In order to tackle this question, we can split
half of the rope into two parts, the part
that is freely hanging and the part that is
on the slope. Examining the freely hanging
part of the rope, the tensile forces on it
are the tension from the part of the rope
on the slope, and the horizontal tension at
the centre of the rope.
Its weight, (mg over 2) f, acts downwards.
The tension up the slope will be at an angle of theta to the horizontal.
Resolving vertically we get that T sin theta
equals mgf over 2. Now we can examine the part
of the rope on the slope. The forces on it
will be the frictional force up the slope, F,
the normal reaction force N, its weight,
(mg over 2)(1-f) downwards, and the tension
from the hanging part of the rope, T. Since
the co-efficient of friction is 1 we know
that F = N. Resolving parallel to the slope
we get that T +(mg/2)(1-f)sin theta = N, which
we also know is equal to (mg(1-f)/2)cos theta
by resolving perpendicular to the slope.
This has eliminated F. From our first equation
we can substitute T in as mgf/(2 sin theta).
The factors of mg/2 will cancel out. We want
to find the maximum f which we can find by
differentiating f with respect to theta and
setting this to zero. It is easiest to do
this implicitly. This will give us an equation
in terms of f and theta which we can solve for theta.
It may be useful to recall the
double-angle formulae sin 2 theta = 2 sin theta cos theta
, and cos 2 theta = cos-squared
theta minus sin-squared theta.
