Coefficient of Friction when Normal Force is Reduced [HS]

[Liam C]

Homework Statement

A 5.0kg block of wood sits on a table with a string running over a pulley suspending a mass. The largest mass that can be suspended without moving the block of wood is 0.75kg.
a) Calculate the coefficient of friction in this case.
b) What type of coefficient of friction has been calculated in this case?

Homework Equations

Fg = mg
FA = ma
FN = ?
Us = coefficient of static friction
Ff = force of friction
Us = Ff / FN

The Attempt at a Solution

m1 = 5.0kg
m2 = 0.75kg
a) Fg = m1 x g
Fg = (5)(9.8)
Fg = -49N

FA = m2 x a
FA = (0.75)(9.8)
FA = 7.35N

FN = 49 - 7.35
FN = 41.65

Ff = FA
Us = Ff / FN
Us = 7.35 / 41.65
Us = 0.18
b) static
According to the answer key, the correct coefficient is 0.15. I suspect I'm calculating the normal force incorrectly, but I was never taught how to do it in this case so I am clueless. How does this applied upward force from the suspended weight affect the force of the normal?

 

[haruspex]

a) Fg = m1 x g
Fg = (5)(9.8)
Fg = -49N

Get into the habit of keeping everything algebraic, not plugging in numbers until the final step. It has many advantages.

FN = 49 - 7.35

In what direction(s) are those forces acting on the 5kg block?

 

[Liam C]

Get into the habit of keeping everything algebraic, not plugging in numbers until the final step. It has many advantages.

In what direction(s) are those forces acting on the 5kg block?

Okay, I will do that from now on.
I think both FN and FA are going up. It doesn't really make sense to subtract one of them from the other when I think about that.

 

[Liam C]

Get into the habit of keeping everything algebraic, not plugging in numbers until the final step. It has many advantages.

In what direction(s) are those forces acting on the 5kg block?

Aha I figured it out! I was under the impression that the normal force would be reduced by the extra weight, but in reality it does not change.
Us = Ff / FN
Us = 7.35 / 49
Us = 0.15
My question now is: why does the force of the normal not change? There is obviously an added upwards force on the system of the block. However, in order for the normal to stay the same there must also be an equal added force downwards. I know that the force of gravity is pulling the weight on the pulley down, but that weight is a separate system.

 

[haruspex]

why does the force of the normal not change? There is obviously an added upwards force on the system of the block

You did not post a diagram, but the only way to make sense of the question is that the string runs horizontally from the 5kg block to the pulley. If it were to run vertically there would be no reason for friction to be involved, and if it were at any other angle it would need to be stated.

 

[Liam C]

You did not post a diagram, but the only way to make sense of the question is that the string runs horizontally from the 5kg block to the pulley. If it were to run vertically there would be no reason for friction to be involved, and if it were at any other angle it would need to be stated.

Interesting. Thank you for the help.