Two blocks sliding past each other

[chiddler]

Homework Statement

A 3.0kg block sits on top of a 5.0kg block which is on a horizontal surface. The two blocks are connected by a string; the string connects the top to the bottom with a pulley. The 5.0kg block is pulled to the right with a force. The coefficient of static friction between all surfaces is 0.55 and the kinetic coefficient is 0.36. (I attached figure)

What is the minimum value of needed to move the two blocks?

Homework Equations

Sum forces = ma

The Attempt at a Solution

So, from the free body diagram I drew up, I wrote this equation for the block on the bottom:

F - T - Ff = ma

F is force, T is tension from string, Ff is friction force. I set ma = 0 since I am finding the lowest amount of force for movement to occur.

To find T, I look at the top block:

T - Ff = ma = 0

T is tension, Ff = 0.36 * 3 * 9.8 = 10.6

So T = 10.6 N; ok now I have T. Now I need to find friction of the bottom block.

Ff = u * Fn (normal force)
Ff = 0.55 * (mg + force from top block)
Ff = 0.55 * (5*9.8 + 3*9.8)
Ff = 43.12

Summing it all together:

F - T - Ff = 0
F - 10.6 - 43.12 = 0
F = 32.5 N. Which is wrong :(

I've spent a good 1.5 hours trying to spot my mistake. I'm angrier than a white supremist in africa right now; please help!

 

[PhanthomJay]

Homework Statement

A 3.0kg block sits on top of a 5.0kg block which is on a horizontal surface. The two blocks are connected by a string; the string connects the top to the bottom with a pulley. The 5.0kg block is pulled to the right with a force. The coefficient of static friction between all surfaces is 0.55 and the kinetic coefficient is 0.36. (I attached figure)

What is the minimum value of needed to move the two blocks?

Homework Equations

Sum forces = ma

The Attempt at a Solution

So, from the free body diagram I drew up, I wrote this equation for the block on the bottom:

F - T - Ff = ma

F is force, T is tension from string, Ff is friction force. I set ma = 0 since I am finding the lowest amount of force for movement to occur.

In your FBD of the bottom block, you are missing a force that acts on the block in the horizontal direction.

To find T, I look at the top block:

T - Ff = ma = 0

T is tension, Ff = 0.36 * 3 * 9.8 = 10.6

So T = 10.6 N; ok now I have T. Now I need to find friction of the bottom block.

Why are you using the kinetic friction coefficient? Is either block moving?

Ff = u * Fn (normal force)
Ff = 0.55 * (mg + force from top block)
Ff = 0.55 * (5*9.8 + 3*9.8)
Ff = 43.12

Yes, this is the friction between the horizontal surface and the lower block.

Summing it all together:

F - T - Ff = 0
F - 10.6 - 43.12 = 0
F = 32.5 N. Which is wrong :(

I've spent a good 1.5 hours trying to spot my mistake. I'm angrier than a white supremist in africa right now; please help!

Please make corrections and resubmit.

 

[chiddler]

In your FBD of the bottom block, you are missing a force that acts on the block in the horizontal direction. Why are you using the kinetic friction coefficient? Is either block moving? Yes, this is the friction between the horizontal surface and the lower block. Please make corrections and resubmit.

Hey thank you for the assistance.

1. I must have forgotten the friction force of the top block exerted on the bottom block. Is that right?

2. I use kinetic friction coefficient because the top block and bottom block will both be moving, when force is exerted, no? I read that kinetic friction coefficient is for an object that is in motion and static is for when it is at rest. Well...they do start at rest. Is that my error?

3. Ok. So now, assuming my #1 is correct, I will rewrite it as:

F - T - Ff - Ff2 = mg = 0

Where Ff2 is the force of friction that the top block exerts on the bottom block. Is this correct?

If so, is Ff2 simply Ff2 = u * Fn = 0.55 * 3 * 9.8

?

 

[PhanthomJay]

Hey thank you for the assistance.

1. I must have forgotten the friction force of the top block exerted on the bottom block. Is that right?

That is correct.

2. I use kinetic friction coefficient because the top block and bottom block will both be moving, when force is exerted, no? I read that kinetic friction coefficient is for an object that is in motion and static is for when it is at rest. Well...they do start at rest. Is that my error?

Yes. They cannot move with respect to each other or with respect to the ground until static friction beteeen them and between the lower block and ground is overcome.

3. Ok. So now, assuming my #1 is correct, I will rewrite it as:

F - T - Ff - Ff2 = mg you mean ma, right?[/color] = 0

Where Ff2 is the force of friction that the top block exerts on the bottom block. Is this correct?

Yes again..

If so, is Ff2 simply Ff2 = u * Fn = 0.55 * 3 * 9.8

?

Yes, again you are correct! Continue...while you're on a roll..

 

[chiddler]

So then:

F - T - Ff - Ff2 = 0

F - 10.6 - 43.12 - (0.55)(3)(9.8) = 48.69 N.

Which is also incorrect lol

Can you see if I missed anything?

 

[PhanthomJay]

So then:

F - T - Ff - Ff2 = 0

F - 10.6 - 43.12 - (0.55)(3)(9.8) = 48.69 N.

Which is also incorrect lol

Can you see if I missed anything?

In calculating T, you used the wrong friction coefficient.

 

[chiddler]

Thanks i got it; very useful!

Too late though :(