The question appears to be incomplete.
I assume that we are to find the coefficient of static friction, μ, between the desk and the book.
Refer to the diagram shown below.
m = the mass of the book
mg = the weight of the book (g = acceleration due to gravity)
N = the normal reaction, which is equal to
N = mg cos(12°)
R = the frictional force that opposes the sliding down of the book. It is
R = μN = μmg cos(12°)
F = the component of the weight acting down the incline. It is
F = mg sin(12°)
Because the book is in static equilibrium (by not sliding down the plane), therefore
F = R
mg sin(12°) = μmg cos(12°)
Therefore, the static coefficient of friction is
μ = tan(12) = 0.213
Answer: μ = 0.21 (nearest tenth)
I assume that we are to find the coefficient of static friction, μ, between the desk and the book.
Refer to the diagram shown below.
m = the mass of the book
mg = the weight of the book (g = acceleration due to gravity)
N = the normal reaction, which is equal to
N = mg cos(12°)
R = the frictional force that opposes the sliding down of the book. It is
R = μN = μmg cos(12°)
F = the component of the weight acting down the incline. It is
F = mg sin(12°)
Because the book is in static equilibrium (by not sliding down the plane), therefore
F = R
mg sin(12°) = μmg cos(12°)
Therefore, the static coefficient of friction is
μ = tan(12) = 0.213
Answer: μ = 0.21 (nearest tenth)