB. Can you explain why it so?

 

[KillerZ]

Block B FBD is same and block A:

I modified my positive directions a little:

Block A:

\sum F_{x} = ma_{x}

-2T + wsin60 -F_{f} = ma_{x}

\sum F_{y} = ma_{y}

N - wcos60 = ma_{y}

Block B:

\downarrow\sum F_{y} = ma_{y}

w - T = ma_{y}

I have said 2SA + SB because:

 

[rl.bhat]

When the block A moves down through a distance x, two segments of the ropes attached to A will also lengthen by x each. Since the total length of the string is constant, B must move up by 2x. Is it not so?