- #1
Lucky13ESP
- 5
- 0
[SOLVED] Kinetic and Static Friction
I've done so many of these but for some reason can't figure this one out. Any help iwll be greatly appreiciated. Here is the problem:
A book of mass M = 0.55 kg rests on a table where the coefficient of static friction μs = 0.48. What is the minimum horizontal force needed to move it?
So I draw my free body diagram and get this
N=m*g
P=μs*N
where N is normal force, P is horizontal force
P = 2.59 N This part is correct
Now for the part I can't solve.
A different force of F = 3 N now acts on the book at an angle of θ = 30 deg above the surface of the table. What is the acceleration of the book if the coefficient of kinetic friction μk = 0.25?
I've tried this part two different ways. First I checked for movement
Fy+mg = N
f = μs*N
Fx - f = ma
Fx-μs(Fy+mg)=ma
3cos30-.48(3sin30+.55*9.81)=ma
-.712 N = ma since friction force > applied force, the book doesn't move so acceleration is 0. However, they don't like that answer.
So now I assumed the book must already be in motion when the force is applied which is why they give me a kinetic friction coefficient.
Fx-μs(Fy+mg)=ma
3cos30-.48(3sin30+.55*9.81)=ma
.874 N = ma
a = 1.59 m/s^2
They don't like this answer either. Can anyone helpshed some light on what I"m doing wrong? Thanks so much in advance.
I've done so many of these but for some reason can't figure this one out. Any help iwll be greatly appreiciated. Here is the problem:
A book of mass M = 0.55 kg rests on a table where the coefficient of static friction μs = 0.48. What is the minimum horizontal force needed to move it?
So I draw my free body diagram and get this
N=m*g
P=μs*N
where N is normal force, P is horizontal force
P = 2.59 N This part is correct
Now for the part I can't solve.
A different force of F = 3 N now acts on the book at an angle of θ = 30 deg above the surface of the table. What is the acceleration of the book if the coefficient of kinetic friction μk = 0.25?
I've tried this part two different ways. First I checked for movement
Fy+mg = N
f = μs*N
Fx - f = ma
Fx-μs(Fy+mg)=ma
3cos30-.48(3sin30+.55*9.81)=ma
-.712 N = ma since friction force > applied force, the book doesn't move so acceleration is 0. However, they don't like that answer.
So now I assumed the book must already be in motion when the force is applied which is why they give me a kinetic friction coefficient.
Fx-μs(Fy+mg)=ma
3cos30-.48(3sin30+.55*9.81)=ma
.874 N = ma
a = 1.59 m/s^2
They don't like this answer either. Can anyone helpshed some light on what I"m doing wrong? Thanks so much in advance.