A book is pushed against a vertical wall at an angle

[purplepig]

Homework Statement

Albert is pushing a 1.8 kg book against a vertical wall at an angle of 30.0 degrees. If the coefficient of static friction is 0.47 and kinetic friction is 0.35, how hard does Albert need to push the book to allow it to slide down the wall at a constant speed?

Homework Equations

Fnet = ma
uK = Fk/Fn (coefficient of kinetic friction)
uK = Fs/Fn (coefficient of static friction)

The Attempt at a Solution

If there was no angle involved, I know that I'd first calculate Fk (force of kinetic friction) as it is equal to Fg (force of gravity). Then using the formula for the coefficient of kinetic friction, I would rearrange to solve for Fn (normal force) since Fn = Fa (force applied) which would get me the answer.

However, I have no idea how to factor in the angle given in this question to solve the problem since I have not done any forces questions involving angles.

Thank you in advance!

 

[haruspex]

Just resolve Albert's push into two forces, a horizontal and a vertical. Do you know how to do that? (Is he pushing down at 30 degrees or up at 30 degrees? Maybe it'll become clear as you solve the problem.)

 

[purplepig]

Thank you for your response haruspex.

The question doesn't state if he's pushing up or down at 30 degrees, so I will assume he is pushing up. By a horizontal and vertical forces, do you mean that the force applied (Fa) and the normal force (Fn) are the horizontal forces and the force of gravity (Fg) and Fk (kinetic friction) are the vertical forces? Because that's how I drew my free body diagram and I have the horizontal forces equalling each other and the vertical forces equalling each other.

 

[haruspex]

By a horizontal and vertical forces, do you mean that the force applied (Fa) and the normal force (Fn) are the horizontal forces and the force of gravity (Fg) and Fk (kinetic friction) are the vertical forces?

No, I meant that if Albert is pushing with force F_a at an angle θ above horizontal then you can resolve that into a horizontal force and a vertical force. I.e. it is equivalent to the sum of a horizontal and a vertical force. Do you not know how to write those in terms of F_a and θ?

 

[purplepig]

No, I meant that if Albert is pushing with force F_a at an angle θ above horizontal then you can resolve that into a horizontal force and a vertical force. I.e. it is equivalent to the sum of a horizontal and a vertical force. Do you not know how to write those in terms of F_a and θ?

Are you saying to break up F_a into x and y components so that F_ax = (F_a)(sin 30°) and F_ay = (F_a)(cos 30°)?

If not, then I'm afraid I don't know how to write it in terms of F_a and θ.

 

[haruspex]

Are you saying to break up F_a into x and y components so that F_ax = (F_a)(sin 30°) and F_ay = (F_a)(cos 30°)?

Yes. Having done that, what are your two equations relating all the forces?

 

[purplepig]

Yes. Having done that, what are your two equations relating all the forces?

Vertical: F_k + F_ay = F_g

Horizontal: F_n = F_ax

Are these the correct equations?

 

[haruspex]

Vertical: F_k + F_ay = F_g

Horizontal: F_n = F_ax

Are these the correct equations?

Yes. You know what F_ax and F_ay are in terms of F_a, and you know the relationship between F_k and F_n. Put all that together.
Note that if you assume Albert is pushing down at 30 degrees you get an invalid solution.