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two blocks of unequal mass are connected by a string over a smooth pulley. one mass of mass 2m rests on the incline while the second mass of mass m is hung over a pulley over the incline if the coefficient of friction is μk, what angle of the incline allows the masses to move at a constant speed?|F1| = |F2|

F1= weight of mass hanging over the pulley

F2=Ʃforce acting on mass (2m) ie.. the x component of force along the direction of the incline + friction

mg=(2m)gsinθ+(μk)(2m)gcosθ

mg=(2mg)(sinθ+(μk)cosθ)

mg's cancel..

(1/2)=(sinθ+(μk)cosθ)

but since i know that.. μk=tanθ,

1/2=sinθ+tanθcosθ = sinθ+sinθ= 2sinθ

sinθ=(1/4)

θ=14.4775°

is the way i substituted tanθ for the coefficient valid? if not, why is it not valid?

F1= weight of mass hanging over the pulley

F2=Ʃforce acting on mass (2m) ie.. the x component of force along the direction of the incline + friction

mg=(2m)gsinθ+(μk)(2m)gcosθ

mg=(2mg)(sinθ+(μk)cosθ)

mg's cancel..

(1/2)=(sinθ+(μk)cosθ)

but since i know that.. μk=tanθ,

1/2=sinθ+tanθcosθ = sinθ+sinθ= 2sinθ

sinθ=(1/4)

θ=14.4775°

is the way i substituted tanθ for the coefficient valid? if not, why is it not valid?

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