Question

Find the accelerations a₁, a₂, a₃ 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 = 10 m/s².

Figure

Answer 1

Given:
μ₁ = 0.2
μ₂ = 0.3
μ₃ = 0.4
Using the free body diagram, we have:



(a) When the 10 N force is applied to the 2 kg block, it experiences maximum frictional force.
Here,
μ₁R₁ = μ₁ × m₁g
μ₁R₁ = μ₁ × 2g = (0.2) × 20
= 4 N (From the 3 kg block)
Net force experienced by the 2 kg block = 10 − 4 = 6 N
∴ $a_1=\frac{6}{2}=3\mathrm{m}/{\mathrm{s}}^2$
But for the 3 kg block (Fig. 3), the frictional force from the 2 kg block, i.e, 4 N, becomes the driving force and the maximum frictional force between the 3 kg and 7 kg blocks.
Thus, we have:
μ₂ = R₂ = μ₂m₂g = (0.3) × 5 kg
= 15 N

Therefore, the 3 kg block cannot move relative to the 7 kg block.
The 3 kg block and the 7 kg block have the same acceleration (a₂ = a₃), which is due to the 4 N force because there is no friction from the floor.
$\therefore a_2=a_3=\frac{4}{10}=0.4\mathrm{m}/{\mathrm{s}}^2$

(b) When the 10 N force is applied to the 3 kg block, it experiences maximum frictional force of (15 + 4) N, i.e., 19 N, from the 2 kg block and the 7 kg block.
So, it cannot move with respect to them.
As the floor is frictionless, all the three bodies move together.
$\therefore a_1=a_2=a_3=\frac{10}{12}=\left(\frac{5}{6}\right)\mathrm{m}/{\mathrm{s}}^2$

(c) Similarly, it can be proved that when the 10 N force is applied to the 7 kg block, all three blocks move together with the same acceleration.
$\therefore a_1=a_2=a_3=\left(\frac{5}{6}\right)\mathrm{m}/{\mathrm{s}}^2$