Question

Find the accelerations $a_1,a_2,a_3$ of the three blocks shown in figure (6-E8) if a horizontal force of 10 N is applied on (a) 2 kg block, (b) 3 kg block (c) 7 kg Block. Take g = $10m/s^2$

Answer 1

Given
${\mu }_1=0.2{\mu }_2=0.3{\mu }_3=0.4$

(a) When the 10N forc is applied on 2 kg block, it experiences maximum frietional force.
$\mu R_1=\mu \times 2g=\left(0.2\right)\times 20=4N$ from the 3 kg block So, the 2 kg block experiences a net force
= 10-4 = 6N
So, $a_1=\frac{6}{2}=3m/s^2$
But for the 3 kg block, (Fig-3), the frictional force from 2 kg block (4N) becomes the driving force and the maximum frictional force between 3 kg and 7 kg blocks.
${\mu }_2=R_2=\left(0.3\right)\times 5kg=15N$
So, the 3 kg block cannot move relative to the 7 kg block, the 3 kg block and the 7 kg block both will have same acceleration $(a_2=a_3)$ which will be due to the 4N force because there is no friction form the floor.
$\therefore a_2=a_3=\frac{4}{10}=0.4m/s^2$
(b) When the 10N force is applied to the 3 kg block, it can experience maximum frictional force of 15 + 4 = 19 N From the 2 kg block and 7 kg block.

So, it can not move with respect to them. As the floor is frictionless, all the three bodies will move together.
$\therefore a_1=a_2=a_3=\frac{10}{12}=\left(\frac{5}{6}\right)$ m/s
(c) Similarly it can be proved taht when the 10N force is applied to the 7 kg block, all three blocks will move together.
Again $a_1=a_2=a_3=\left(\frac{5}{6}\right)$m/s