Question

Find the accelerations $a_1,a_2,a_3$ of the three blocks shown in figure. If a horizontal force of $10N$ is applied on (i) $2kg$ block (ii) $3kg$ block, (iii) $7kg$ block (Take $g=10m/s^2$). (i.e. find $a_1,a_2,a_3$ in all three cases)

Answer 1

i) For $2kg$ block;

$10-f_1=2a_1$

$10-{\mu }_1\left(2g\right)=2a_1$

$10-0.2\times 20=2a_1$

$6=2a_1$

$a_1=3m/s^2$

for $3kg$ block;

$f_1-f_2=3a_2$

${\mu }_1\left(2g\right)-{\mu }_2\left(5g\right)=3a_2$

$0.2\times 20-0.3\times 50=3a_2$

$4-15=3a_2$

$a_2=-\frac{14}{3}$ which is not possible

So, $f_2$ act only equal to $f_2$ and $3kg$ block

$3kg$ block move together with $7kg$ lock.

Now, we have;

$F_1=7a$

$a=\frac{4}{7}m/s^2$

$a_2=a_3=\frac{4}{7}m/s^2$

ii) we know that,

$f_{2max}={\mu }_2\left(5g\right)=0.3\times 50=15N$

$f_{1max}={\mu }_1\left(2g\right)=4N$

Therefore, pseudo force acting on the $2kg$ block is less than $f_{1max}$

So, all the three blocks will move together with speed,

$a_1=a_2=a_3=a$

$F-m_{1}a=(7+3+2)a$

$10-2a=12a$

$14a=10$

$a=\frac{5}{7}m/s^2$

iii) All the blocks move together;

$a_1=a_2=a_3=a$

$F-m_{1}a-m_{2}a=(7+3+2)a$

$10-2a-3a=12a$

$17a=10$

$a=\frac{10}{7}m/s^2$