2.3 Solutions
71
left, Fig. 2.22. Since the external force in the horizontal direction is zero, the
component of momentum along the x-direction must be conserved:
Fig. 2.22
(M + m)
dx
dt
− m
ds
dt
cos α = 0
( 1 )
Since the wedge is smooth, the only force acting down the plane is mg sin α
m
d 2 s
dt 2 −
d 2 x
dt 2 cos α
= mg sin α
(2)
Differentiating (1)
(M + m)
d 2 x
dt 2 − m cos α
d 2 s
dt 2 = 0
( 3 )
Solving (2) and (3)
d 2 s
dt 2 =
(M + m)g sin α
M + m sin
2
α
(acceleration of m)
d 2 x
dt 2 =
mg sin α cos α
M + m sin
2
α
(acceleration of M)
2.15 The lighter body of mass m 1 = m moves up the plane with acceleration a 1 and
the heavier one of mass m 2 = 3 m moves down the plane with acceleration
a 2 . Assuming that the string is taut, the acceleration of the two masses must
be numerically equal, i.e.
a 2 = a 1 = a. Let the tension in the string be T .
The equations of motion of the two masses are
F 1 = m 1 a 1 = ma = T − mg sin θ
(1)
F 2 = m 2 a 2 = 3ma = 3mg sin θ − T
(2)
Adding (1) and (2)
a =
g
2
√
2
(3)