Question

two blocks A and B of mass 10 kg and 40 kg are connected by an ideal string as shown in the figure neglect the masses of pulleys and effect of friction in the pulley between the blocks and the inclines.(wedge is fixed)

Answer 1

As per constraint relation if acceleration of block B is "a" then acceleration of block A is "2a"

Now as per force equation of B

$m_bgsin45 - 2T = m_ba$

$T - m_agsin45 = m_a*2a$

now solving above two equations for acceleration of the blocks

$m_bgsin45 - 2m_agsin45= (m_b + 4m_a)*a$

$a = \frac{(m_bsin45 - 2m_asin45)g}{m_b + 4m_a}$

now plug in all values

$a = \frac{(20\sqrt2 - 10\sqrt2)9.8}{40 +40}$

$a = 6.6 m/s^2$

so acceleration of block B is 6.6 m/s^2 and acceleration of block A is 13.2 m/s^2

Answer 2

Answer: Friction in the pulley is $6.6\frac { m }{ { s }^{ 2 } }$

Given data states that the constraint relation is followed that is acceleration of block B with mass as $m_2$ is a implying the acceleration of block A with mass $m_1$ be 2a. Thereby applying the equation of force, we get,

for block A , $m_2 g sin 45 - 2 T = m_2 a$

and similarly for block A be, $T - m_1 g sin 45 = m_1 \times 2 a$.

Both equation be solved and we get,

                  $m_2 g sin 45 -2 m_1 g sin 45 = ( m_2 + 4 m_1 ) \times a$

          $\Rightarrow a=\frac { g(m_{ 2 }sin45-2m_{ 1 }sin45) }{ (m_{ 2 }+4m_{ 1 }) }$

          $\Rightarrow a=\frac { 9.8(20\sqrt { 2 } -10\sqrt { 2 } ) }{ (40+40) }$

          $\Rightarrow a=6.6\frac { m }{ { s }^{ 2 } }$