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Objects in Equilibrium/Center of Gravity

  1. Oct 18, 2004 #1
    Guess who? Yeah it's me yet again. LOL

    I am a bit lost with the following problem.

    Here's the problem: A uniform plank of length 4.5m and weight 223N rests horizontally on two supports, with 1.1m of the plank hanging over the right support (see the drawing). To what distance x can a person who weighs 450N walk on the overhanging part of the plank before it just begins to tip?

    Drawing

    Ok, so......I don't know what to do here. As far as forces go there are two forces acting up on the plank by those two supports. The plank itself has a weight force pointing down and the man has a weight force pointing down.

    I thought perhaps I should use the center of gravity equation but I don't really understand what that would do for me.

    Any help to get me started in the right direction would be greatly appreciated. :smile:
     
  2. jcsd
  3. Oct 18, 2004 #2

    Pyrrhus

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    Beautiful Problem!! requires creativity!

    First Equilibrium Conditions

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = 0 [/tex]

    [tex] \sum_{i=1}^{n} \vec{\tau}_{i} = 0 [/tex]

    Torque's Trick: Pick an origin system which eliminates the most forces or unknowns.

    Now Let's see the idea for this concept.... We got the a contact force in each support point, and a weight of the planck in its center of gravity, and the weight of the person [another fancy word for Block].

    Imagine the person walking that panck, there will be a maximium time [period, moment], where the contact force on the first support will be 0 thus marking the way for our max x distance the person will move, Think about it, when the person passes that x distance the planck will balance on the second support, so if the person is exactly at that distance our first support normal force [contact force] will be 0.
     
    Last edited: Oct 18, 2004
  4. Oct 19, 2004 #3
    Ok, so this is going to require me to make a chart I see. So my origin point, should be probably be at the part of the plank where the first support meets it. If that were the case then my chart would be as follows, or at least I think. I'll label the first support force F_n1 and the second support force F_n2, the weight of the plank W_p and the weight of the man W_m. H being the horizontal force and V being the vertical force.

    So my forces being:
    F_n1 = 0
    F_n2 = ?
    W_p = 223N
    W_m = 450N
    H = 0
    V = 0

    Length of the plank : 4.5m
    Length of the plank to the second supprt : 3.4m
    Length of the extension : 1.1m

    So now I would have to draw my line of action and lever arms for each of the forces correct? Then I would be able to find F_n2.

    Am I at least headed in the right direction?
     
    Last edited: Oct 19, 2004
  5. Oct 19, 2004 #4

    Pyrrhus

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    Use the Torque trick i posted, you only need the second condition of equilibrium to solve this problem.
     
  6. Oct 19, 2004 #5
    Oh, I think I see. My origin would be the center of gravity of the plank?

    I'm trying to use what you say to use, but I just keep drawing blanks. I should probably be drawing a free body diagram, shouldn't I?

    Lord, if this was only a normal torque problem I think I could get it. LOL
     
  7. Oct 19, 2004 #6

    Pyrrhus

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    Well let's analyse what we know and we don't know.

    We know the weight of the Planck, We don't know the contact forces, and we know the weight of the person. The contact force on the first support must be 0 for x max, so we are left with an unknown force.

    Where should our origin for our second condition should be??

    The second support, because for any point you pick if the object is at equilibrium then the sum of all the torques will be 0.
     
  8. Oct 19, 2004 #7
    Ok, which is what I had in an earlier post. When I listed my forces. I just had my origin wrong. But I don't see how this is going to help me find a distance. Unless I have to make my chart again, finding the x and y componants.
     
  9. Oct 19, 2004 #8

    Pyrrhus

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    Visualize it in this origin...
     
  10. Oct 19, 2004 #9

    Pyrrhus

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    I think you may have a Torque misconception,

    You know torque is

    [tex] \vec{\tau} = \vec{r} \times \vec{F} [/tex]

    so a torque's magnitude is

    [tex] |\vec{\tau}| = |\vec{F}||\vec{r}|sin\theta [/tex]

    where theta is the angle between the force and the radius vector.

    Also

    [tex] \tau = Fd [/tex]

    where d is the lever arm (distance perpendicular to the line of action of the force). So if you have an unknown force on the second support, and you put your origin there, the lever arm for the Torque of that force will be 0, so on your second equilibrium equation set up for x max you will only have 1 unknown (x).
     
    Last edited: Oct 19, 2004
  11. Oct 19, 2004 #10
    Ok well, this is officially the dumbest I have felt in a long time. It's funny how physics has a constant habit of making me feel this way.

