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Block suspended on a string

  1. Sep 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A .66 kg block is suspended from the middle of a 1.14 m long string (.57m segments). The ends of the string are attached to the ceiling at points separated by 1 m, and the block can slip along the long string.

    The acceleration of gravity is 9.81 m/s2.

    1. What angle does the string make with the ceiling?

    2. What is the tension in the string?

    3. The .66 kg block is removed and two .33 kg blocks are attached to the string such that the lengths of each string segments are equal (.38m). What is the tension in the string segments attached to the ceiling on the right?

    4. What is the tension in the horizontal segment?

    2. Relevant equations

    F = ma

    tan (theta) = y-component / x-component

    3. The attempt at a solution

    All I could come up with was figuring out the Newton force of the block:

    F = ma
    F = .66 * 9.8
    F = 6.468

    I do not know where to continue from there. I am thinking I have to do something with the .57 m string pieces but am unsure on what to do with them. I am just focusing on the first part for now, I just decided to post the other parts in case I need help with them down the road.

    Thanks for any help.
     
  2. jcsd
  3. Sep 11, 2009 #2

    Doc Al

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    Staff: Mentor

    Figure out the angle that the strings make by drawing a triangle. (The top side is 1 m, the other two sides are 0.57 m.)

    Then consider the forces acting on the mass. Hint: Consider the vertical components.
     
  4. Sep 11, 2009 #3
    Okay, I will try that tomorrow morning. Thanks for the hint.
     
  5. Sep 12, 2009 #4
    Since I know three sides and no angles, should I cut the triangle in half to give me a 90 degree angle at the top and solve from there? Or is there a way to solve for an angle given all the sides.
     
  6. Sep 12, 2009 #5
    I tried twice to get the angle and came up with the wrong answer:

    I split the triangle in half so the top was .5 and the hypotenuse was .57.

    a2 + b2 = c2
    .52 + b2 = .572
    .25 + b2 = .3249
    b2 = .0749
    b = .27

    sin(theta) = .27 / .57
    theta = sin-1(.47)
    theta = 28.27

    I also tried to multiply 28.27 times two and subtract that from 180 to get that larger angle at the bottom of the string and still got the incorrect answer.

    Am I not supposed to use sin or am I supposed to take a different approach?
     
  7. Sep 12, 2009 #6

    Doc Al

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    Staff: Mentor

    Your approach is perfectly good. If this is some online system, perhaps it wants an answer to 3 significant figures. (Don't round off the side to .27 .)
     
  8. Sep 12, 2009 #7
    Yeah, it is an online system. I will try using 3 sig figs and report back after dinner. Thanks.
     
  9. Sep 12, 2009 #8
    Okay, I got the first part correct. The accepted answer was 28.694. These online systems are ridiculous. I am going to work on the other parts tonight and see if I can figure them out. I will post here if I have any troubles. Thanks again.
     
  10. Sep 13, 2009 #9
    I figured out the tension in the string just fine but I am having a problem with part 3. I just cannot figure out where to start at. Should I use the same method as the first part or is there something I am missing?
     
  11. Sep 13, 2009 #10
    Now you need to draw a different diagram.

    In this one, the string and the ceiling will form a trapezoid.
     
  12. Sep 13, 2009 #11

    Doc Al

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    Staff: Mentor

    Yes, you'll use the same general method, but as PhaseShifter says you'll have a different diagram with different angles.
     
  13. Sep 13, 2009 #12
    The diagram is already drawn for me (which is nice) but I do not know exactly where to cut the trapezoid to form my triangle. I am guessing that the top half will be .333, the hypotenuse will be .38 and the other part will be left up to me to solve.
     
  14. Sep 13, 2009 #13
    Okay, I solved for the angle:

    a2 + b2 = c2
    .3332 + b2 = 0.382
    .110889 + b2 = .1444
    b2 = .033511
    b = .183

    sin(theta) = .183 / 0.38
    theta = sin--1(.183 / 0.38)
    theta = 28.789

    This angle is similar to the first part but I am not sure if I can use the same equation to solve for T because even though the tension is equal on the other side, I don't know if it is equal on the horizontal segment.

    The equation I used was T = mg / 2sin(theta)
     
  15. Sep 13, 2009 #14
    Alright, so using T = mg / 2sin(theta) did not work. I am unsure on how to set up the force diagram from this point.
     
  16. Sep 13, 2009 #15

    Doc Al

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    Staff: Mentor

    No need to guess. You know the distance between the masses and the total length of the top of the trapezoid.
     
  17. Sep 13, 2009 #16
    I solved for the angle which I believe is correct. But I am having trouble coming up with an expression for the tension.
     
  18. Sep 13, 2009 #17

    Doc Al

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    Staff: Mentor

    You solved for the angle using a distance that you guessed. (It's close, but why guess?)
    You have two tensions to find so you'll need two equations: Analyze vertical and horizontal force components.
     
  19. Sep 16, 2009 #18
    Sorry I haven't replied back in a while, I have been studying for a different class and had to put Physics on the back-burner. I solved the problem though, thanks for all the help.
     
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