# Homework Help: 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