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Homework Help: 'Motion Near the Earth

  1. Apr 13, 2007 #1
    1. The problem statement, all variables and given/known data

    In the diagram below, a force of 150 N is applied at 27* above the horizontal on a block with a mass of 35 Kg. The block is connected to another block of mass 5.0 Kg by means of a string passing over a pulley.The coefficient of kinetic friction between the 35 Kg mass and the surface is 0.18.

    Find: a) The acceleration of the system
    b) The tension in the string

    This is where I am having issues. I've been out of school for the past few days due to illness and now today an assignment that my partner was so kind to give me last night is due today. This is the only tension question on the assignment and I've never seen or done a tension question before, so now I'm asking for help in understanding how to do it. I looked at the formulas at the top of this sub-forum and didn't see anything that would help, so I would like to know if anyone could assist me in learning tension.

    I've already looked at the book, but our teacher has only assigned 10 questions from the book during this semester so far, so the methods the book uses is very confusing and there is nothing relevant to this question that I've found. Any help would be greatly appreciated! :)

    - M

    P.S. I'm aware of the improper title, I hit enter before I could adjust it.

    Attached Files:

    Last edited: Apr 13, 2007
  2. jcsd
  3. Apr 13, 2007 #2
    NP. For simple start, Tension in these problems is a stress in rope or cable or caused by another body trying to pull it apart.

    It is somewhat like spring force except if you try to push rope towards center, it deforms--it cannot store compressive energy like spring, and we generally assume the material is ideal so that it doesn't stretch, so more properly it conveys forces and doesn't add or subtract.

    Easiest case is man hanging from end of rope which is tied to big eye bolt in ceiling rafter. T=m*g The force is same at both ends of rope so in fact physics of situation not different from man hanging from eyebolt.
  4. Apr 13, 2007 #3
    Alright, so could I then say that in this case T=(m)*(Fapplied*cos27)? and if that is true how can the block have an acceleration if the forces are balanced?

    Thanks for the help, it's greatly appreciated.

    And to find acceleration could I not somehow subtract the force of the 5 Kg block from the 35 Kg block and then just solve as a basic kinematics question??
    Last edited: Apr 13, 2007
  5. Apr 13, 2007 #4
    well the diagram is still pending, so I'm sort of guessing at what it looks like,
    generally the idea is to develop at least 2 eqns (if there are 2 unknowns) involving T and a. In other words there likely is a term like
    T-5kg*g=ma where we are looking at the force sum on the 5kg block
    Last edited: Apr 13, 2007
  6. Apr 13, 2007 #5
    I should have made clearer the notion that T can be the result of multiple forces like in a tug of war, so for two body problems like this one, it is an unknown.
  7. Apr 13, 2007 #6
  8. Apr 13, 2007 #7
    thanks, and that is what I envisioned, so on big block, 3 forces

    Tension and frictional force directed to left and Fappl to right.
    On small block tension and weight. Try to set up two eqns using Newtons second and solve for 2 unknowns
  9. Apr 13, 2007 #8
    Alright, I'm still rather confused about Tension. I understand how to set up the question and solve. Is the tension of the string on the small block the same as that on the big block? and if it's that is true could I not just use T= m*g as you stated above to find this tension?

    Then weight is also W=m*g, so could I not say the forces on this block are equal to 2(m*g)? and then the equation for the forces on the big block being:

    (150cos27)-[(2(m*g)) + (0.18)(Fnormal)]

    and then from there determine the acceleration from forces?
    Last edited: Apr 13, 2007
  10. Apr 13, 2007 #9
    yes and no, the tension is the same throughout. what you can say is what I did in post 4. Or explicitly, what if the small block is being accelerated vertically then, T=m(g-a) That make sense? If it were static then it would be just mg.
  11. Apr 13, 2007 #10
    I see, I understand now.

    Thank you so much!
  12. Apr 13, 2007 #11
    no problem, just a hint, your frictional force will be less than mg*mu as a component of Fappl is lifting upward. Just want to potentially head that off so you can focus on the Tension aspects and not get frustrated by possible oversight.
  13. Apr 13, 2007 #12
    alright, once again thanks a million.
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