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Quiz of the Month (August 2005)

Hector C. Parr

***

SOLUTION TO LAST MONTH'S QUIZ

  1.  12.43 cm
  2.  5.97o
  3.  690 cm

Notes.
  1. Minimum height = 30 cm.
     Maximum height = 30 sqrt(2)
     30 sqrt(2) - 30 = 12.426 cm.

  2. Greatest angle occurs when front wheel centre is at maximum
      height and rear wheel centre at minimum height.
     Now sin p = 7/120. So p = 3.344o
     and sin q = (30 sqrt(2) - 23)/120. So q = 9.316o
     The cross-bar turns through the same angle as the line of
      centres, and q - p = 5.972o
   (Several erroneous methods give answers quite close to the correct value.
    So we accept only answers within 0.01o of this value.)

  3. Distance must be a multiple of 46 cm and an odd multiple of 30 cm.
     By inspection, this first occurs when distance = 690 cm.

THIS MONTH'S QUIZ

A rectangular up-and-over garage door is 6 feet high. On each side of the door is a pin B, 2 feet from the bottom, which slides in a vertical groove. On both sides of the door the points D, 2 feet from the top, are connected by light ropes of length 2 feet to the points F at the top of the grooves 6 feet above the ground. From each of the pins B a vertical cord is taken over a pulley and carries a counterweight as shown. The door weighs 20 kg and its centre of gravity, C, is equidistant from its top and bottom edges.

1. Describe accurately in words the path followed by the point E as the door opens and closes.

2. Describe accurately in words the path followed by the point A as the door opens and closes.

3. Neglecting all frictional forces, find the weight of each counterweight if the door is to remain stationary in any position between fully open and fully closed.

***

© Hector C. Parr (2005)

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