(^162) Aptitude Test Problems in Physics
The second equation (the condition of equality to
zero of the sum of all moments of force about point
0) implies that
N 1 = 2N 2. (1)
Combining these conditions, we get (see Fig. 165)
T = N 1 — N2 = N2. (2)
At the moment of snapping of the right string,
the accelerations of the loads of mass m 1 and m 2
will be vertical (point 0 is stationary, and the rod
is inextensible) and connected through the relation
a 2 = 2a 1. ( 3 )
Let us write the equations of motion for the loads
at this instant:
mfg — N; = miai , m 2 g + Na = m 2 a 2 ,
where N; and N; are the normal reactions of the
rod on the loads of mass m 1 and m 2. Since N;
and N; = N2, we have
m fg — 2N 2 = mial, m 2 g + N2 = 2m 2 a 1.
Hence we can find N2, and consequently (see
Eq. (2)) the tension of the string
T N 2 — menz^
g.
mi 4m 2
1.59. Let the ring move down from point A by a
distance Ax during a small time interval At elapsed
Fig. 166
after the beginning of motion of the system
and acquire a velocity v (Fig. 166). The velocity of
translatory motion of the hoops at this moment
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