Forces and Friction of sliding blocks

[shawpeez]

I've been looking at this question for the last two hours, here's the question and what I've done so far

The picture shows 3 blocks, A B and C, block C is ontop of block A. Block B is connected to block A by a rope, and is hanging over the edge.

here it is,

Blocks A and B have weights of 44N and 22N, respectively. (a) Determine the minimum weight of block C to keep A from sliding if s between A and the table is 0.20. (b) Block C suddenly is liffted off A. What is the acceleration of block A if k between A and the table is 0.15?


This is what I've done

Block A + C
mg= 44N
m = 44/g
= 4.49 + X

Fnet(y) = N - Mg = ma

N = Mg
=(4.49 + x)g

Fnet(x) = T - f = ma
T - f = (4.49 + x)a

Block B
mg = 22N
m= 22/g
= 2.24

Fnet = T - mg = ma
T - (2.24)(9.8) = 0
T = 22N

From this i assumed that when there is no acceleration the Tension in the rope is 22N, therefore the static friction when there is no movement in Block A+C is 22N

Fmax = s Fn
22N = 0.2 (mg)
22N = 0.2 (4.49 + x)(9.8)
22 = 8.8 + 1.96x
x= 6.7 Kg Therefore the min. weight of C is 6.7kg times 9.8m/s^2 = 66N

im not sure if I did this right, so I would appreciate if someone could let me know. Thanks

 

[ehild]

Therefore the min. weight of C is 6.7kg times 9.8m/s^2 = 66N
im not sure if I did this right, so I would appreciate if someone could let me know.

It is correct.

ehild