(^508) „„„„„ A Textbook of Engineering Mechanics
Since this body is moving downwards with an accleration (a) therefore force acting on the
body
= m 1 a ...(ii)
Equating the equations (i) and (ii),
m 1 g – T = m 1 a ...(iii)
Now consider the motion of body 2 of mass m 2 , which is moving horizontally. We know that
only force acting on it
= T (horizontal) ...(iv)
Since this body is moving horizontally with an acceleration (a), therefore force acting on
this body
= m 2 a ...(v)
Equating the equations (iv) and (v),
T = m 2 a ...(vi)
Adding equation (iii) and (vi),
m 1 g = m 1 a + m 2 a = a (m 1 + m 2 )
∴
1
12
mg
a
mm

• 
  

  Substituting this value of a in equation (vi),
  112
  2
  12 12
  mg mm g
  Tm
  mm mm
  =× =
  ++
  Example 25.4. Find the acceleration of a solid body A of mass 10 kg, when it is being pulled
  by another body B of mass 5 kg along a smooth horizontal plane as shown in Fig. 25.5.
  Fig. 25.5.
  Also find the tension in the string, assuming the string to be inextensible. Take g = 9.8 m/s^2.
  Solution. Given : Mass of body A(m 2 ) = 10 kg : mass of body B(m 1 ) = 5 kg and acceleration
  due to gravity (g) = 9.8 m/s^2.
  Acceleration of the body A
  We know that the acceleration of the body A,
  1 2
  12
  59.8
  3.27 m/s
  510
  mg
  a
  mm
  ×

  
  

  ++
  Ans.
  Tension in the string
  We know that tension in the string,
  12
  12
  510 9.8
  32.7 N
  510
  mm g
  T
  mm
  ××
  == =
  ++
  Ans.