In the arrangement shown in Fig. pulleys are small light and frictionless, threads are inextensible and mass of blocks A, B, and C is $m_{1}$ =5 kg, $m_{2}$ = 4 kg, and $m_{3}$ = 2.5 kg, respectively. Co-efficient of friction for both the planes is $μ$ = 0.50. Calculate acceleration of which block when system is released from rest.

a_{1} = 2 m s^{- 2}; B, a_{2} = 0; a_{3} = 2 m s^{- 2}

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a_{1} = 4 m s^{- 2}; B, a_{2} = 0; a_{3} = 2 m s^{- 2}

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a_{1} = 4 m s^{- 2}; B, a_{2} = 0; a_{3} = 3 m s^{- 2}

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a_{1} = 5 m s^{- 2}; B, a_{2} = 0; a_{3} = 2 m s^{- 2}

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Solution

The correct option is B $a_{1} = 4 m s^{- 2}; B, a_{2} = 0; a_{3} = 2 m s^{- 2}$
let the acceleration of $A, B & C$ be $a_{1}, a_{2}$ and $a_{3}$ .
For block $A$ ,

$T + w_{1} g sin 37^{\circ} - w_{1} g cos 37^{\circ} \times μ = m, a$
$T + 10 = 5 a_{1} - 1$

$\frac{m_{3} g}{2} = \frac{2 \cdot 5 g}{2} = 12 \cdot 5 N$

$M a x i m u m f o r c e f o r b l o c k B f i m a x = μ_{mag}$

$= 0.5 \times 4 g$
$= 20 N$

so block B will remain in $ses t, f_{max} > T_{max}$

$a_{2} = 0 \to 2$

for block C
$m_{3} g - 2 T = m_{3} a_{3}$
$25 - 2 T = 2 \cdot 5 a_{3} - 3$

by constraint relation

$a_{1} + a_{2} = 2 a_{3} - 4$

$o n s o l v i n g a b o v e 4 e q u a t i o n w e g e t a_{1} = 4 m / s^{2}$

$a_{2} = 0 m / s^{2}$
$a_{3} = 2 m / 5^{2}$

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Q. Two blocks of masses

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and

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(see fig.) are initially at rest on the floor. They are connected by a light inextensible string, passing over a light frictionless pulley. An upward force

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and

a_{2}

of the

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and

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masses, respectively, when

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110 N (g = 10 {m/s}^{- 2})

In the shown diagram m $_{1}$ = m $_{2}$ = 4 kg and m $_{3}$ = 2 kg.Coefficient of friction between m $_{1}$ and m $_{2}$ is 0.5. The mass m $_{1}$ is given a velocity v and it just stops at the other end of the mass m $_{2}$ in 2 sec. Let a $_{1}$ , a $_{2}$ and a $_{3}$ be the acceleration m $_{1}$ , m $_{2}$ and m $_{3}$ respectively, then: