In this question we have a non-uniform bar
of mass m hung horizontally between two walls
using two light ropes attached to the ends.
These ropes will have tensions T-one and T-two
at angles of phi and theta to the vertical
respectively. The bar's weight, mg, acts downwards,
a distance x from the right hand end. The
bar has length l. Resolving forces horizontally
we have T-one sin phi equals T-two sin theta,
which gives a relationship between the two
tensions. We can then take moments about a
point on the bar, choosing the point to be
the centre of mass so that the weight doesn't
feature in our equations. This gives a relation
between T-one, T-two and x: T-one (l minus
x) cos phi equals T-two x cos theta. By making
T-two the subject of our first equation and
substituting this into our second equation
we can cancel the factor of T-one and get
x as a function of length and the angles,
all of which we have been given values for
in the question.
