How Do You Calculate Forces and Acceleration in a Two-Block Pulley System?

[33639]

I'm having difficulty with this question

In the diagram, the coefficient of kinetic friction between the 2.00 kg and the 3.00 kg blocks is 0.300. The horizontal surface and the pulleys are frictionless, and the masses are released from rest.
(a) Draw a free body diagram of each block.
(b) Determine the acceleration of each block.
(c) Find the tension in the strings.

I've also been given a diagram:
http://i1289.photobucket.com/albums/b505/homework2012/q5_zpsed497c95.jpg

and here's my FBD: http://i1289.photobucket.com/albums/b505/homework2012/q5_zps75004cd9.jpg

- Am i missing any forces on my FBD Diagram? If yes which ones?
- For part b what would forces would Fnet be equal to, to find the acceleration?
I though it would be Fnet = Fg3 + Ff
Where Fg3 is the 10kg ball
- How many force of friction is acting on the system? Is it just 1 between the 2kg and 3kg blocks? If not one, where are the other Ff?

 

[Doc Al]

- Am i missing any forces on my FBD Diagram? If yes which ones?

You forgot the friction acting on the bottom block.

 

[33639]

You forgot the friction acting on the bottom block.

Do you also include that in the FBD diagram for the 3kg block? Because i though it was already accounted for by the 2kg block?

What would Fnet now be equal to to find Tension?

 

[Doc Al]

Do you also include that in the FBD diagram for the 3kg block?

Of course. You must include all forces acting.

Because i though it was already accounted for by the 2kg block?

Realize that the blocks exert the friction force on each other. Newton's 3rd law.

What would Fnet now be equal to to find Tension?

Apply Newton's 2nd law to each block, then solve those equations together.

 

[Nickie]

Hi there,

I understand that you are having difficulty with this question. I would be happy to provide some guidance and help you understand the concept better.

Firstly, your free body diagrams look correct. You have correctly identified the forces acting on each block - weight (mg), normal force (N), and tension (T). However, there are a few things to consider:

1. In your diagram, you have labeled the weight of the 3kg block as 10kg. This should be corrected to 3kg.

2. The force of friction (Ff) acts in the opposite direction of motion. In this case, the 3kg block is moving to the right, so the force of friction is acting to the left. Therefore, Ff should be labeled in the opposite direction on your diagram.

3. For part (b), you are correct in thinking that the net force (Fnet) is equal to the sum of all the forces acting on the block. In this case, the only forces acting on the block are weight (mg) and friction (Ff). So, the equation would be Fnet = mg - Ff. This will give you the acceleration of the block.

4. There is only one force of friction acting on the system, between the 2kg and 3kg blocks. This is because the surface and the pulleys are frictionless, as stated in the question.

I hope this helps you with your question. If you have any further doubts or questions, please do not hesitate to ask. It is important to fully understand the concept before attempting to solve the problem. Good luck!