Engineering Mechanics

Chapter 25 : Motion of Connected Bodies „„„„„ 515

We know that the normal reaction on the inclined surface due to body of mass m 2 (as shown in
Fig. 25.11).

= m 2 g cos α
∴ Frictional force = μ m 2 g cos α
This frictional force will act in the opposite direction to the motion of the body of mass 2.
First of all, consider the motion of body 1 of mass m 1 , which is coming down. We know that
the forces acting on it are m 1 .g (downwards) and T (upwards). As the body is moving downwards,
therefore, resultant force
= m 1 g – T ...(i)
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 – T = m 1 a ...(iii)
Now consider the motion of the body 2 of mass m 2 , which is moving upwards on inclined
surface. We know that the forces acting on it, along the plane, are T (upwards), m 2 .g sin α (down-
wards) and force of friction μ.m 2 .g (downwards). As the body is moving upwards, therefore resultant
force

= T – m 2 g sin α – μ m 2 g cos α ...(iv)
Since this body is moving upwards along the inclined surface with an acceleration (a) there-
fore force acting on this body
= m 2 a ...(v)
Equating the equations (iv) and (v),
T – m 2 g sin α – μ m 2 g cos α = m 2 a ...(vi)
Adding equations (iii) and (vi),
m 1 g – m 2 g sin α – μ m 2 g cos α = m 1 a + m 2 a
g (m 1 – m 2 sin α – μ m 2 cos α) = a (m 1 + m 2 )

( 12 2 )

12

gm msin mcos

a

mm

−α−μ α

=

+

From equation (iii) we find that

T = m 1 g – m 1 a = m 1 (g – a)

Substituting the value of a in the above equation,

() 12 2

1

12

gm msin mcos

Tmg

mm

⎡⎤−α−μ α

=−⎢⎥

⎢⎥⎣⎦+

12 2

1

12

sin cos

1

mm m

mg

mm

⎡⎤−α−μ α

=−⎢⎥

⎢⎥⎣⎦+

1212 2

1

12

mm mmsin mcos

mg

mm

⎡⎤+−+ α+μ α

= ⎢⎥

⎢⎥⎣⎦+

12 ()

12

mm g1sin cos

mm

+α+μ α

=

+