# Overcoming Static Friction

1. Aug 7, 2012

### HHenderson90

1. The problem statement, all variables and given/known data
A child pulls a string attached to a block of mass m = 1.5 kg at an angle θ = 35.5° with an increasing amount of force. When she pulls with a force F = 7.39 N the block just begins to move. What is the coefficient of static friction between the block and floor?

2. Relevant equations
F_Static=U_static*F_normal

3. The attempt at a solution
I first simply called the force she pulls with F_Static, and divided it by the Normal Force (M*G)
This was wrong.
I then thought that maybe only the force in the x-direction would be considered the static force, I found that by FCos(theta). I then divided this by the force normal, and again no luck.
I then thought the force applied by the child in the y direction might be subtracted from the force normal? This didn't work either.
What do I do?

2. Aug 7, 2012

### Dick

Your second and third approaches combined sound right. The force in the y direction should subtract from the normal force and the force in x direction is the force contributing to making the block slide. It would be a good idea to show your work and the numbers you got.

3. Aug 8, 2012

### HHenderson90

I ended up getting it correct by combing my second and third approach. Found the x-component of the force applied by 7.39Cos(35.50 which was 6.0163N and I called this F_Static. I found the F_N by M*A which was 15*9.81=14.715
Subtracting the y-component of the applied force (7.39sin(35.5)) I got the F_N as 10.4336.
F_Static= U_s*F_N
Dividing F_Static by F_N I got the answer of .577.

There is a second part to the problem that I am having trouble with now.
1. The problem statement, all variables and given/known data
Once the block starts to move, it accelerates at a rate of 1.13 m/s2. What is the coefficient of kinetic friction between the block and floor?

2. Relevant equations
Fnet=M*A
F_Kinetic=Uk*Fn

3. The attempt at a solution

I think my issue here is what constitutes the Fnet.
I found Fnet by multiplying M*A 1.5*1.13=1.695N
I called this the force applied, considering that the x component would be equal to the force of static friction.
Finding the x component- 1.695cos(35.5)=1.38N
I then took the Fn from before, and subtracted the y component from this force
14.715-1.695sin(35.5)=13.73N
I then used the Fk=UkFn
1.38=Uk13.73=.100

4. Aug 8, 2012

### Dick

Your applied x and y forces and normal force are exactly the same as in the first problem. What's different is that the horizontal force Fx and the force of friction Fk combine to accelerate the block. So (Fx-Fk)=m*a. Use that to find Fk.