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Homework Help: Two masses on inclined plane with frictionless pulley

  1. Nov 18, 2013 #1
    1. The problem statement, all variables and given/known data

    A 1.00-kg aluminum block and a 4.00-kg copper block are connected by a light string over a frictionless pulley. The two blocks
    are allowed to move on a fixed steel block wedge (of angle 31.0°). The aluminum block is on the left side, on a flat plane. On the right is the copper block, sitting at a 31 degree angle. Picture is attached for clarification.
    Coefficient of static friction for aluminum on steel: .61 Copper on steel: .53
    Coefficient of kinetic friction for aluminum on steel: .47 Copper on steel: .36

    2. Relevant equations
    ƩFx=Fgx-tension-frictional force=(m2)(acceleration)
    Normal force=(massofcopperblock)(gravity)(cosθ)

    3. The attempt at a solution
    I was able to get the components of the force of gravity on the copper block: ≈33.6 N perpendicular to the plane, and ≈20.19 N parallel to the plane. Also, the normal force is 33.6 N. On the aluminum block, the normal and gravitational forces are both 9.8 N. From here, I am unsure how to incorporate the angle, the frictional forces, and how these will factor into solving for tension and acceleration of the bl

    I seriously appreciate any help. It really means a lot for you to take the time to help a stranger out, I thank you immensely.

    Attached Files:

  2. jcsd
  3. Nov 18, 2013 #2


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    Hi APPhysic, welcome to PF:smile:

    I see the equation for the cooper block, but miss the one for the aluminium block. What is its acceleration with respect to the acceleration of the cooper block? What forces act on it?

  4. Nov 18, 2013 #3

    Thanks for the welcome! I believe the aluminum block has tension, normal & gravitational force, and the force of friction acting on it. And won't the copper and aluminum block have the same acceleration or no?
  5. Nov 18, 2013 #4
    Yes, the accelerations would be the same in this system.
  6. Nov 19, 2013 #5


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    Yes, the acceleration of both blocks is the same parallel to the surfaces of the wedge. The normal force comes in by determining the force of friction. So what is the equation ƩF=ma for the aluminium block? .

  7. Nov 19, 2013 #6
    Well, for the x component, ƩFx=T-fk=m1a1

    And for y, ƩFy=n-m1g

    So then, can I say that T-μkm1g=m1a1 ?

    m1 in this case being the aluminum block
  8. Nov 19, 2013 #7


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    Calculate the friction forces, write up the equations for both blocks. Add them, the tension will cancel, and you can get the acceleration.

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