Solve the following :

For $2 \mathrm{Kq}$;

$N_{1}=2 g=20 \mathrm{~N} ;$

$f f_{1}=\mu_{1} N_{1}=0.2 \times 20=4 \mathrm{~N}$

For $3 \mathrm{Ka}$ :

$N_{2}=N_{1}+3 g=20+30=50 N$

$f f_{2}=\mu_{2} N_{2}=0.3 \times 50=15 \mathrm{~N}$

For $7 \mathrm{Kg}$;

$N_{3}=N_{2}+7 g=50+70=120 \mathrm{~N}$

$f f_{3}=\mu_{3} N_{3}=0$

(a)

The bond between $2 \mathrm{Kg}$ and $3 \mathrm{Kg}$ breaks as applied force is more than limiting friction. Acceleration of $2 \mathrm{Kg}$ block $\Rightarrow F_{N}=10-f f_{1}$

$m \times a_{1}=10-4$

$2 \times a_{1}=6$

$a_{1}=3 \mathrm{~m} / \mathrm{s}^{2}$

Bond between $3 \mathrm{Kg}$ and $7 \mathrm{Kg}$ will not be broken by $4 \mathrm{~N}$ force so both will move together. Acceleration of $3 \mathrm{Kg}$ and $7 \mathrm{Kg}$ block $\Rightarrow F_{N}=4-0$

$m a_{2}=4$

$(3+7) a_{2}=4$

$a_{3}=a_{2}=0.4 \mathrm{~m} / \mathrm{s}^{2}$

(b) The bond between $3 \mathrm{Kg}$ and $7 \mathrm{Kg}$ will not be broken so they will definitely move together. Now individual acceleration

For $2 \mathrm{Kg}$;

$f f_{1}-0=2 \times a_{1}$

$4=2 \times a_{1}$

For $3 \mathrm{Kg}$ and $7 \mathrm{Kg}$

$F-f f_{1}-f f_{3}=(3+7) a_{2}$

$10-4-0=10 a_{2}$

$a_{2}=0.6 \mathrm{~m} / \mathrm{s}^{2}$

Acceleration of $2 \mathrm{Kg}$ block cannot be greater than $3 \mathrm{Kg}$ block. So, all will move together.

$F_{N}=m \times$ acceleration

$10-0=(2+3+7) a$

$a=\frac{5}{6} m / s^{2}$

So, $a_{1}=a_{2}=a_{3}=\frac{5}{6} \mathrm{~m} / \mathrm{s}^{2}$

(c) The bond between $3 \mathrm{Kg}$ and $7 \mathrm{Kg}$ will not be broken as applied force $(10 \mathrm{~N})$ is less than $f f_{2}=15 \mathrm{~N}$

. So, both of them moves together.

Now, individual acceleration

For 2Kg;

$f f_{1}-0=2 \times a_{1}$

$4=2 \times a_{1}$

$a_{2}=0.6 \mathrm{~m} / \mathrm{s}^{2}$.

Acceleration of $2 \mathrm{Kg}$ block cannot be greater than $3 \mathrm{Kg}$ block. So, all will move together.

$F_{N}=m \times$ acceleration

$10-0=(2+3+7) a$

$a=\frac{5}{6} m / s^{2}$

So, $a_{1}=a_{2}=a_{3}=\frac{5}{6} \mathrm{~m} / \mathrm{s}^{2}$.