Calculating Forces and Kinetic Friction for a System of Connected Blocks

[Sagrebella]

Hi everyone,

I'm having a difficult time figuring out this problem. Could someone give me some pointers? I set up the equations and the free body diagrams (hopefully they're correct) ; all I have to do now is solve for F.1. Homework Statement

A block of mass m is on top of a block of mass
M = 5m.
The two blocks are connected by a string that passes over an ideal pulley. The bottom block is also tied to a string - exerting a force F, after the blocks have been set in motion, causes the blocks to move at constant velocity.

Use
g = 10 N/kg
for this problem. The coefficient of kinetic friction between all surfaces in contact is 0.400.

Homework Equations

F_N1= normal force of small block
F_N2= normal force of large block
F_k1 = force of kinetic friction on small block
F_k2 = force of kinetic friction on large block
F_T = force of tension

Bock 1 (small)

F_N1 - m₁g + F_k1 - F_T = 0

Block 2 (large)

F_N2 - m₂g - F_N1 + F - F_k1 - F_T - F_k2 = 0

EDIT: so sorry, I missed the second part of the question!

[/B]
If the value of m is 260 g, what is the value of F?

The Attempt at a Solution

[/B]
- I set both equations equal to zero because they are moving at constant velocity, and hence constant acceleration, which makes the total force equal to zero
- Next, I set both equations equal to each other through the tension (F_T)
- I replaced the term F_k1 and F_k2 with the equations for kinetic friction and then attempted to solve ...

My answer came out to 10 N, but this is wrong. What can I do to get the right answer?

Thanks

Pictures shown below for more detail ...

 

[TSny]

In setting up your equations, it looks like you are treating the forces as though they are all parallel to one another.

When you add vectors, such as forces, you must use vector addition.

(Your free body diagrams look very good.)

 

[Sagrebella]

In setting up your equations, it looks like you are treating the forces as though they are all parallel to one another.

When you add vectors, such a forces, you must use vector addition.

(Your free body diagrams look very good.)

Thanks for the complement. Unfortunately, I'm not exactly sure what you mean. Did I not add the forces like vectors? I indicated whether they are moving in the positive or negative direction in the equation and in the diagram (I even drew a little coordinate plane next to the FBDs in order to differentiate the positive and negative directions).

 

[TomHart]

Did I not add the forces like vectors?

Normally, the way people work problems like this is they break it down into x and y components. You are adding vertical forces with horizontal forces, which doesn't make a lot of sense. That, I believe, is what @TSny meant by treating all forces as if they were parallel.

One other comment of mine: I didn't see a value for mass m. Maybe I missed it. It looks like you are using 0.26 kg. I just didn't see that in the problem statement.

 

[TomHart]

As I mentioned, adding vertical and horizontal forces together doesn't make a lot of sense. But since there is no acceleration in the vertical direction, those forces should sum to zero and not affect the answer. However, it looks like your F_N2 does not account for the weight of the smaller block sitting on top of the larger block. The normal force, F_N2, has to equal the weight of both blocks. If it doesn't, there would be acceleration.

So in summary, if you simply eliminate all of the vertical forces in your equations, I suspect you will get the correct answer.

 

[Sagrebella]

Normally, the way people work problems like this is they break it down into x and y components. You are adding vertical forces with horizontal forces, which doesn't make a lot of sense. That, I believe, is what @TSny meant by treating all forces as if they were parallel.

One other comment of mine: I didn't see a value for mass m. Maybe I missed it. It looks like you are using 0.26 kg. I just didn't see that in the problem statement.

yes, so sorry, I forgot to include the second part of the problem If the value of m is 260 g, what is the value of F?

and, I think I understand what you're saying. Would the vertical forces simply not be included in the equation because they cancel out (normal force and force of gravity are equal and opposite) ?

 

[TomHart]

Would the vertical forces simply not be included in the equation because they cancel out (normal force and force of gravity are equal and opposite) ?

Yes, they will cancel out in this case. And you know that because there is no acceleration in the y direction. I have seen people add x and y components in the same equation by designating an i and j vector for x and y direction. But for me, it makes it clearer for me if you separate out vertical and horizontal components.

So for me, I sum the forces in the x direction and sum the forces in the y direction, keeping them completely separate.

 

[Sagrebella]

Yes, they will cancel out in this case. And you know that because there is no acceleration in the y direction. I have seen people add x and y components in the same equation by designating an i and j vector for x and y direction. But for me, it makes it clearer for me if you separate out vertical and horizontal components.

So for me, I sum the forces in the x direction and sum the forces in the y direction, keeping them completely separate.

Ok, thank you, that makes sense; hopefully I got your point. Here is my revised work; does it look better now?

Note: Block A is the small block, Block B is the big block

 

[TSny]

Ok, thank you, that makes sense; hopefully I got your point. Here is my revised work; does it look better now?

BEAUTIFUL!

 

[Sagrebella]

BEAUTIFUL!

Thanks! but is the answer correct?

 

[TSny]

Thanks! but is the answer correct?

It agrees with my answer. (Of course, that might not mean much.)

 

[TomHart]

It agrees with my answer. (Of course, that might not mean much.)

I got the same answer so it may mean less than it did before. :)

 

[Sagrebella]

It agrees with my answer. (Of course, that might not mean much.)

wonderful! And the answer is indeed correct ( I just inputed it into my online homework). Thank you both for your help. I look forward to your assistance with future problems.