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A few more problems [equilibrium, torque, &c]

  1. Nov 21, 2006 #1
    Hi, got stumped again, but the sooner I figure out how to do these, the easier the others will be in the future.

    for the 240N weight
    [tex]\SigmaF_x=0=T-240N[/tex]
    so we know what the tension is for the rope. for the large pulley, going clockwise:
    [tex]\Sigma\Tau=0=(0.25m)(240N)-(1)(W)[/tex] so the weight was pretty simple to calculate (60N)

    The second part is where complications arose; the second equation became:
    [tex]\Sigma\Tau=0=(0.25m)(240N)-(1)(60N)(cos(\theta))-(2)(20n)(cos(\theta))[/tex]
    I solved for theta and got 55.7 degrees, the book said 53.2...



    For a separate problem, I need clarification about what the definitions for slipping and tipping are in the context of equilibrium, and I'm pretty sure I can take it from there.



    I got as far as [tex]\SigmaF_x=0=F_1+F_2-1960N[/tex]
    two variables, the equation for torque (counterclockwise) about the center of mass should handle this:
    [tex]\Sigma\Tau=0=(P_2)x(F_2)-(P_1)x(F_1)=F_2\frac{\sqrt{2}}{2}-F_1\frac{\sqrt{2}}{4}[/tex]
    This doesn't get the results I need (1370N for F1 and 590N), so I think I'm computing the torque wrong, since the answers both add to 1960 (thank goodness!!!)

    I labeled the tension, natural force exerted from the pivot, and weight of the beam.
    [tex]\Sigma\F_x=n-Tcos(\theta), \Sigma\F_y=Tsin(\theta)-w[/tex]
    Since the girder can't accelerate in the X direction, the point cannot be closer to the pivot on the girder than L/2.
    From there, the torque about the pivot would be
    [tex]\Sigma\Tau=(\lambda+\frac{L}{2})T* arctan(\frac{\lambda}{h})-\frac{L}{2}W[/tex]
    The answer was L/2+h^2/L, and I couldn't get to that answer. Not sure how to go from here...

    Thanks again for the help
     
    Last edited: Nov 21, 2006
  2. jcsd
  3. Nov 21, 2006 #2

    Doc Al

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    problem 11-59

    Your method and final equation are correct. Redo your calculation.
     
  4. Nov 21, 2006 #3

    Doc Al

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    problem 11-61

    I don't understand how you are calculating the torques. How did you determine the distance from the center of mass to the lines of the two vertical forces? (Note that they are lifting at the bottom of the crate, while the center of mass is in the middle of the crate.)
     
    Last edited: Nov 21, 2006
  5. Nov 21, 2006 #4
    I found the position vectors relative to the center of mass for each point of contact for each force and took the cross products. That's how I did it before and I got the right answer, not sure why it doesn't work now.
     
  6. Nov 21, 2006 #5
    Must have been a typo on my calculator; thanks.
     
  7. Nov 22, 2006 #6

    Doc Al

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    That's a good approach. What did you get for the position vectors and the angles they make with the applied forces?
     
  8. Nov 22, 2006 #7
    I got [tex](\frac{-1}{4}, \frac{-1}{2})\sqrt{2})[/tex] and [tex](\frac{1}{2}, \frac{1}{4})\sqrt{2})[/tex]. I got them by adding and subtracting half of the position vector of the bottom edge (3/4 sqrt(2) for both i and j) by the position of the midpoint on the bottom edge (1/8, -1/8)sqrt(2). I re-did the calculation on Maple just to double-check myself, and got the same vectors. As for the angles, I didn't take that into consideration, as the cross-product alone is the definition of the torque.
     
  9. Nov 22, 2006 #8

    Doc Al

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    Here's what I'd do. I'd use a coordinate system with the center of the crate at the origin and the x-axis parallel to the bottom of the crate. Given that, I get the following position vectors:
    r_1 = (-.25, -.625)
    r_2 = (-.25, +.625)

    And the following force vectors:
    [tex] \vec{F}_1 = (\sqrt{2}/2, \sqrt{2}/2) F_1[/tex]
    [tex] \vec{F}_2 = (\sqrt{2}/2, \sqrt{2}/2) F_2[/tex]
     
  10. Nov 22, 2006 #9
    That did it, not sure exactly why my previous system was flawed though, careless error perhaps?
     
  11. Nov 23, 2006 #10
    I still need some help with the last one and the one about the definitions for slipping and tipping are in the context of equilibrium. Thanks
     
  12. Nov 24, 2006 #11

    Doc Al

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    problem 11-83

    The pivot point is free to exert force, as needed, in both the x and y directions. Also, what do you mean by "natural" force? I'm not familiar with that term. (Is it the same as "normal" force?)

    Two problems:
    (1) Arctan is just the angle, I presume you mean the sine of that angle.
    (2) Why assume that the weight is in the middle? The problem states that all sorts of things are hanging from the girder, so who knows where the net weight acts?

    I'd say that that answer is not quite right, but close to what I get. In any case, rewrite your torque equation in terms of [itex]\lambda[/itex], L, and h. (What's the net torque equal? What's constant?) Then solve for the value of [itex]\lambda[/itex] that will minimize the tension. Hint: Find an expression and take its derivitive.
     
  13. Nov 24, 2006 #12

    Doc Al

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    Why don't you post that problem so we have a context for your questions?

    "Slipping" is the sliding of two surfaces against each other. Often a problem will ask for some condition that will just barely prevent slipping--meaning that the force of static friction preventing slipping is at its maximum value.

    "Tipping" often means that some applied force creates enough torque to tip something over (about its bottom edge, perhaps)--meaning that the applied torque must overcome any torque in the opposite direction, often from the weight of the object.
     
  14. Nov 25, 2006 #13
    I fixed that, using the cosine, and it worked. It must have been an input oversight on my part.
     
  15. Nov 25, 2006 #14
    A conveyor system for loading bales which are 0.5m high, 0.25m wide, 0.80m long, mass 30kg, coefficient of static friction between the bale and the conveyor is 0.6, and the system moves with constant speed. What angle [tex]\beta[/tex] between the horizontal and the belt would cause the bale to slip and tip?
    What would the angles be if the coefficient was 0.4?

    For slipping, when [tex]\Sigma F_x=0[/tex], I got 31 and 22 degrees, in line with the book's solutions, so I can't be too far off.
     
  16. Nov 27, 2006 #15

    Doc Al

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    Good, so it looks like you understand slipping. As for tipping, figure out what angle of incline would make the weight--acting through the center of mass--fall outside the base of the bale. (This angle will depend on the orientation of the bale on the conveyer.)
     
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