sled weighing 80 N rests on a plane inclined at 20° to the horizontal ?
Between the sled and the plane the coefficient of static friction is 0.26, and the coefficient of kinetic friction is 0.14.
(a) What is the minimum magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane?
(b) What is the minimum magnitude F that will start the sled moving up the plane?
(c) What value of F is required to move the block up the plane at constant velocity?
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(a)
static friction = 0.26(80cos20) = 19.5 N
component of weight parallel to the incline = 80sin20 = 27.4 N
F = 27.4 - 19.5 = 7.82 N
(b)
F = 27.4 - 0.14(80cos20) = 16.8 N
(c)
constant velocity means zero acceleration and hence zero resultant force.
therefore, F = 16.8 N
you only have A right
reaction stress = 80*cos 20 Frictional stress = Rf * 0.25 till there is any stream Gravitatioal stress in the path of the incline = 80 * sin 20 internet min stress = F = Gf - Ff