## Overcoming Static Friction

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?
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 Recognitions: Homework Help Science Advisor 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.
 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 This answer is wrong.

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