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This question is driving me nuts. (counter-balancing forces)

  1. Oct 2, 2008 #1
    Hi, im new. I need someone to help me out with this question, because ive spent the last hour and a half trying to figure this out. I have an exam tomorrow and this question might be asked. Here it is:

    A man's leg and cast weigh 210 N (center of the mass). The counterbalance weighs 105 N. Determine the weight, w2 and the angle needed so that no force is exerted on the hip joint by the leg plus the cast.

    There is a diagram showing a man's leg being raised at a 40 degree angle by a wire and pulley on the right, on the end of it is a 105 N weight. On the other side (left), there is another wire pulley with an unknown weight, w2, attached to it at an unknown angle as well. Here is the diagram:


    How do I even begin this question?
  2. jcsd
  3. Oct 3, 2008 #2
    First sum up all of the forces that are acting in the system. Make sure you get your positive and negative symbols correct.

    Hint:Convert every force to an x and y plane equivelant.

    The key to solving this is remembering that the sum of all forces must equal 0 because it is in equilibrium. Using this, you should be able to form two simultaneous equations. Let me know how you go.
  4. Oct 3, 2008 #3
    okay. i pretty much had a hunch that this what they wanted.

    w1 triangle:

    r1 = 105/sin40 = 163.4
    y1 = 105
    x1 = cos40 x 163.4 = 125.2

    the x1 value should be the same as the x2 value, if T-T=0 so:

    x2 = 125.2

    and this is where im stuck. I suppose i could use 210 as the height of the triangle, which gives me 244, but then what? I receive an angle approximate to 59.2 degrees by taking the inverse cosine of 125/244. However, then i get w2 as 210 N, which is clearly wrong. Insight?
  5. Oct 3, 2008 #4
    You are right in thinking that the two x values must be equal (and opposite).

    Remember I said sum up all the forces. You did that for x, now for y.

    You have the weight of his leg pushing down, the y tension of w1 pushing up and the y tension of w2 pushing up too.

    Therefore you can get the equation:
    210 = 105 + w2
  6. Oct 3, 2008 #5
    That's exactly what I was thinking, Rake, but the answer for this question is stating that the w2 is not 105, but actually 163.8. I can't figure out where they got that value, because I kept getting 105 for w2.
  7. Oct 3, 2008 #6
    I'm sorry I was up all night, perhaps seek someone's second opinion because I'm sure I overlooked some tiny little thing.
  8. Oct 3, 2008 #7
    Oh hang on, how did I not see this. The force of 210N is not going straight down from his leg. Well essentially it is, but his hip can't just drop straight down, it's a joint. Therefore there will be a torque created.
    But I wouldn't suppose there's a length given of his leg?
  9. Oct 3, 2008 #8
    Nope. No length. Lol. This is why this question is driving me nuts.
  10. Oct 3, 2008 #9
    Heheh, Serway strikes again. Be careful, if you don't use a standard angle for many problems, the signs can become quite confusing.

    This is in a chapter strictly concerning forces, so not torque problems yet.

    Your idea seems correct, so it may be a matter of checking your work.

    [tex]w_x_1 + w_x_2 =0[/tex]

    [tex]220 = (105*\sin(140)) + w_y_2

    Now use [tex]|w_2| = \sqrt{w_x + w_y} [/tex] to find your magnitude and [tex]\theta = \tan^{-1} \frac{y}{x}[/tex]
    Last edited: Oct 3, 2008
  11. Oct 3, 2008 #10
    I figured it out. I was thinking the side of the triangle with the weight, or the y of the triangle was 105 when it was actually the r. it makes sense now, since that's where the tension should have been.

    So if r = 105, cos40*105 = 80.43 for x, which in turn is 80.43 on x2 because of T-T'=0.

    If I get the sin40*105, the y is 67.5.
    210 (force acting downward) - 67.5 = 142.5 for y2.
    142^2 + 80.43^2 = 26775. Square that and you come to your answer of what the weight must be: 164 N.
  12. Oct 3, 2008 #11
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