Start by drawing the two rods AB and CD. The
question says that both these rods are uniform,
which means that their weight mg acts in the
middle of each rod, and as the rods are of
length 6a this means that the weight acts
a distance 3a from each end. The hinge, H,
is 2a away from the bottom end of each rod,
which means that the weight acts a distance
a away from the hinge. We're also given in
the question that the distance between A and
C is 2a. We also need to add the forces on
the rods due to the weights hung from them
at B and D. We're also going to draw on two
angles here which are both the same theta
because we know that all sides of this triangle
at the bottom are the same. Then I'm going
to add normal reaction forces and frictional
forces where A and C touch the ground. The
two rods are in equilibrium which means if
you look at the entire system of the two rods
the sum of the forces will be zero. If we
look at the vertical forces, the forces downwards
will equal the forces upwards. For each rod
we can also say that the sum of the moments
acting is zero, which means if we take the
moments about any point the clockwise moments
and the anti-clockwise moments will balance
out. We don't know what forces are acting
on each rod at the hinge, but if we take moments
about this point, we don't need to worry about
these forces as they will have zero moment
about this point. Taking moments about this
point on rod AB gives us 5mg times 4a cos
theta plus mga cos theta plus N_A 2a cos theta
equals F_A 2a sin theta. Rearranging this
gives the equation shown here. Similarly,
you can look at the horizontal forces and
the moments about H on rod CD. Also, remember
that the maximum magnitude of a frictional
force is given by mu multiplied by the magnitude
of the normal reaction force at that point.
Combine all of these to solve the problem.
