Engineering Mechanics

(^510) „„„„„ A Textbook of Engineering Mechanics
Adding equation (iii) and (vi),
m 1 g – μ m 2 g = m 1 a + m 2 a
g (m 1 – μ m 2 ) = a (m 1 + m 2 )
∴
() 122
12

• m/s

gm m

a

mm

μ

=

+

From equation (vi) we find that

T = m 2 a + μ m 2 g = m 2 (a + μg)

Now substituting the value of a in the above equation,

() 12

2

12

gm m

Tm g

mm

⎡⎤−μ

=+⎢⎥μ

⎢⎥⎣⎦+

1212

2

12

mm mm

mg

mm

⎡⎤−μ +μ +μ

= ⎢⎥

⎢⎥⎣⎦+

12 ()

12

mm g 1

mm

+μ

=

+

Note. For smooth surface, if we substitute the value of μ = 0 in the above equations for a and

T, the relations obtained will be the same as we derived in the last article.

Example 25.5. Two blocks shown in Fig. 25.7, have masses A = 20 kg and B = 10 kg and the

coefficient of friction between the block A and the horizontal plane, μ = 0.25.

Fig. 25.7.

If the system is released, from rest, and the block B falls through a vertical distance of 1m,

what is the velocity acquired by it? Neglect the friction in the pulley and the extension of the string.

Solution. Given : Mass of block A (m 2 ) = 20 kg ; Mass of block B (m 1 ) = 10 kg ; Coefficient

of friction between block A and horizontal plane (μ) = 0.25 ; Initial velocity (u) = 0 (because the

system is released from rest) and vertical distance (s) = 1 m

Let v = Final velocity of the block A.

We know that acceleration of the block A,

( 12 ) ()

12

9.8 10 0.25 20

10 20

gm m

a

mm

−μ −×

==

++

= 1.63 m/s^2

and v^2 = u^2 + 2as = 0 + 2 × 1.63 × 1 = 3.26

∴ v = 1.81 m/s Ans.