Chis Petrie’s “grandad” problem

Many yeas ago, his grandad suggested to the young Chris Petrie that he calculate the angle of the dangling rod in the diagram. It is rumoured that the grandad added that it would keep Chris quiet on a rainy day. Was this a case of “Spare the rod and spoil the child”?

In order to make things easy, we are to assume that the two strings are weightless, and the magic knots allow complete freedom or movement.

In order to make things difficult, all four lengths in the diagram are in general different.
This solution is due to Robin Corbett who pointed out that when a body is in equilibrium subject to three forces, those forces must all act through a single point, otherwise the body would experience a couple.

The three forces here are the tension is each of the two strings, and the vertical force of gravity operating through the centre of mass of the rod (shown as a red dot, and assumed to be in the centre of the rod whose length is 2 l).

The problem is thus reduced to one of geometry − albeit rather messy geometry.

To be continued . . . . . . .

Let the inclination of string 1 to the horizontal be θ1, and that of string 2 θ2. Let the rod be at an angle of β to the horizontal.

Robin’s analysis with DH’s notation

Applying the sine rule in the big triangle gives:
           x2 + s2
sin (θ1 – β) 		  =   s1 + x1
sin (θ2 + β) 		  =   2 l
sin (θ1 + θ2)

      L   =   x1 cos θ1 + x2 cos θ2
             =   ( 2 l sin (θ2 + β)
sin (θ1 + θ2) 		– s1 ) cos θ1   +   ( 2 l sin (θ1 – β)
sin (θ1 + θ2) 		– s2 ) cos θ2
             =   2 l sin (θ2 + β) cos θ1
sin (θ1 + θ2) 		  +   2 l sin (θ1 – β) cos θ2
sin (θ1 + θ2) 		  –   s1 cos θ1   –   s2 cos θ2
             =   2 l ( sin (θ2 + β) cos θ1   +   sin (θ1 – β) cos θ2 )
sin θ1 cos θ2 + sin θ2 cos θ1 		  –   s1 cos θ1   –   s2 cos θ2

DH’s analysis based on Robin’s original inspiration

Applying the sine rule in the two big triangles gives:
           l
cos θ1 		  =   s1 + x1
cos β
           l
cos θ2 		  =   s2 + x2
cos β

      x1 cos θ1 + x2 cos θ2 = L
      x1 sin θ1 = x2 sin θ2
           x1 is the upwards continuation of s1, and similarly for x2.

Cross multiply and add:
      l cos β   =   s1 cos θ1 + x1 cos θ1
      l cos β   =   s2 cos θ2 + x2 cos θ2
      2 l cos β   =   s2 cos θ2 + s1 cos θ1 + L
           which is actually fairly obvious just by resolving lengths horizontally.
and also substract:
      s1 cos θ1 + x1 cos θ1   =   s2 cos θ2 + x2 cos θ2

Resolving vertically:
      s1 sin θ1 = 2 l sin β + s2 sin θ2

Applying the sine rule in the two big triangles using h for the vertical distance of the centre of mass below the top vertex:
           l
cos θ1 		  =   h
sin (θ1 – β)
           l
cos θ2 		  =   h
sin (θ2 + β)

      l sin (θ1 – β) cos θ2   =   h cos θ1 cos θ2
      l sin (θ2 + β) cos θ1   =   h cos θ1 cos θ2

      \ sin (θ1 – β) cos θ2   =   sin (θ2 + β) cos θ1
      [sin θ1 cos β – cos θ1 sin β] cos θ2   =   [sin θ2 cos β + cos θ2 sin β] cos θ1
      sin θ1 cos θ2 cos β – cos θ1 cos θ2 sin β   =   sin θ2 cos θ1 cos β + cos θ2 cos θ1 sin β
      cos β (sin θ1 cos θ2 – sin θ2 cos θ1 )   =   sin β (cos θ1 cos θ2 + cos θ2 cos θ1)

