Find the force to push a book sideways on a table

[y90x]

Homework Statement

Place a book flat on a table and press down on it with your hand. Now suppose the hand-book and table- book values of are 0.50 and 0.40, respectively; the book's mass is 0.90 kg and your downward push on it is 10 N. How much horizontal force is needed to keep the book moving at a constant speed if your hand is stationary with respect to the table?

Homework Equations

F=ma

The Attempt at a Solution

The given values for friction kinda threw me off but this is my attempt to it
Fa= 10 N (force applied)
Fg=mg = .90•9.8 = 8.82 N
Fn = mg+ Fa = 18.82 N
Ff = Uk•Fn = .40•18.82 = 7.528

Since it’s going at a Constant speed , it means there’s no acceleration and all the forces are equal to Each other .
So the horizontal force needed wouldn’t it be the same as the Friction force (Ff) .
The correct answer should be 12.5 N

Where did I go wrong ?

 

[haruspex]

Homework Statement

Place a book flat on a table and press down on it with your hand. Now suppose the hand-book and table- book values of are 0.50 and 0.40, respectively; the book's mass is 0.90 kg and your downward push on it is 10 N. How much horizontal force is needed to keep the book moving at a constant speed if your hand is stationary with respect to the table?

Homework Equations

F=ma

The Attempt at a Solution

The given values for friction kinda threw me off but this is my attempt to it
Fa= 10 N (force applied)
Fg=mg = .90•9.8 = 8.82 N
Fn = mg+ Fa = 18.82 N
Ff = Uk•Fn = .40•18.82 = 7.528

Since it’s going at a Constant speed , it means there’s no acceleration and all the forces are equal to Each other .
So the horizontal force needed wouldn’t it be the same as the Friction force (Ff) .
The correct answer should be 12.5 N

Where did I go wrong ?

pay attention to the statement that your hand does not move. Something else is pushing the book sideways.

 

[y90x]

pay attention to the statement that your hand does not move. Something else is pushing the book sideways.

Oh, so I just include the friction of the hand
Ff= Uk(Fn) =.5•10N = 5
Then add both friction forces together
Okay thanks !