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Block-Pulley Tension Problem

  1. May 20, 2009 #1
    1. The problem statement, all variables and given/known data
    A 9 kg block on a frictionless table is connected by a light massless cord through a frictionless pulley to a 3 kg block hanging over the edge of the table. The blocks are free to accelerate. What is the tension in the cord?

    2. Relevant equations

    sum of forces = F tension + F gravity + F normal = 0
    The blocks are connected by massless string so we consider them as one object.
    F = ma

    3. The attempt at a solution

    Since the blocks are not moving they are in static equilibrium. The net forces on the block unit = 0

    I know I need to glean something from the fact that the blocks are free to accelerate but I am not sure what it is.

    Also, if the table is frictionless what is opposing the force of tension on the cord to keep it from sliding off the table.

    I know this is a basic problem but I can not work it!

    If someone could please push me in the right direction I would appreciate it very much.
  2. jcsd
  3. May 20, 2009 #2
    The block's aren't moving, are you sure? What the question is telling you is the setup, then it's effectively saying, "We're now letting them go" after holding them in place, and asking you to describe what happens (either in acceleration & tension).

    Tip: Try doing the forces on 1 of them, then the forces on the other, and comparing them (the tension is the same throughout all of the string remember)
  4. May 20, 2009 #3
    So, since the tension is equal in the string if I solve for force of tension for either block I will get the same answer. Correct?

    So, if this is a dynamics problem, I use net force is equal to m times a

    All forces acting on block on table are:

    Force tension + force of gravity using weight of both blocks and + normal force using only the mass of the block on the table.

    I get F of tension = 12kg times 9.8 - 9 times 9kg = 29.4N

    BUT I know I am missing something because the correct answer is 22.0 N

    Can you direct me further, please?
  5. May 20, 2009 #4
    Remember here, the only force 'powering' this movement is the downward force of the 3kg block! Not all 12 of it, so although as you say you can treat is as one large 12kg one, i much prefer doing it seperateley.

    For example; the forces on the 3kg block are Gravity (down) and Tension (up), as you said, equalling mass * acceleration, try that for both of them (you'll need to use simultaneous equations)

    Edit: The 'normal force' has no place in this question, as friction is not involved (smooth surface & pulley), which i suppose is in itself an answer to your former question;

    "Also, if the table is frictionless what is opposing the force of tension on the cord to keep it from sliding off the table. " nothing.
    Last edited: May 20, 2009
  6. May 20, 2009 #5
    Solving for tension for each block which must be equal since both sides of a rope etc., I know has the same tension:

    Force on block on table:

    f tension - f gravity = ma = 9a f tension = f gravity + 9a

    Force on block over edge:

    f tension - f gravity = ma = 3a f tension = f gravity + 3a

    then if I set these equal to each other:

    9 times 9.8 + 9a = 3 times 9.8 + 3a

    Doesn't this just prove the acceleration is 9.8 downward???? which is simply gravity!~!
  7. May 20, 2009 #6
    No it doesn't, the point is the acceleration is NOT 9.8, it can't be.

    I'll go through a bit of the calculation so you can see how it works.

    It's not in equilibirum, we agree so the force on it is equal to m*a, what are the forces on it? (The 3kg one), there's no horizontal forces, only down (gravity) and tension in the rope (upwards), so let's just put that algaebraically.

    3g - T = 3a

    Now let's look at the 9kg one, what are the forces on it? There's no friction here so there's only 1 force horizontally --->> (that direction) which is the tension of the string.

    T = 9a

    Do you see what i've created doing that? The tension of the string is the same at every point, and the acceleration of the blocks are equal, you can't assume anything here in terms of gravity.

    Can you take it from here or are you still a bit unsure?

    Edit: Sorry, re-reading your post you were on the right lines, but remember with the block over the edge, it's not Tension - Gravity = ma it's Gravity - tension, gravity > tension, otherwise it would be pulled up!
  8. May 20, 2009 #7
    Thanks so much for your help! I know I am being dense but this one just confused me.

    So, we have:

    3g - T = 3a and T = 9a

    THerefore, 3g - 9a = 3a

    12 a = 3 g

    a = 2.45 m/s^2

    Now we use the weight of block on table 9 times 2.45 = 22.05 N

    We could not use the weight of other block because we want tension between the two blocks. Is that right? NOT tension of any string that might be below hanging block, etc
  9. May 20, 2009 #8
    The tension of the string between the two blocks, correct, that calculation is correct as well.

    I think it always helps to think about these problems (and any relatively simple mechanic problems) always in terms of horizontal forces & vertical forces separately, the answers seem to fall out if you do that. (Also in the case of this 1 block and then the other)
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