We have 3 independent(?) equations in three unknowns θ1, θ2 and β:
      2 l cos β   =   s2 cos θ2 + s1 cos θ1 + L
      s1 sin θ1 = 2 l sin β + s2 sin θ2
      cos β (sin θ1 cos θ2 – sin θ2 cos θ1)   =   2 sin β cos θ1 cos θ2

A thought:
           tan β   =   s1 sin θ1 – s2 sin θ2
s1 cos θ1 + s2 cos θ2 + L 		  =   sin θ1 cos θ2 – cos θ1 sin θ2
2 cos θ1 cos θ2 		  =   sin (θ1 – θ2)
2 cos θ1 cos θ2
Re arrange our three equations with a view to using the middle one to eliminate θ1 from the other two:
      s1 cos θ1   =   2 l cos β – s2 cos θ2 – L
      s1 sin θ1 = 2 l sin β + s2 sin θ2          also       s12 – s12 cos2 θ1   =   (2 l sin β + s2 sin θ2)2
      cos β sin θ1 cos θ2   =   cos θ1 (2 sin β cos θ2 + cos β sin θ2)

Elininate θ1 by substitution the first equation into the right-hand version of the second equation:
      s12 – (2 l cos β – s2 cos θ2 – L)2   =   (2 l sin β + s2 sin θ2)2
and use both first and second equations to eliminate θ1 from the third equation:
      cos β cos θ2 (2 l sin β + s2 sin θ2)   =   (2 l cos β – s2 cos θ2 – L) (2 sin β cos θ2 + cos β sin θ2)
Multiply out the squares in the first of these two:
      s12 – (2 l cos β – L)2 + 2 (2 l cos β – L) s2 cos θ2 – s22 cos2 θ2   =   4 l2 sin2 β + 4 l s2 sin β sin θ2 + s22 sin2 θ2
      s12 – s22 – (2 l cos β – L)2 + 2 (2 l cos β – L) s2 cos θ2   =   4 l2 sin2 β + 4 l s2 sin β sin θ2
We now have 2 equations:
      cos β cos θ2 (2 l sin β + s2 sin θ2)   =   (2 l cos β – s2 cos θ2 – L) (2 sin β cos θ2 + cos β sin θ2)
      s12 – s22 – (2 l cos β – L)2 + 2 (2 l cos β – L) s2 cos θ2 – 4 l2 sin2 β   =   4 l s2 sin β sin θ2
It just remains to square the second equation to get a quadratic in cos θ2,
the solution of which can be substitued into the first equation to yield a (high-order!) polynomial equation in either sinβ or cosβ.

Another thought:
      β   =   β(l, L, s1, s2)
If the two strings are swapped over (or if you just look from the other side), the angle β will be the same magnitude, but in the opposite direction, i.e.:
      β(l, L, s1, s2)   =   – β(l, L, s2, s1)

Yet another thought, can we simplify?:
      cos β cos θ2 (2 l sin β + s2 sin θ2)   =   (2 l cos β – s2 cos θ2 – L) (2 sin β cos θ2 + cos β sin θ2)
      2 l cos β cos θ2 sin β + s2 cos β cos θ2 sin θ2   =   2 l cos β (2 sin β cos θ2 + cos β sin θ2)   –   (s2 cos θ2 + L) (2 sin β cos θ2 + cos β sin θ2)
                      =   4 l cos β sin β cos θ2 + 2 l cos2 β sin θ2   –   2 s2 sin β cos2 θ2 – 2 L sin β cos θ2   –   s2 cos θ2 cos β sin θ2 – L cos β sin θ2
      0   =   2 l cos β sin β cos θ2 + 2 l cos2 β sin θ2   –   2 s2 sin β cos2 θ2 – 2 L sin β cos θ2   –   2 s2 cos θ2 cos β sin θ2 – L cos β sin θ2

It might then be interesting to investigate the normal modes of oscillation of the system.