Wb = m*g = 13kg * 9.8N/kg = 127.4 N. =
Wt. of box.
Fb = 127.4N @ 33 Deg. = Foce of box.
Fp = 127.4*sin33 = 69.4 N. = Force parallel to incline.
Fv = 127.4*cos33 = 106.8 N. = Force perpendicular to incline.
1. Fn = Fp - Fk = ma
69.4 - Fk = 13 * 0.20 = 2.6
-Fk = 2.6 -69.4 = -72
Fk = 72 N = Force of kinetic friction.
2. u*Fv = Fk
127.4u = 72
u = 72 / 127.4 = 0.565 = Coefficient of
kinetic friction.
A 13.0 kg box is released on a 33 degree incline and accelerates down the incline at 0.20 m/s^2.
1. Find the friction force impeding its notion?
2. What is the coefficient of kinetic friction?
*Are we dealing with Kinetic or Static Friction...this will effect coefficient correct?
*The equation coefficient*m*g*cos angle DOES NOT give you the friction force
3 answers
Correction:
-Fk = 2.6 - 69.4 = -66.8Fk = 66.8 N.
2. u*Fv = Fk
127.4u = 66.8
u = 66.8 / 127.4 = 0.524 = uk.
-Fk = 2.6 - 69.4 = -66.8Fk = 66.8 N.
2. u*Fv = Fk
127.4u = 66.8
u = 66.8 / 127.4 = 0.524 = uk.
OOPs!
2. u*Fv = Fk
1o6.8u = 66.8
u = 66.8 / 106.8 = 0.625.
NOTE: Fv = 106.8 N.
2. u*Fv = Fk
1o6.8u = 66.8
u = 66.8 / 106.8 = 0.625.
NOTE: Fv = 106.8 N.