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That classic 3-blocks question

  1. May 21, 2005 #1
    OK 3 blocks attached by 3 strings. stick figure picture follows:

    M1---(T1)---M2---(T2)---M3---(T3)---(Gigantic Hand)

    T3 = 65.4N
    M1=11.2 kg
    M2=24 kg
    M3=30.8 kg

    Ok, I'm not going to post what the question asks because I'm not really concerned with getting the homework problem right. I just need help understanding the system. I have no idea what to place where. I know I have to put ma=ma somewhere, but it's saturday and I left my brain at school. :zzz:
     
  2. jcsd
  3. May 21, 2005 #2
    Hehe. Ok, I'm guessing the hand is pulling the whole system, which means the whole system has some net acceleration. This can be found directly from Newton's second law. The net acceleration of the body is the sum of the external forces acting on it divided by its mass. Clump the mass of the three together and divide by the force the hand exerts to find the net acceleration.

    The most useful thing in these types of problems is to know what tension really is. It is a force exerted on a string by objects on either end. The amount of force that BOTH ends share will be the tension in the string, so for example,

    If a string has a mass pulling it to the left at 10N and another mass pulling it on the right at 5N, what is the tension in the string?

    Both ends of the string have a minimum of 5N acting on them, so that is the tension. The other 5N from the left will result in a net force on the entire system.

    Onto your problem
    You are given the tension in the third rope, and the three masses. If you read above, you can figure out that the force applied by the hand minus the force from the other three blocks is 65.4N

    The net acceleration is

    [tex] \vec{a} = \frac{\vec{F}}{m_{net}} = \frac{65.4}{66} = 0.99 \ or \ about \ 1m/s^2[/tex]

    The force on M1 by Newton's 2nd Law is

    [tex] \vec{F}_{m1} = m\vec{a} = (11.2kg)(1m/s^2) = 11.2N [/tex]

    Where we rounded the acceleration. The only force acting on M1 is the force pulling it to the right. There are no forces to its left.

    Force on mass 2 by Newton's second law is
    edit - fixed
    [tex] T_{2} - T_{1} = ma = 24N \ with \ T_1 = 11.2N [/tex]


    The net force on block 2 is the sum of the hand pulling it to the right (the net force on the system) and M1 pulling it to the left.


    Can you do the third mass?
     
    Last edited: May 21, 2005
  4. May 21, 2005 #3
    So I do Fm3=(m3*a)-Fm2 right? then I get 18N.

    But 18N + 11.2N + 12.8N doesn't equal 65.4? Is that ok?

    And then we find the tension of the string in the middle by subtracting one force from the other? For example, what would the tension be between M1 and M2?

    Thanks for helping on a saturday! :!!)
     
  5. May 21, 2005 #4
    Remember, M3 is dragging BOTH M1 and M2.
     
  6. May 21, 2005 #5
    I'm still confused.

    Tension is just the net force between the two bodies right? So what's the relationship between the forces acting on the bodies and the forces acting between the bodies?

    I think what I dont understand is what the numbers mean. If those bodies are accelerating at 1 m/s^2, then why do I have to subtract those forces of M1 and M2? It's accelerating, it's moving regardless of how many blocks are behind it.
     
  7. May 21, 2005 #6

    OlderDan

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    While I expect you know what you meant to say, there are a couple of things you did say that you need to look at again.

    The only way this could happen is if there is a tension gradient in the string, and that could only happen if the string had mass, which is not the case in this problem. A massless string experiencing different forces at both ends would have infinite acceleration. In this problem, and in your calculations, the tension in each string is constant and serves as a mechanism for two masses to exert equal and opposite forces upon one another.

    Your statement suggests that 65.4N is the difference between the force applied by the hand and forces to the left, which fits with your description of tension, but that description is not correct. I don't think you want to talk about 65.4N as being the difference between the force of the hand and other forces. 65.4 N is one force (or more correctly one action-reaction pair of forces), the force applied by the hand (and the reaction force T3), and that is all. If the "object" experiencing the force is the group of three masses, then your statement is correct, but only because the the 65.4N is the only external force acting, so the "force from the other three blocks" means external forces acting on those blocks which is zero. If other forces (friction for example) were acting on the blocks, and the blocks were considered as one object, the 65.4N would still be 65.4N and T3, but the net external force would be reduced and the acceleration would be less. In this frictionless problem, if you are looking at M3, T3 minus the force from the other blocks is not 65.4N; T3 is 65.4N and T3 - T2 is the net force acting on block M3.

    I think what the OP needs to see explicitly is something that is contained in your analysis, but not so obviously stated up front, which is that each block is acted upon by one or two forces and that the blocks share a common acceleration

    T1 = M1 a
    T2 - T1 = M2 a
    T3 - T2 = M3 a
    (T3 = force applied by gigantic hand = 65.4N)

    Add the first three equations to arrive at your starting point.

    T3 = (M1 + M2 + M3) a

    a = T3/(M1 + M2 + M3)

    Then step through the first three equations to find the tensions as you have outlined.

    If there were friction forces f1, f2 and f3 involved, the equations would become

    T1 - f1 = M1 a
    T2 - T1 - f2 = M2 a
    T3 - T2 - f3 = M3 a
    (T3 = force applied by gigantic hand = 65.4N)

    EDIT

    This is not correct

    [tex] \vec{F}_{m2} = m\vec{a} - \vec{F_{m1}} = (24kg)(1m/s^2) - (11.2N) = 12.8N[/tex]

    It should be

    [tex] T2 - T1 = (24kg)(1m/s^2)[/tex]

    [tex] T2 = (24kg)(1m/s^2) + T1 [/tex]
     
    Last edited: May 21, 2005
  8. May 21, 2005 #7
    Ok, the net acceleration of each block as we established earlier is 1m/s^2.

    The net force on each block is therefore the block's mass multiplied by 1m/s^2.

    Net force on M1: 11.2N
    Net force on M2: 24N
    Net force on M3: 30.8N

    #1 is easy since the force is just in one direction. The force on it is 11.2N, and this is teh tension in the string that is pulling it.

    #2, The net force is 24N, but it has two forces pulling it, the tension from M1, and the pull from the hand. The sum of these is 24N.

    [tex] F_{net} = ma = 24N [/tex]

    The force experienced by M2 is due to the tension in the rope between it and M3.
    [tex] T_{m32} - T_{m12} = ma = 24N [/tex]

    [itex] T_{m12} = 11.2N [/itex], so [itex] T_{m3} [/itex] (tension between M2 and M3) is 35.2N

    edit: I've made a mistake as OlderDan pointed out, so this information might not be correct.
     
    Last edited: May 21, 2005
  9. May 21, 2005 #8

    OlderDan

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    See my edit in my post about T2. I gotta go- I'm sure you will sort this out.
     
  10. May 21, 2005 #9
    Okay the mistake I made is what I thought it was initially, I was actually in the process of amending it when you posted. My post before this is the correct solution.

    I apologize for my mistake.
     
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