Rotational Equilibrium

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Discussion

Equilibrium of Extended Bodies

Translational and Rotational Quantities Compared
  translational connections rotational
equilibrium
∑ F = 0 ⇒ 

 ⎧∑ F+x = ∑ F−x
⎨∑ F+y = ∑ F−y
⎩∑ F+z = ∑ F−z		    
∑ τ = 0 ⇒ 

 ⎧∑ τ+x = ∑ τ−x
⎨∑ τ+y = ∑ τ−y
⎩∑ τ+z = ∑ τ−z

Stability of Equilibrium

Stability of Equilibrium
equilibrium
condition		 stable
equilibrium		 unstable
equilibrium		 neutral
equilibrium
∑ F = 0
no net force		 ∑ F(x + Δx) ∝ −Δx
restoring force		 ∑ F(x + Δx) ∝ +Δx
repelling force		 ∑ F(x + Δx) = 0
no force
dU/dx = 0
local extrema		 d2U/dx2 > 0
concave up		 d2U/dx2 < 0
concave down		 d2U/dx2 = 0
flat
  center of gravity is below pivot center of gravity is above pivot center of gravity is at the pivot
  small displacement raises center of gravity small displacement lowers center of gravity small displacement does not raise or lower center of gravity

Summary

 

∑ F = 0 ⇒ 

   ⎧
⎪
⎨
⎪
⎩		   ∑ F+x = ∑ F−x  
∑ τ = 0 ⇒ 

   ⎧
⎪
⎨
⎪
⎩		   ∑ τ+x = ∑ τ−x
∑ F+y = ∑ F−y ∑ τ+y = ∑ τ−y
∑ F+z = ∑ F−z ∑ τ+z = ∑ τ−z
No Net Force in Any Direction   No Net Torque About Any Axis
 
The center of mass is computed from the mass distribution. The center of gravity is computed from the weight distribution.
   
rcm =  ∑ miri  = (x, y, z) rcg =  ∑ Wiri  = (x, y, z)
∑ mi ∑ Wi
Discrete Collection of Objects
   
rcm =  1 ⌠⌠⌠
⌡⌡⌡		 r dm =  1 ⌠⌠⌠
⌡⌡⌡		 r dV rcg =  1 ⌠⌠⌠
⌡⌡⌡		 r dW =  1 ⌠⌠⌠
⌡⌡⌡		 r dV
m V W V
Continuous Distribution of Matter

Problems

practice

  1. Write something.
    • Answer it.
  2. Write something else..
    • Answer it.
  3. The modified photographs below show a counterweighted, steel drawbridge in the closed (span down) and open (span up) positions. The elements used to solve this problem are highlighted in color: pivot points in red, lever arms in yellow, and forces in green.
         
     
    Bridge Closed -- Span Down [magnify]   Bridge Open -- Span Up [magnify]
         
    • The red triangles represent the pivot points for …
      • the counterweight assembly on the left and
      • the bridge span on the right.
    • The yellow lines highlight the lever arms.
      • The line running along the roadbed is divided into intervals of equal length, L.
      • The other lines have lengths that can be determined with geometric reasoning if we assume that the angles between beams are all multiples of 45° (that is, 0°, 45°, 90°, 135°, 180°).
    • The green arrows show the relevant forces.
      • Wb and Wc are the weights of the moveable bridge span and counterweight, respectively.
      • T1 and T2 are the tensions in the linkage when the bridge is closed and open, respectively.
      • When the bridge is closed the moveable span is balanced so that there is no normal force on the far end.
    Determine the following quantities in terms of the weight of the bridge span.
    1. The weight of the counterweight and the tension when the bridge is closed.
    2. The tension when the bridge is open and the torque needed to keep the span open.
    Solutions …
    1. Answer the first two parts by stating the equilibrium conditions when the span is down.
         
