Coefficient of Kinetic Friction...

Feldoh

This really isn't a homework question -- but it does involve my homework. Say you've got a box sliding down an incline of x degrees at a constant speed, I somehow got that that coefficient of kinetic friction is equal to tan(x). Will this always hold true for objects moving at a constant speed on an incline in two dimensions?

Basically I solved for the normal force:
[tex]F_{Net}=0[/tex]
[tex]F_N-F_g_y=0[/tex]
[tex]F_N=mgcos(x)[/tex]

Then for the coefficient of friction:
[tex]F_{Net} = 0[/tex]
[tex]F_f-F_g_x = 0[/tex]
[tex]F_f = F_g_x[/tex]
[tex]\mu_kF_n = F_g_x[/tex]
[tex]\mu_k = (F_g_x)/(F_N)[/tex]
[tex]\mu_k = mgsin(x)/mgcos(x)[/tex]
[tex]\mu_k = tan(x)[/tex]

Where [tex]F_{g_x}[/tex] is the horizontal component of the force of the gravitational field, and [tex]F_{g_y}[/tex] is the vertical component.

I was just wondering, it's sort of situational but it is a shortcut none-the-less if it does actually work.

G01

Yes, this equation should hold true for situations like you have described above. Of course if you have more complicated scenarios, such as one with acceleration, this won't hold.

Feldoh

Awesome, thanks for the fast reply.

G01

No problem.