    I am really trying hard to use what you are telling me. Really I am....but for my second equilibrium equation you say to set up for x max leaving me with one unknown (x). But as to what equation you are talking about, I have no idea. To get an equilibrium equation it was my understanding that you had to set up a chart with all the forces and what not, then add the columns to give you an equation.

    Other than that, the only other equation I know of is the center of gravity equation.

    I'm sorry to be acting so ignorant about this, but I really am trying to understand this but for some reason none of this is making any sense to me.
     
  12. Oct 19, 2004 #11

    Pyrrhus

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    I will keep explaining til you understand...

    You see, you don't need both conditions to solve a problem, you could use one or another relation other than the conditions to solve the problems.

    Well, i will do the complete work of the two conditions for this problem.
    Right and up is positive

    Applying

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = 0 [/tex]

    On x-axis

    [tex] \sum F_{x} = 0 [/tex]

    On y-axis

    [tex] \sum F_{y} = N_{1} + N_{2} - m_{planck}g - m_{person}g = 0 [/tex]

    On our first condition if we put our first contact force (N1 = 0) then we could find the value the N2 force should have, but does that really answers our question??

    [tex] \sum F_{y} = N_{2} - m_{planck}g - m_{person}g = 0 [/tex]

    Applying the second condition

    [tex] \sum_{i=1}^{n} \vec{\tau}_{i} = 0 [/tex]

    I will put a subindex for the origin (CG = Center of Gravity, A = First support, B = Second Support) d = 1.1 m or the length of the planck past the second support.

    [tex] \sum \tau_{cg} = (\frac{l}{2}-d)N_{2} - (\frac{l}{2}-d + x)m_{person}g - \frac{l}{2}N_{1} = 0[/tex]

    Since the N1 must be 0

    [tex] \sum \tau_{cg} = (\frac{l}{2}-d)N_{2} - (\frac{l}{2}-d + x)m_{person}g = 0 [/tex]

    [tex] x \approx 0.569 m [/tex]

    We could use our value found from N2 from the first condition and solve for x, and it will be ok, but why not do it directly. If we put our origin in the second support we should have applying the second condition.

    [tex] \sum \tau_{B} = (\frac{l}{2}-d)m_{planck}g - (l-d)N_{1} - xm_{person}g = 0 [/tex]

    Because N1 should be 0

    [tex] \sum \tau_{B} = (\frac{l}{2}-d)m_{planck}g - xm_{person}g = 0 [/tex]

    [tex] x \approx 0.569 m [/tex]

    I wanted you to solve this problem in the direct way, i was suggesting, well it doesn't matter as you go doing equilibrium problems you will notice what condition(s) to apply to solve the problem.
     
    Last edited: Oct 19, 2004
  13. Oct 20, 2004 #12
    Ok this part I understand:

    [tex] \sum F_{y} = N_{2} - m_{planck}g - m_{person}g = 0 [/tex]

    That's basically what I was saying earlier, about adding my forces in my y column. That I get.

    However, I don't know where you get this from....

    [tex] \sum \tau_{cg} = (\frac{l}{2}-d)N_{2} - (\frac{l}{2}-d + x)m_{person}g = 0 [/tex]


    There isn't any equation in my book like that or anything remotely close. So I don't have a clue how I would have known how to come up with an equation such as this one.

    Btw, did I mention that I really appreciate you helping me with this? Because I do. Even though it may seem like your talking to a wall sometimes. lol
     
  14. Oct 20, 2004 #13

    Pyrrhus

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    I said til you understand....

    [tex] \sum_{i=1}^{n} \vec{\tau}_{i} = 0 [/tex]

    [tex] \vec{\tau} = \vec{r} \times \vec{F} [/tex]

    [tex] \tau = Fd [/tex]

    Force * Lever arm, the sign for each torque it's because of the Right hand rule.
     
  15. Oct 20, 2004 #14
    Ok, so if I'm thinking correctly the lever arm in the above equation then would be the part wich has [tex] (\frac{l}{2}-d)[/tex] ? And if that's the case why is the full length of the plank divided by 2? Oh........it's divided by 2 to get the center of the board. Am I on the right track with that?
     
  16. Oct 20, 2004 #15

    Pyrrhus

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    Yes. In a more tecnically way, It's because in this problem the gravitational field is considered uniform and the density of the objects is uniform too, so the center of gravity is in the center of mass of an object. The center of mass of our planck which could be look like a rod is Lenght/2.
     
  17. Oct 20, 2004 #16
    Ok! Cool, I think I understand now.

    Thanks again for all your help. :smile:
     
  18. Oct 20, 2004 #17

    Pyrrhus

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    No problem, :smile:
     
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