(a) Find the acceleration of the blocks shown in Fig.
(b) Find the friction force between all contact surfaces.

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Solution

Let us first calculate the limitation and kinetic friction between various surfaces. Between 3 kg and ground:
$f_{l_{1}} = f_{k_{1}} = μ_{1} (2 + 3) g = 0.1 \times 5 g = 5 N$
Between 2 kg and 3 kg:
$f_{l_{2}} = f_{k_{2}} = μ_{2} g = 0.2 \times 2 g = 4 N$
Let us first assume that both blocks move together with common accleration a.
$a = \frac{5 + 10 - f_{k_{1}}}{2 + 3} \Rightarrow a = \frac{15 - 5}{5} = 2 m s^{_{2}}$
Now let us see how much friction force is required between 2kg and 3kg for common acceleration a.
$5 - f_{2} = 2 a \Rightarrow 5 - f_{2} = 2 \times 2 \Rightarrow f_{2} = 1 N$
Since $f_{2} < f_{l 2}$ ,
1. Both blocks will move together with common acceleration $a = 2 m s^{- 2}$
2. Friction between 2kg and 3 kg = $f_{2}$ = 1N Friction between 3 kg and ground = $f_{k_{1}}$ = 5N

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