Initially friction between mass A & table is just sufficient to prevent the motion. If an additional 1.25 kg is added to the 3.75 kg mass, find the acceleration of the masses.

Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 04

Years : DEC 2015

As initially system is in equilibrium, applying conditions of equilibrium.

Block A

$T - µ 25 g=0 \\ 3.75g - µ 25(9.81)=0 \\ µ=0.15 $

Block B

$3.75(g)-T=0 \\ T= 3.75g$

Now, after addition of mass to block B, system starts motion

Let acc of blocks be 'a'

Applying NSL to block A

$\sum Fx= max \\ T - 0.15(25g)=25a --------------(i) $

Applying NSL to block B

$\sum Fy= may \\ 5g - T=5a------(ii) $

Solving (i) & (ii)

$T=47 N$ & $a=0.409 m/s2 $

Acceleration of masses is 0.409 m/s2