(^520) „„„„„ A Textbook of Engineering Mechanics
Since the body is moving downwards with an acceleration (a), therefore force acting on
this body
= m 1 a ...(ii)
Equating equations (i) and (ii),
m 1 g sin α 1 – T = m 1 a ...(iii)
Now consider the motion of the body 2 of mass m 2 which is moving upwards along the
inclined plane BC. We know that the forces acting on it, along the plane, are T (upwards) and
m 2 .g sin α 2 (downwards). As the body is moving upwards, therefore resultant force
= T – m 2 g sin α 2 ...(iv)
Since the body is moving upwards with an acceleration (a) therefore force acting on
this body
= m 2 a ...(v)
Equating equations (iv) and (v),
T – m 2 g sin α 2 = m 2 a ...(vi)
Adding equations (iii) and (vi)
m 1 g sin α 1 – m 2 g sin α 2 = m 1 a + m 2 a
g (m 1 sin α 1 – m 2 sin α 2 ) = a (m 1 + m 2 )
∴ a =
()112 2
12
gmsin msin
mm
α− α

• 
  

  From equation (iii), we have
  T = m 1 g sin α 1 – m 1 a = m 1 (g sin α 1 – a)
  Substituting the value of a in above equation,
  ()112 2
  11
  12
  sin sin
  sin
  gm m
  Tmg
  mm
  ⎡⎤α− α
  =α−⎢⎥
  ⎢⎥⎣⎦+
  112 1112 2
  1
  12
  mm mmsin sin sin sin
  mg
  mm
  ⎡⎤α+ α− α+ α
  = ⎢⎥
  ⎢⎥⎣⎦+
  12 () 1 2
  12
  mm gsin sin
  mm
  α+ α

  
  


• 
  

  Note. For vertical surface for the mass m 1 , if we substitute the value of α 1 = 0 and α 2 = α in
  the above equations for a and T, the relations obtained will be the same as derived in Art. 25.5.
  Example 25.11. Two smooth inclined planes whose inclinations with the horizontal are 30º
  and 20º are placed back to back. Two bodies of mass 10 kg and 6 kg are placed on them and are
  connected by a light inextensible string passing over a smooth pulley as shown in Fig 25.15.
  Fig. 25.15.
  Find the tension in the string. Take g = 9.8 m/s^2.