N, which is not the case.Find Coefficient of Static Friction for 6.65 kg Box

[strugglin-physics]

A force of 21.2 N is required to start a 6.65 kg box moving across a horizontal concrete floor. What is the coefficient of static friction between the box and the floor?

I've drawn my free body diagram and found that the y motion is N^F,B-W^E,B=0 I also know that the Force is larger than the friction
21.2 N - friction = 6.65a acceleration is not zero in this case right cause if it were zero than the friction would be 21.2

 

[vsage]

Speaking in components parallel to the floor, u*Fn + Fapplied = 0 for a box to start moving at a constant (but slow) velocity. Or..

u * m * g <= -21.2N in this case because the force applied needs to be a teeny bit larger than static friction where u is the coefficient of static/kinetic friction.

 

[TenaliRaman]

really don't understand ur problem or possibly the notation u have used
N^F,B-W^E,B=0 ??

Anyways ...
F <= mu_s*N
and this resists ur applied force
consider the equilibrium and finish ur work

-- AI

 

[strugglin-physics]

Excellent, thanks for the help. What I was missing was the propper equation for the problem. This is the other half of the problem and I know it uses a slightly different equation but I don't know what that is.

If the 21.2 N force continues, the box accelerates at 0.400 m/s2. What is the coefficient of kinetic friction?

(sorry about the notation thing I should've used lower case letters instead of coefficents.)

 

[vsage]

same equation except sum of the forces isn't 0 anymore:

Fapplied + Ffriction = Fresultant

21.2N + u*Fnormal = m*0.4

careful with the sign on Fnormal.

 

[strugglin-physics]

Isn't Fnormal just 21.2N*-9.8?

 

[strugglin-physics]

R U still around? =(

 

[e(ho0n3]

If the box isn't moving horizontally, then the normal force N is equal to the weight of the box mg.