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Static friction and equilibrium - problem with geometry

  1. Nov 28, 2003 #1

    ShawnD

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    Here is the question
    http://myfiles.dyndns.org/pictures/statics10-2.png
    I'll just do the part where the wire is long to show you what I have problems with.

    To answer the problem I drew a FBD for member AB. Here are my variables and what they mean:
    T - the tension in the wire at angle z
    By - force of friction pushing down on member AB at point B
    Bx - horizontal reaction force pushing left on AB at point B

    The only horizontal forces on the member are from T and Bx. I can say this:
    Bx = Tcos(z)

    Since the friction of By is 0.35 times the reaction force Bx, I can say this:
    By = 0.35Bx

    Since point A and B are equal distance from the centre of mass for member AB, and since the moments on AB must cancel out, I can say this:
    By = Tsin(z)

    2 equations are equal to By so I'll rewrite them:
    0.35Bx = Tsin(z)

    now substitute in once more:
    0.35Tcos(z) = Tsin(z)

    now factor out the T:
    0.35cos(z) = sin(z)

    now make it equal zero:
    0 = sin(z) - 0.35cos(z)

    put it into my calculator and I get this:
    z = 19.29 degrees off the X axis



    Ok now here comes the problem. I have attained an angle for my cable but how do I find how long my cable is? At the angle I have, I don't know the horizontal length of AB, I don't know the vertical, I don't know it's angle.
    I've tried using the angle between AB and AC but that seems extremely hard; I can't even get that method started. I've tried changing the axies to line up with AB but that makes the By and Bx forces very tricky.
    One thing I noticed about the forces on AB is that the vertical forces at A and B make a couple, and the horizontal forces at A and B make a couple. If I sum moments around the centre of mass It shows that the horizontal couple and the verticle couple are equal and opposite (they cancel out). Going that approach would probably give me an answer in the end but the algebra would be very ugly.

    Is there some easy way to solve this problem that I am overlooking?
     
  2. jcsd
  3. Nov 28, 2003 #2
    ShawnD,
    I think your answer can't be right since you didn't use the given lengths.
     
  4. Nov 28, 2003 #3

    ShawnD

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    You don't need to know the lengths to figure out the forces.
     
  5. Nov 28, 2003 #4
    Yes, I see. The 1st part can be done solely with angles. But it's a tough one. I have some ideas, but no answer. Just let me tell my ideas, maybe it helps.

    First off, I think you're wrong in the formula for Bx.
    I think
    [tex]
    B_x = T \sin{z}.
    [/tex]
    Consequently, I think your expression for By is also false. But I don't think it's just T cos(z). The rod's angle must come in here, IMO.

    Next, I tried to find an expression for T.
    Let's denote by β the angle the rod makes with the horizontal.
    Then, the torque on the rod WRT B is
    [tex]
    t = mg \cdot 1m \cdot \cos{\beta}.
    [/tex]
    Thus, the force attacking at A and trying to rotate the rod is
    [tex]
    F_{rot} = \frac{mg}{2}\cos{\beta}.
    [/tex]
    Let's denote by α the angle at A, then the string tension is
    [tex]
    T = F_{rot} \cos{(90°-\alpha)} = F_{rot} \sin{\alpha} = \frac{mg}{2}\sin{\alpha}\cos{\beta}.
    [/tex]
    We could now argue that, in order for the system to be static, the frictional force must overcome total downward force:
    [tex]
    .35 T \sin{z} \geq mg - T \cos{z}.
    [/tex]
    This leads to the ugly equation
    [tex]
    \sin{\alpha}\cos{\beta}(.35\sin{z}+\cos{z}) \geq 2.
    [/tex]
    Looks suspect to me: You usually have something smaller than 1 on the RHS in equations like this. Where's my error?
    However, the next step probably involves the law of sines:
    [tex]
    \frac{\sin{z}}{2m} = \frac{\cos{\beta}}{l} = \frac{\sin{\alpha}}{1m}
    [/tex]
    where l is the length of the string.
    Well, we could build from this an equation that contains only l (or β), but it looks ABSOLUTELY nasty.

    I guess my idea (torque WRT B) was not so good from the start. Or the simple solution is hidden in all this. Where are the experts?
     
    Last edited: Nov 28, 2003
  6. Nov 28, 2003 #5

    krab

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    This seems wrong. There is also a torque from Bx about the rod's centre. Better to just find all downward forces.

    [tex]
    B_y+T\sin z=W
    [/tex]

    where W is the weight of the rod.
     
  7. Nov 28, 2003 #6

    Doc Al

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    Staff: Mentor

    I don't understand this equation. If you are doing torques about the center, what happened to Bx? (oops... krab beat me to it!)
    Given the angle "z" you know the angle ACB. Try using the law of cosines for the triangle ABC.

    Remember, you are asked to find a range of lengths.
     
  8. Nov 28, 2003 #7

    ShawnD

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    Re: Re: static friction and equilibrium - problem with geometry

    Hey yer right. I never noticed that.



    arcnets, Bx does not equal Tsin(z) because Bx is a horizontal force and Tsin(z) is a vertical force. Remember that since the coefficient of friction is not 1, these cannot be the same.

    krab, for By = Tsin(z), I think I based that on a wrong assumption; that doesn't mean it's wrong though. Instead of looking at the moments (which makes too many assumptions) look at the verticle forces on the bar that goes straight up and down, the bar that the cable is hanging from and AB is leaning against. This bar has 2 visible vertical forces on it: the vertical force from the cable AC and the vertical force from the bar AB. cable AC pulls down on the bar and beam AB pushes up on the bar. Since those are the only 2 forces on that bar, they must be equal to each other.
     
