Minimum Force of 86N to Move Two Blocks

[bsvh]

1. Homework Statement
A 3.0-kg block sits on top of a 5.0-kg block which is on a horizontal surface. The 5.0-kg block is pulled to the right with a force $\vec{F}$. The coefficient of static friction between all surfaces is 0.63 and the kinetic coefficient is 0.38. What is the minimum value of F needed to move the two blocks?

Homework Equations

$\|\vec{F}_{fs}\| = \mu_s \|\vec{N}\|$[/B]

The Attempt at a Solution

$\sum F_{bottom} = F_{external} - T - F_{fb}$
$F_{fb} = \mu_s(m_{bottom}+m_{top})g + m_{top}g$
$\sum F_{top} = T - F_{ft}$
$F_{ft} = \mu_sm_{top}g$
$\sum F = F_{external} - T - F_{fb} + T - F_{ft} = F_{external} - F_{fb} - F_{ft} = 0$
$F_{external} = F_{fb} + F_{ft} = \mu_sg((m_{bottom}+m_{top}) + 2m_{top} )$

After plugging in all the numbers:

$F_{external} = 86N$

86N is the right answer, but it seems odd to me because I essentially counted the friction between the two surfaces twice. Is it just the case that the frictional force between two surfaces moving across each other in opposite directions is twice that of when one of the surfaces isn't moving?

 

[Doc Al]

First off, a diagram would be most helpful. I was able to guess how things are arranged, but realize the only time you mention the pulley is in the thread title.

86N is the right answer, but it seems odd to me because I essentially counted the friction between the two surfaces twice.

You used it twice only because the friction between the blocks acts on each block, and you analyzed each block separately.

Is it just the case that the frictional force between two surfaces moving across each other in opposite directions is twice that of when one of the surfaces isn't moving?

No, why would you think that?

 

[bsvh]

First off, a diagram would be most helpful. I was able to guess how things are arranged, but realize the only time you mention the pulley is in the thread title.

You are right. I added one.

No, why would you think that?

Because the total amount of friction between the two blocks is the friction between the bottom block and the top block, and the friction between the top block and the bottom block. If one of the blocks couldn't move you would only have the one force of friction.

 

[Doc Al]

Because the total amount of friction between the two blocks is the friction between the bottom block and the top block, and the friction between the top block and the bottom block. If one of the blocks couldn't move you would only have the one force of friction.

The force of friction is an interaction between two surfaces. The two surfaces exert equal and opposite friction forces on each other, regardless of motion.

 

[bsvh]

The force of friction is an interaction between two surfaces. The two surfaces exert equal and opposite friction forces on each other, regardless of motion.

Does this mean that the forces of friction would cancel each other out if there wasn't a pulley, because they are equal and opposite? Edit: I'm thinking not because we analyse each block individually.

Also, when you say regardless of motion, what exactly do you mean?

Let's say that there is only the bottom block without the pulley. In this case, the friction between this block and the ground would be $\mu_smg$. This is force of friction the ground exerts on the block, correct? Is there a force of friction that the block is exerting on the ground?

Edit: Maybe this will help with my original question.

Is the total friction of these two cases the same, assuming the ground is frictionless in the top case? Maybe I shouldn't use the term friction. In other words, would the required force acting on the blocks to get A and B moving in the top case, and A moving in the bottom case, be the same?

 

[NascentOxygen]

Friction is governed by the relative motion of the surfaces.

One body can exert a force on another only if the other body exerts an equal and opposite reaction force. This applies to all forces, friction is no exception.

You can't push on something with force F unless it pushes back with an equal opposing force. Try it.

 

[Amani5576]

View attachment 173613

1. Homework Statement
A 3.0-kg block sits on top of a 5.0-kg block which is on a horizontal surface. The 5.0-kg block is pulled to the right with a force $\vec{F}$. The coefficient of static friction between all surfaces is 0.63 and the kinetic coefficient is 0.38. What is the minimum value of F needed to move the two blocks?

Homework Equations

$\|\vec{F}_{fs}\| = \mu_s \|\vec{N}\|$[/B]

The Attempt at a Solution

$\sum F_{bottom} = F_{external} - T - F_{fb}$
$F_{fb} = \mu_s(m_{bottom}+m_{top})g + m_{top}g$
$\sum F_{top} = T - F_{ft}$
$F_{ft} = \mu_sm_{top}g$
$\sum F = F_{external} - T - F_{fb} + T - F_{ft} = F_{external} - F_{fb} - F_{ft} = 0$
$F_{external} = F_{fb} + F_{ft} = \mu_sg((m_{bottom}+m_{top}) + 2m_{top} )$

After plugging in all the numbers:

$F_{external} = 86N$

86N is the right answer, but it seems odd to me because I essentially counted the friction between the two surfaces twice. Is it just the case that the frictional force between two surfaces moving across each other in opposite directions is twice that of when one of the surfaces isn't moving?

I think that when you were calculating Force of friction for the bottom you were trying to say:

$F_{fb} = \mu_s ((m_{bottom}+m_{top})g + m_{top}g)$