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Please check my work (friction on an incline plane)

  • Thread starter iScience
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  • #1
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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?
 
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Answers and Replies

  • #2
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Why should the force from the hanging mass be equal to the horizontal component of the force on the other mass? I would assume the string goes along the incline.

mg=(2m)gsinθ+(μk)(2m)gcosθ
This looks like the mass m is hanging now, instead of the mass 2m. How did you get that formula?

The result will certainly depend on μk, a specific number cannot be right as result.

Please add explanations to your formulas, otherwise it is hard to understand what you did and what was wrong.
 
  • #3
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i fixed the m and 2m error.

i set mg equal to the rest of that because the problem wanted me to find the theta when the net force is zero.

so starting here,

mg=(2m)gsinθ+(μk)(2m)gcosθ

factoring mg..

mg=(2mg)(sinθ+(μk)cosθ)

mg's cancel..

(1/2)=(sinθ+(μk)cosθ)

now i'm supposed to solve for theta but i don't know how.. :(
 
  • #4
34,071
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You have to start earlier. Did you draw a sketch? Did you identify all forces acting on the system?
Which forces have to be in equilibrium?

(sinθ)^2+(cosθ)^2=1 might be handy somewhere in the solving process, but first you need correct equations for the forces.
 
  • #5
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  • #6
Doc Al
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here is the free body diagram with the arbitrary direction-sign specification (although that part wasn't needed)

http://i.imgur.com/BCmluRZ.png
You left out the tension force and the friction.

Hint: If the speed is constant, what must be the tension?
 
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  • #7
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i essentially set the two tensions equal to one another. where..

T1=mg
T2=(2m)gsinθ+(μk)(2m)gcosθ

does this depend on the direction of the blocks? since if the blocks move to the right..

mg=(2m)gsinθ+(μk)(2m)gcosθ

and if the blocks move to the left..

mg=(2m)gsinθ-(μk)(2m)gcosθ

so this is actually [itex]\hat{v}[/itex] dependent?
 
  • #8
Doc Al
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Solve it both ways and see if you get reasonable answers.
 

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