How Does Force Angle Affect Book Acceleration with Friction?

[hri12]

Homework Statement

A book of mass M = .55 kg rests on a table where the coefficient of static friction us = .45. A force, F = 4 N acts on the book at an angle of 15 degrees above the surface of the table. What is the acceleration of the book if the coefficient of kinetic friction uk = .23?

Homework Equations

ΣFx = ma
ΣFy = ma

Ff = Fn(uk)

The Attempt at a Solution

I drew free body diagrams.

ΣFx = Fx - Ff
Fx - Ff = ma(x)
Fx - Fn(uk) = ma(x)

ΣFy = ma
Fn - Fy - mg = ma(y)

I'm not sure where to go from here. I know that I need to use Pythagoras theorem when I have the values of a(x) and a(y) but I don't know how to get that far.

Please help, I've been working on this for 2+ hours.

 

[SammyS]

Homework Statement

A book of mass M = .55 kg rests on a table where the coefficient of static friction us = .45. A force, F = 4 N acts on the book at an angle of 15 degrees above the surface of the table. What is the acceleration of the book if the coefficient of kinetic friction uk = .23?

Homework Equations

ΣFx = ma
ΣFy = ma

Ff = Fn(uk)

The Attempt at a Solution

I drew free body diagrams.

ΣFx = Fx - Ff
Fx - Ff = ma(x)
Fx - Fn(uk) = ma(x)

ΣFy = ma
Fn - Fy - mg = ma(y)

I'm not sure where to go from here. I know that I need to use Pythagoras theorem when I have the values of a(x) and a(y) but I don't know how to get that far.

Please help, I've been working on this for 2+ hours.

What do you suppose the value of a_y is ?

 

[hri12]

I know it's not zero because:
1. I already tried with zero and got the wrong answer.
2. It's being pulled diagonally upwards, so both a(x) and a(y) should have a nonzero value.

 

[hri12]

Edit: the answer should be 5.20 m/s^2

 

[hri12]

Never mind. Got it. Thanks everyone!

SOLUTION
ΣFx = Fx - Ff
Fx - Ff = ma(x)
Fx - Fn(uk) = ma(x)

ΣFy = ma
Fn + Fy - mg = ma(y)

a(y) = 0 (not sure why...)
Fn + Fy - mg = 0
Fn = mg - Fy

Fx - (mg - Fy)(uk) = ma(x)
a(x) = (Fx - (mg - Fy)(uk) )/m
a(x) = 5.2

((5.2)^2 + (0)^2)^(1/2) = 5.2

 

[SammyS]

Never mind. Got it. Thanks everyone!

Excellent !

SOLUTION
ΣFx = Fx - Ff
Fx - Ff = ma(x)
Fx - Fn(uk) = ma(x)

ΣFy = ma
Fn + Fy - mg = ma(y)

a(y) = 0 (not sure why...)

Then this is worth looking at further.

a_y can possibly be positive, but it can't be negative. Right ?

If a_y > 0, then the normal force will be zero. Right? Then F_y - mg = a_y > 0 . Is F_y - mg > 0 ?

If that's not the case, then your solution with a_y = 0 is correct.

 

[hri12]

Why will the normal force be zero? If anything the normal force will be Fn = ma(y) + mg -Fy...

I still don't understand.

 

[SammyS]

Why will the normal force be zero? If anything the normal force will be Fn = ma(y) + mg -Fy...

I still don't understand.

I may not have stated that too clearly.

Why is there any normal force? It's there because the surface of the table keeps the book from accelerating downward through the surface. So the surface exerts whatever force is needed for this to be the case.

The normal force cannot be negative. If it were, it would be pulling down on the book.

If a_y is positive, then the book is no longer on the surface of the table thus no normal force

 

[hri12]

So basically because the book isn't moving in the y direction? That makes sense.

Thank you so much btw for all of your help. :)