2.3 Solutions
73
Fig. 2.24
(c) Combining (1), (2) and (3) and simplifying
a =
g
√
3m 2 − m 1
M + 2(m 2 + m 1 )
2.17 Let m 1 = 2.1 kg and m 2 = 1.9 kg, Fig. 2.25. As the pulley is weightless the
tension is the same on either side of the pulley. Equations of motion are as
follows:
Fig. 2.25
m 1 a = m 1 g − T
(1)
m 2 a = T − m 2 g
(2)
Adding (1) and (2)
(m 1 + m 2 )a = (m 1 − m 2 )g
∴ a =
(m 1 − m 2 ) g
m 1 + m 2
=
(2.1 − 1.9)9.8
2.1 + 1.9
= 0.49 m/s
2
73
Fig. 2.24
(c) Combining (1), (2) and (3) and simplifying
a =
g
√
3m 2 − m 1
M + 2(m 2 + m 1 )
2.17 Let m 1 = 2.1 kg and m 2 = 1.9 kg, Fig. 2.25. As the pulley is weightless the
tension is the same on either side of the pulley. Equations of motion are as
follows:
Fig. 2.25
m 1 a = m 1 g − T
(1)
m 2 a = T − m 2 g
(2)
Adding (1) and (2)
(m 1 + m 2 )a = (m 1 − m 2 )g
∴ a =
(m 1 − m 2 ) g
m 1 + m 2
=
(2.1 − 1.9)9.8
2.1 + 1.9
= 0.49 m/s
2