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

 

[quicknote]

I would say that your approach to solving this problem is mostly correct. You have correctly identified the forces acting on each block and used Newton's laws of motion to determine the minimum weight of block C to prevent block A from sliding. I would also suggest that you double check your calculations to ensure that all units are consistent and that you have correctly accounted for the acceleration due to gravity (9.8 m/s^2). Additionally, it would be helpful to label all of your variables and equations clearly to make your work easier to follow.

For part (b) of the question, you have correctly determined that the tension in the rope is 22N when there is no acceleration. However, I would suggest using the equation Fnet = ma to solve for the acceleration of block A, rather than assuming that the tension remains constant. This will give you a more accurate answer.

Overall, your approach to solving this problem is sound and demonstrates a good understanding of the concepts of forces and friction. Keep up the good work!