Find the acceleration of the block of mass M in the situation of figure (6-E10). The coefficient of friction between the two blocks is $μ_{1}$ and that between the bigger block and the ground is $μ_{2}$ .

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Solution

Let the acceleration of the block M is 'a' towards right. So, the block 'm' must go down with an acceleration 2a.
As the block m' is in contact with the block 'M' it will also hae acceleration 'a ' towards right, so it will experience two intertial forces as shown the free body diagram. From free body diagram...1

$R_{1} - m a = 0 \Rightarrow R_{1} = m a A g a i n, 2 m a + T - M g + μ_{1} R_{1} = 0 \Rightarrow T = m g - (2 + μ_{1}) m a \dots$
From free body diagram - 2,
$T + μ_{1} R_{1} + M g - R_{2} = 0$
Putting the value of $R_{1}$ from
R_2 = T + \mu _1 ma + mg \\

Putting the value fo T from
(ii),
$R_{2} = (M g - 2 m a - μ_{1} m a) = μ_{1} m a + M g + m g ∴ R_{2} = M g + m g - 2 m a$
Again, from the free body diagram- 2,
$T + T - R - M a - μ_{2} R_{2} = 0 \Rightarrow 2 T - M a - m a - μ_{2} (M g + m g - 2 m a) = 0$
Putting the values of $R_{1}$ and $R_{2}$ From (i) and (iii)
$2 T = (M + M) a + μ_{2} (M g + m g - 2 m a) . . .$
From equation (ii) and (iv), we have
2T = 2mg - 2(2 + \mu_1) ma\\
= (M + m) a + \mu_2 (Mg + mg - 2ma)\\
\Rightarrow 2mg - \mu_2 (M + m) g\\

$= a [M + m - 2 μ_{2} m + 4, + 2 μ_{1} m]$
$\Rightarrow a = \frac{[2 m - m_{2} (M + m)] g}{M + m [5 + 2 (μ_{1} - μ_{2})]}$

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