- #1
Feldoh
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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.
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.
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