Engineering Mechanics

Chapter 25 : Motion of Connected Bodies „„„„„ 511

Example 25.6. A block of wood A of mass 10 kg is held on a rough horizontal table. An
elastic string connected to the block passes over a smooth pulley at the end of the table and then
under a second smooth pulley carrying a body B of mass 5 kg as shown in Fig. 25.8.

Fig. 25.8.
The other end of the string is fixed to a point above the second pulley. When the 10 kg block is
released, it moves with an acceleration of g/9. Determine the value of coefficient of friction between
the block and the table.

Solution. Given : Mass of block A (m 2 ) = 10 kg ; Mass of body B (m 1 ) = 5 kg and accelaration

of block A =.
9

g

Let T = Tension in the string in N, and
μ = Coefficient of friction between block and table.
We know that the normal reaction on the horizontal surface due to body of mass 10 kg
R = 10 g
∴ Frictional force = μR = μ × 10 g = 10 μg
First of all consider the motion of block A, which is moving horizontally. We know that the
forces acting on it are T (towards right) and frictional force 10 μg (towards left). As the block is
moving towards right, therefore resultant force

= T – 10 μg ...(i)

Since the block is moving with an acceleration of (g/9) therefore force acting on it

10

10

99

gg

=×= ...(ii)

Equating the equations (i) and (ii)

T – 10μg =

10

9

g

Multiplying both sides by 2,

or 2 T – 20μg =

20

9

g

...(iii)

Now consider the motion of the block B, which is coming downwards. A little consideration
will show the acceleration of this block will be half of that of the block A i.e. g/18.We know that the
forces acting on it are mg = 5 g (downwards) and 2T (upwards). Therefore resultant force
= 5g – 2T ...(iv)
Since the block is moving with an acceleration of (g/18) therefore force acting in it

= ma =

5

5

18 18

gg

×= ...(v)