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Find the acceleration of the block 1 kg and 2 kg, such that Pulley A is having friction, and the coefficient of friction is $\mu =0.5$ . The string covers an angle of 180

With the pulley, A. Assume that there is no sliding between the pulley A and string.

(Use $g=9.8 \ m/s^2$ and $e^\pi =4.81$ )
Option: 1
$a_1= 1.946 \ m/s^2 ,\ a_2=3.893 \ m/s^2$

Option: 2
$a_1= 2.946 \ m/s^2 ,\ a_2=5.893 \ m/s^2$

Option: 3
$a_1= 4.946 \ m/s^2 ,\ a_2=9.893 \ m/s^2$

Option: 4
None of these

Answers (1)

best_answer

Let the acceleration of (2) is a

First, we have to find the relation between the acceleration of pulley A and Block (2)

So $a_p=\frac{a_a+a_2}{2}\\ -a_p=\frac{0-a}{2}\\ a_p=\frac{a}{2}....(1)$

Now for the block (1)

$m_1g-T_1-T_2=1*a_p$

By using equation (1)

$m_1g-T_1-T_2= \frac{a}{2} ... (2)$

For block 2

$T_1= 2a ... (3)$

Also, there is friction between pulley and string

So $T_2= T_1*e^{\mu \theta }\\ T_2= T_1*e^{0.5*\pi }\\ T_2=4.81 T_1 ... (4)$

Using equations (2), (3) and (4)

We get $m_1g-T_1-(4.81)T_1=\frac{a}{2}\\ m_1g- (5.81)T_1=\frac{a}{2} ....(5)$

And $T_1= 2a ... (3)$

Multiply with 4 on both sides of equation (5)

We get $4*m_1g- ( 23.24)T_1=2a$

Using equation (3)

$4*m_1g-=( 24.24)T_1 \\ 4g=( 24.24)T_1\\ T_1=1.61 N T_2=7.78 N$

Similarly

$T_1=2a\\ a=3.893 m/s^2$

So for a block of 2 kg

$a_2=3.893 m/s^2$

for a block of 1 kg

$a_1=\frac{3.893}{2}=1.946 m/s^2$

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Gaurav

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