Solutions (^141)
elongation Al of the string will be such as if it
were acted upon by the force mg/2 at the point of
suspension and at the lower end, and the string
were weightless; hence
Al =
2k •
1.40. We assume that the condition mi + m 2 >
m 3 + m 4 is satisfied, otherwise the equilibrium is
impossible. The left spring was stretched with the
force T 1 balancing the force of gravity ma of the
load: T 1 = m 2 g. The equilibrium condition for the
load m 3 was
msg + T2 — Ften = 0 ,
where T2 is the tension of the right spring, and
Ft is the tension of the rope passed through the
pulley (see Fig. 14). This rope holds the loads of
mass m^1 and ma, whence
Ften = (ml + m2) g.
We can express the tension T2 in the following way:
T2 = (m1 m2 — m 3 ) g.
After cutting the lower thread, the equations of
motion for all the loads can be written as follows:
= mig + T1 — Ften, msas = m2g T1,
Maas = T2 + mag — Fun, -=- ma — T2.
Using the expressions for the forces T1, T2, and
F2er, obtained above, we find that
al = a 3 = a 3 =-.- 0, a4— (ms^4 - m4 — M1— MO g
- m4
1.41. Immediately after releasing the upper pulley,
the left load has a velocity v directed upwards,
while the right pulley remains at rest. The accel-
erations of the loads will be as if the free end of
the rope were fixed instead of moving at a con-
mg