      left axis right axis
      ∑τcounterclockwise  =  ∑τclockwise ∑τcounterclockwise  =  ∑τclockwise
      (1 L)(Wc)(sin 90°)  =  (√2 L)(T1)(sin 90°) (√2 L)(T1)(sin 90°)  =  (4 L)(Wb)(sin 90°)
      1 Wc  =  √2 T1 √2 T1  =  4 Wb
       
      combine equations
      Wc  =  4 Wb
      T1  =  2√2 Wb = 2.83Wb
           
    2. Answer the second two parts by stating the equilibrium conditions when the span is up.
         
      left axis right axis
      ∑τcounterclockwise  =  ∑τclockwise ∑τcounterclockwise  =  ∑τclockwise
      (1 L)(4 Wb)(sin 45°)  =  (√2 L)(T2)(sin 135°) (√2 L)(T2)(sin 45°) +   =  (4 L)(Wb)(sin 135°)
      2√2 LWb  =  LT2 LT2 +   =  2√2 LWb
       
      combine equations
      T2  =  2√2 Wb = 2.83Wb
      τ  =  0
           
      Because of the clever way the linkage folds, the bridge is maintained in balance throughout operation. There is no change in the tension and no extra torque is needed to keep it open. In accordance with Newton's second law of motion, some extra torque is needed to start it moving, but not much. The bridge and counterweight, which together weigh more than a million kilograms, is opened an closed with a relatively small electric motor. Something like 50 or 60 kW (75 hp) is powerful enough.

      The bridge in this photo is a part of the La Salle Causeway, which spans the Cataraqui River in Kingston, Ontario. It is an example of a "Strauss heel trunnion bascule bridge".
      • "Strauss" for Joseph B. Strauss, chief engineer and owner of the Strauss Bascule Bridge Company in Chicago;
      • "Heel" since it tips back like a foot balanced on its heel;
      • "Trunnion" for the two large, weight-bearing axles (the term originally referred to the axle upon which cannons and artillery were balanced);
      • "Bascule" from the French word for seesaw (when one side goes up, the other goes down).
  4. Write something completely different.
    • Answer it.

conceptual

  1. How can all three types of equilibrium -- stable, unstable, and neutral -- be demonstrated using …
    1. an egg
    2. a cone
    3. a torus (a shape like a donut or a bagel)
  2. Explain the reasoning behind each of the following general rules of design.
    1. Aircraft carriers are designed to be stable in the ocean.
    2. Fighter planes are designed to be unstable in flight.

statistical

  1. center-of-population.txt
    The center of population of the United States as defined by the Census Bureau is the same as that of the center of gravity of a collection of point masses on a plane. It is the point at which a weightless, smooth, spherical shell in the shape of the "lower 48" states and the District of Columbia would balance if weights of identical size were placed on it -- each weight representing the location of one person. On such an imaginary surface, north-south distances between parallels of latitude (ϕ) are identical and their angular measure in degrees may be used as units of displacement. In contrast, east-west distances between meridians of longitude (λ) are not constant but vary with latitude from a maximum at the equator to zero at the poles. Multiplying by the cosine of the latitude will correct for this convergence of the meridians at the poles. In addition, small areas of the country are used as data points rather than individual human beings, which reduces the computational burden. (In 1960 43,000 areas were used but by 2000, this number had risen to more than 8,000,000. By 2020 or 2030, the number of areas will probably equal the number of residents.)

    Thus, the center of population of the US computed by the Census Bureau is the point whose latitude (φ) and longitude (λ) satisfy the equations …
             
    φ =  ∑ wiφi   λ =  ∑ wiλicos φi
    ∑ wi ∑ wicos φi
             
    Where ϕi, λi, and wi are the latitude, longitude, and population of the census areas included in the calculation.
    1. The data on the accompanying tabs-delimited text file give the population and the effective latitude and longitude in degrees of the fifty states and the District of Columbia from the 2000 census. Using this data, determine the coordinates of the population center of the United States at this time. (Be sure to exclude Alaska and Hawaii from your calculations, but do include the District of Columbia.)
    2. In what state is this point located? Which county? What is the nearest incorporated community (city, village, or town)? What is the nearest street intersection?
    3. Go to this location and await further instructions.

investigative

  1. Determine the mass of a ruler using a known weight to balance it over a pivot.

worksheets

  1. The Physics Teacher has published several articles containing free body diagram worksheets. They are available free to members of the American Association of Physics Teachers (AAPT). Everyone else has to pay.
    1. Free-body diagrams revisited — II. James E. Court. The Physics Teacher. Vol. 37, No. 8 (November 1999): 490-495. Note: pages 490-493 are relevant to this topic.

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