Solutions 149
sin n/6 = 1/2 into this equation, we obtain the
ratio:
M 2-3sina —4
m sing aThe velocity of the body at the moment of sep-
aration is
u=v sin a— 01
2 2 •
1.48. The rod is under the action of three forces:
the tension T of the string, the force of gravity mg,
and the reaction of the wall R = N Ffr (N is
the normal reaction of the wall, and Ffr is friction,
Ffr < RN). When the rod is in equilibrium, the
sum of the moments of these forces about any point
is zero. For this condition to be satisfied, the line
of action of the force R must pass through the
point of intersection of the lines of action of T
and mg (the moments of the forces T and mg
about this point are zero).
Depending on the relation between the angles a
and 13, the point of intersection of the lines of ac-
tion of T and mg may lie (1) above the perpen-
dicular AM 0 to the wall (point M 1 in Fig. 155);
(2) below this perpendicular (point M (^2) ); (3) on the
perpendicular (point M 0 ). Accordingly, the friction
is either directed upwards along AC (Fin), or down-
wards along AC (Fin), or is equal to zero. Let us
consider each case separately.
(1) The equilibrium conditions for the rod are
T cos a Frri — mg = 0, N— T sin a = 0 (1)
(the sums of the projections of all the forces on the
x- and y-axes respectively must be zero), and the
moments of forces about point A must also be zero:
mgl Ti
mgcli.Td 2 , or 2 sin 0= sin (a-Fp), (2)