  9. Nov 28, 2003 #8
    Re: Re: Re: static friction and equilibrium - problem with geometry

    Ah, OK, I see. I thought z was the angle at C, but in your view it's the angle that the string makes with the horizontal. So, just exchange sin z and cos z in my post. The rest still holds, I think.
     
  10. Nov 28, 2003 #9

    Doc Al

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    Re: Re: Re: static friction and equilibrium - problem with geometry

    This reasoning is invalid. First, what makes you think that the two forces you mention are the only forces on the wall? Second, don't you think the wall exerts a net upward force on the wire plus rod system? (Something must be holding that system up! )
     
  11. Nov 28, 2003 #10

    ShawnD

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    The base of the wall provides no horizontal or vertical force but it can support a moment.

    You've never taken statics have you?
     
  12. Nov 28, 2003 #11

    Doc Al

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    Not sure what you mean by "the base of the wall". But the following is true: the net horizontal force of the wall on the "wire & rod" is zero. Maybe that's what you're thinking. The net vertical force of the wall on that system is equal to the weight of the system. (That's krab's equation.)
    Lucky you aren't in my class! You'd better hit the books.
     
  13. Nov 28, 2003 #12

    krab

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    Re: Re: Re: static friction and equilibrium - problem with geometry

    No. There are 3 forces. The 2 you mentioned, plus the weight.
     
  14. Nov 28, 2003 #13

    ShawnD

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    the weight of the slanted beam does not directly act on the vertical beam he is talking about.
     
  15. Nov 28, 2003 #14
    Torque problems are usually tedious nevertheless simple, let me just guide you through this.

    First find the center of mass. As you can see the varying the length of AC will vary at which position the center of mass is located. Figure this out for yourself.

    Gravity will act on this point of the center of mass. This is where your knowledge of torque is relevant. The weight will not be the same on A as on B. It will be greater on A than on B.

    In any equilibrium situation (different lengths of AC) the tension of the cable and the wall serve to counteract the forces of gravity. At both points A and B gravity can be partioned to its X and Y components. Our standard coordinate will have X as the length of the bar and Y perpendicular to the bar. Find the X and Y components on point A and B.

    On point A the tension will counteract both X and Y components and at point B the wall will counteract both X and Y components. The important point is that varying the length of AC will vary the center of mass which will cause the weight on point A to be different than on point B.
     
  16. Nov 28, 2003 #15

    ShawnD

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    That explanation is very logical but how can I get any answers that way? The way you say to do it leaves too many variables.
    1. The angle of cable AC
    2. The horizontal length of beam AB (to find the torque caused by gravity)
    3. The tension in cable AC
    4. The length of cable AC

    That's just too many variables.
     
  17. Nov 29, 2003 #16

    joc

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    i've been working on this problem...and it's basically a complicated trigonometry problem. you have to use several equations:

    1) [tex]F_f\leq\mu\F_N[/tex]
    2) balance of vertical and horizontal forces
    3) balance of torques
    4) the sine rule (for both lengths and forces i think)

    i'm trying to do it without any numerical values, working purely with variables. hopefully i'll have something to show later...
     
  18. Nov 29, 2003 #17
    I had an idea how to find Bx (the force component directed horizontally against the wall at B).

    Let's calculate the torque WRT C. Yes, C!

    Let's denote the rod's center-off-mass by M.
    Let's denote the distance CM by x.
    Let's denote the angle BCM by ε.

    Then, the torque WRT C is
    [tex]
    t = mgx \sin{\varepsilon}
    [/tex]
    At point B, this torque acts at right agles against the wall, and the distance CB is 1m. thus:
    [tex]
    B_x = \frac{t}{1m} = \frac{mgx \sin{\varepsilon}}{1m}
    [/tex]
    Let's use the law of sines:
    [tex]
    \frac{\sin{\varepsilon}}{1m} = \frac{\sin{\beta}}{x}
    [/tex]
    where β is the angle at B.
    Thus
    [tex]
    x = \frac{\sin{\beta}}{\sin{\varepsilon}}\cdot 1m
    [/tex]
    Thus
    [tex]
    B_x = \frac{mg \sin{\varepsilon}}{1m}\cdot \frac{\sin{\beta}}{\sin{\varepsilon}}\cdot 1m
    = mg \sin{\beta}
    [/tex]
    Any errors in this? If not, I'll try and continue.
     
  19. Nov 29, 2003 #18

    ShawnD

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    OK your work is a little confusing because I don't understand what you are doing.

    what is WRT?
    is distance CM is a straight line or is that a horizontal distance?
     
  20. Nov 29, 2003 #19
    Continued from my last post...

    I believe that we can ignore the tension in this problem. Since the rope does not exert an effort upon the bar we can just concentrate on point B. Remember that if we are to partition the weight on point B to its X and Y components we can find that the X component will serve to go against the frictional as welldecrease the normal. The Y component will serve as the normal as well as to go against the frictional. To see this for yourself find the horizontal and verticle components of both X and Y. You can see what happens as well increase AC. The normal starts to decrease and the frictional starts to increase.

    Don't make it too complicating. If you are writing out too many equations it is most likely that you are doing something wrong. Keep it simple.
     
  21. Nov 29, 2003 #20
    By WRT I mean "with respect to". A torque needs a point (pivot) with respect to which it is calculated. Sorry for any confusion - I'm not a native speaker.
    Straight line. Not horizontal.
    Do you think my expression for Bx is correct?
     
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