Start by drawing a free-body force diagram
for each section of the hinged rod. The first
section is flat, running from A to B, and
has a length of 2a. The weight kmg acts downards
a distance a away from point A. The force
on the rod due to the wall is split into two
components, a component M acting vertically,
and a component N acting horizontally.
At the hinge, the force is again split into two
components, a force x acting horizontally
and a force y acting vertically. The other
part of the rod runs from B to C, and again
has a length 2a. This part of the rod hangs
at an angle theta to the vertical.
The weight of this part of the rod, mg, acts downwards
a length a along this. The forces at the hinge
act in exactly the opposite direction to the
forces on the other rod because they form
an action-reaction pair, so x acts horizontally
to the right, and y acts vertically downwards.
At the floor, there is a normal reaction force
acting upwards, R, and a frictional force
Fr, acting horizontally. We know that the
magnitude of Fr has got to be less than or
equal to mu, the coefficient of friction,
times by R, the normal reaction force at this point.
Looking at the part of the rod between
A and B, the forces must balance because this
rod is in equilibrium. Looking vertically
this means that M plus y is equal to kmg.
The total moments about any point must also
be equal to zero, so taking moments about
the centre of mass gives us a clockwise moment
of M times a, which is equal to the anti-clockwise
moment of y times a. The forces N, x and the
weight all act through the centre of mass and
so give zero moment about this point. Cancelling
out and re-arranging these we can find that
the size of y is equal to kmg divided by two.
Looking at the other part of the rod, the
forces must also balance because the rod is
in equilibrium. Vertically this gives y plus
mg equals R, and taking moments about the
point B we get a clockwise moment due to the
weight of mga sin theta, a clockwise moment
due to the frictional force of Fr times 2a times cos theta,
and these are balanced out by an anticlockwise moment due to R of R times 2a sin theta.
We can subsitute the value of
y that we obtained from part AB into these
equations and we can also look at the situation
when the frictional force is at its maximum
and is therefore equal to mu R, and substitute
this into these equations. This will allow
you to calculate a value for mu.
