Find the accelerations $a_{1}, a_{2}, a_{3}$ of the three blocks shown in figure (6-E8) if a horizontal force of $10 \mathrm{~N}$ is applied on (a) $2 \mathrm{~kg}$ block, (b) $3 \mathrm{~kg}$ block, (c) $7 \mathrm{~kg}$ block. Take $\mathrm{g}=10^{\mathrm{m}} / \mathrm{s}^{2}$.
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, $a_{1}=a_{2}=a_{3}=\frac{5}{6} \mathrm{~m} / \mathrm{s}^{2}$.
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