Question

A block of mass 5 kg is placed on a smooth table
and tied from both ends by two masses 3 kg and
2 kg by means of light, inextensible strings
passing through pulleys as shown. The tension in
the string connecting 5 kg and 3 kg is (g = 10 m/s2)
​

Answer 1

Answer:-

$\mathsf{T_1 = 27 N}$

Given :-

$m_1 = 3 kg$

$m_2 = 5 kg$

$m_3 = 2 kg$

g = 10 m/s²

To find:-

The tension force in the string connecting mass of 5 kg and 3 kg.

Solution:-

Let the tension force between mass of 5 kg and 3kg block be $T_1$ and tension force between 5kg and 2 kg block be $T_2$

Now,

Take $m_1$ block as system.

Force acting on $m_1$ block are :-

1) Tension force upwards.

2) Gravitational force downward.

Since, acceleration is downward,

→$\mathsf{m_1g - T_1 = m_1a}-----1)$

For block$m_2$,

acceleration is in right side,

→$\mathsf{T_1 - T_2 = m_2 a}-------2)$

For block $m_3$ acceleration is in upward direction.

→$\mathsf{T_2 - m_3g = m_3 a} ------3)$

Adding these all equations,

→$\mathsf{m_1g -T_1 + T_1 - T_2 +T_2 -m_3g = m_1 a + m_2a + m_3a}$

→$\mathsf{m_1 g -m_3g = m_1a + m_2a + m_3a}$

→$\mathsf{g(m_1 -m_3) = a(m_1 + m_2 + m_3)}$

→$\mathsf{ a = \left(\dfrac{g(m_1-m_3)}{m_1 + m_2 + m_3}\right) }$

Put the given values,

→$\mathsf{ a =\left( \dfrac{10 (3-2)}{3+5+2}\right)}$

→$\mathsf{a = \dfrac{10 \times 1}{10}}$

→$\mathsf{ a = \dfrac{10}{10}}$

→$\mathsf{a = 1 m/s^2 }$

The tension force acting between 5kg and 3 kg block is :-

→$\mathsf{m_1g -T_1 = m_1a}$

→$\mathsf{m_1g -m_1a = T_1}$

→$\mathsf{ m_1(g-a) = T_1}$

→$\mathsf{3 (10-1) = T_1 }$

→$\mathsf{3 \times 9 = T_1 }$

→$\mathsf{T_1 = 27 N}$

hence,

Tension force on string will be 27 N.

Answer 2

Hope that you will be able to understand