If the coefficient of friction between A and inclined plane is 0.25 determine the tension on the strains and acceleration of A and B.

$$m_A = 3kg , m_B=5 kg , \mu =0.25$$

FBD of the blocks A and b are as shown.

Let T be the tension in the string. Let the acceleration of the system of blocks A and B be ‘a’

For block B,

$\sum F=ma \\ \therefore 5g-T=5a \\ \therefore T=5g -5a \rightarrow (1)$

Since block A does not move perpendicular to the inclined plane,

$\sum F_y=0 \\ \therefore N-3g \cos 45=0 \\ \therefore N=39 \cos 45 $

Friction force $=(F_d)=\mu \times N \\ = 0.25 \times 3g\cos 45 \\ =0.75g \cos 45 \rightarrow (2) $

Also , for block A $\sum F=ma \\ T+3g\sin 45 -F=3a \\ \therefore 5g -5a +3g \sin 45 -0.75 g \cos 45 =3a $ (from 1 & 2)

From (1) ,$ T=5g=5- 58.0794 \\ \therefore g(5+3\sin 45 -0.75\cos 45 )= 8a \\ \therefore a\dfrac {g(5+3\sin 45-0.75\cos 45)}8 =8.0822 m/s^2 \\ =8.6390 N$

Hence,

Tension on the string $=8.6390 N$

Acceleration of A and B $= 8.0822m/s^2 $