2.3 Solutions
65
(b) Tension T 1 = Force acting on m 1
T 2 = m 1 a =
m 1 F
(m 1 + m 2 + m 3 )
(2)
where we have used (1).
Applying Newton’s second law to m 2
m 2 a = T 2 − T 1
or T 2 = m 2 a + T 1 = (m 1 + m 2 )a
T 2 =
(m 1 + m 2 )F
(m 1 + m 2 + m 3 )
(3)
where we have used (1) and (2).
2.2 (a) The equations of motion are
ma = mg − T
(1)
Ma = T − μMg
(2)
Solving (1) and (2)
a =
(m − μM)g
m + M
(3)
T =
Mm
M + m
(1 + μ)g
(4)
Thus with the introduction of friction, the acceleration is reduced and tension is increased compared to the motion on a smooth surface (μ = 0).
2.3 F max = (m 1 + m 2 )a
(1)
The condition that m 1 may not slide is
a = μg
(2)
Using (2) in (1)
F max = (m 1 + m 2 )μg
65
(b) Tension T 1 = Force acting on m 1
T 2 = m 1 a =
m 1 F
(m 1 + m 2 + m 3 )
(2)
where we have used (1).
Applying Newton’s second law to m 2
m 2 a = T 2 − T 1
or T 2 = m 2 a + T 1 = (m 1 + m 2 )a
T 2 =
(m 1 + m 2 )F
(m 1 + m 2 + m 3 )
(3)
where we have used (1) and (2).
2.2 (a) The equations of motion are
ma = mg − T
(1)
Ma = T − μMg
(2)
Solving (1) and (2)
a =
(m − μM)g
m + M
(3)
T =
Mm
M + m
(1 + μ)g
(4)
Thus with the introduction of friction, the acceleration is reduced and tension is increased compared to the motion on a smooth surface (μ = 0).
2.3 F max = (m 1 + m 2 )a
(1)
The condition that m 1 may not slide is
a = μg
(2)
Using (2) in (1)
F max = (m 1 + m 2 )μg