# Acceleration of a Block

1. Oct 13, 2009

### KillerZ

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

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.

2. Relevant equations

$$\sum F = ma$$

3. 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?

Last edited: Oct 13, 2009
2. Oct 13, 2009

### rl.bhat

What is the value of Ff and N?

3. Oct 13, 2009

### KillerZ

Wouldn't they be:

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

and

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

4. Oct 13, 2009

### 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) in to 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?

5. Oct 13, 2009

### 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:

6. Oct 13, 2009

### 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?