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Help with Pulley Problem

  1. Jan 24, 2007 #1
    1. The problem statement, all variables and given/known data

    This is a problem off of http://www.physics247.com/physics-homework-help/pulley.php It is question number 6 but with an additional part and slight modification by my teacher. I will copy and paste it here but it may be best to view the link for the diagram.

    Two blocks, one 0.8 kg and the other 2.0 kg are connected by a massless string over a frictionless pulley. The coefficient of kinetic friction is 0.14, and the downward ramp angle is 30 degrees (NOT 60 degrees as shown in the diagram).

    a) Determine the acceleration of the blocks.

    b) Calculate the tension of the string.

    c) If the string broke, for what minimum value of the coefficient of static friction would the 2kg block not begin to slide?

    m1 = 0.8kg
    m2 = 2.0kg
    uk = 0.14

    2. Relevant equations

    Fnet = ma
    Ff = umg
    Fg = mg
    Fn = Fg

    3. The attempt at a solution

    I did a FBD and came up with this equation
    Fnet = Fgx2.0 - Ff2.8 + Fg0.8 + Fgy2.0 - Fn0.8 I'm unsure if this is correct
    Which meant 2.8a = 2gsin60 - u2.8g + 0.8g + 2g - 0.8g
    But now i'm not sure what to do because my forces in the y direction dont equal zero.
  2. jcsd
  3. Jan 24, 2007 #2
    Well you have two unknowns the acceleration [itex] \textbf{a} [/itex] of the blocks and the tension force [itex] \textbf{T} [/itex], both can be considered 1-dimensional, hence 2 unknowns. What's the equation of motion for each of the blocks?
  4. Jan 24, 2007 #3
    well for the 0.80kg mass, we have:

    Fnet = Fg - Fn = 0 in the y direction
    Fnet = Fgx2.0 - Ff
    (0.8)a = (2)(9.8)(sin60) - (0.14)(0.8)(9.8) Or so I believe...

    for the 2.0kg mass we have:

    Fnet = Fgx - Ff + Fgy - Fn
    (2.0)a = (2)(9.8)(sin60) - (0.14)(2.0)(9.8)

    I have a feeling im doing something wrong though.
  5. Jan 24, 2007 #4
    You're first equation with the y-direction (vertical to the surface) is correct, you can use that to determine the force of friction on block 1 (0.80 kg). But you have to use the tension force T. The only forces in te x-direction on block 1 is the force of friction ("backwards") and the force of tension (oppisite direction of friction), the normal force and the force of gravity cancel in the y-direction (as you also have in your first equation).

    In you're second equation you seem to mix up the forces in x- and y-direction. You need to consider them seperatly. The forces in the y-direction (vertical to the surface) cancel each other out again. Forces on the x-direction are the component of the gravity in the direction of motion, the force of friction in the oppisite direction of the motion and the force of tension.
    To make it a little clear, then let m1 is the mass of the first block (0.80kg) and m2 be the mass of block 2 (2.0 kg), and [itex] \theta [/itex] be the angle 60 degrees, the equation of motion for the two block in the x-direction could be expressed as

    [tex] m_1\ddot{x} = T - f_1 [/tex]

    [tex]m_2\ddot{x} = F_{g_x} - f_2 - T[/tex]

    where [itex] f_1 [/itex] is the force of friction of the first block and [itex]f_2[/itex] is on te second block. Try figuring how to find the forces and then you can solve the system for the unknowns.
    Last edited: Jan 24, 2007
  6. Jan 24, 2007 #5
    I'm confused, if m1a = T - Ff then
    0.8a = T - (0.14)(0.8)(9.8) in the x direction.

    This means that I would need the tension to solve for a and vice versa.
  7. Jan 24, 2007 #6
    Indeed, but you have there 2 equations with 2 unknowns. The unknowns being the acceleration [itex] \ddot{x} [/itex] and [itex]T[/itex]. You can solve for both unknowns. Try adding the 2 equations and you should be free from the tension force, you can then solve for the acceleration.
    Besides calculating the tension force is problem b).
  8. Jan 24, 2007 #7
    Ah I see now, so then for part a)

    a(m1 + m2) = Fgx - Ff1 - Ff2
    a = [ (2)(9.8)(sin60) - (0.14)(9.8)(0.8 - 2.0) ] / (0.8 + 2.0)
    = 6.65 m/s/s

    It is sin60 right? I'm not too sure if thats it, for some reason I just know to always use sin.

    For part b) then

    a(m1 - m2) = 2T + Ff2 - Ff1 - Fgx Would the 2T be cluttered as the total tension or would I need to divide by two in the end? This is basically solved now...

    for part c) ... Would I set umg = Fgx?
    Then u = Fgx / mg
    = sin60?
  9. Jan 25, 2007 #8
    a) Yes sin60 is correct for Fgx. But your expression for the friction force f2 doesn't look correct, you know that [itex] f = \mu N[/itex] where N is the normal force from the surface. What is the normal force on block 2?

    b) When you have the acceleration a, you can just use one of the equations to solve for T.

    c) It's correct that you can solve f2 = Fgx (the sum of the forces should be 0 on horizontal direction if the block is stay on the surface without moving). But again you use the same expression for f2 as in a), which isn't correct, because what is the normal force from the surface on block 2?
    Last edited: Jan 25, 2007
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