Question

Three blocks, of masses m, = 2.0, m = 4.0 and
m, = 6.0 kg are connected by strings on a
frictionless inclined plane of 60°, as shown in the
figure. A force F = 120 N is applied upwards along
the incline to the uppermost block, causing an
upward movement of the blocks. The connecting
cords are light. The values of tensions T, and T, in
the cords are​

Answer 1

Answer :-

$T_1 = 19.98 N$

$T_2= 60.06 N$

Given :-

$m_1 = 2kg ,m_2 = 4kg ,m_3 = 6kg \\ F = 120 N$

To find :-

The value of $T_1 ,T_2$

Solution:-

• Take block $m_1$as system.

The force applied on block is :-

• Tension force upward.
• mg component as downward.

→$\mathsf{ T_1 - m_1g Sin\theta =m_1a }$

→$\mathsf{T_1 - m_1g Sin60^{\circ} = m_1a -----1)}$

• For block $m_2$

→$\mathsf{T_2 - T_1 - m_2g Sin60^{\circ} = m_2a------2)}$

• For block $m_3$

→$\mathsf{ F - T_2 - m_3g Sin60^{\circ} = m_3a -----3) }$

• Adding eq.1 ,eq.2 and eq.3

→$\mathsf{T_1 - m_1g Sin60^{\circ} + T_2 - T_1 - m_2g Sin60^{\circ} + F - T_2 - m_3g Sin60^{\circ} = m_1a + m_2a + m_3a}$

→$\mathsf{F - m_1g Sin 60 - m_2gSin60 -m_3g Sin60 = a ( m_1 + m_2 + m_3)}$

→$\mathsf{F - g Sin60 ( m_1 +m_2 + m_3 ) = a (m_1 + m_2 + m_3)}$

→$\mathsf{120 - 10 \times \dfrac{\sqrt{3}}{2}( 2 + 4 + 6) = a ( 2 +4+6) }$

→$\mathsf{120 - 5 \sqrt{3}( 12 ) = 12a}$

→$\mathsf{120 -60\sqrt{3} = 12 a}$

→$\mathsf{16. 07= 12 a}$

→$\mathsf{a = \dfrac{16.07}{12}}$

→$\mathsf{a = 1.33m/s^2 }$

hence,

The acceleration of the system is

The acceleration of the system is 1.33 m/s².

• Put the value of a in eq.1 and eq.2.

→$\mathsf{T_1 - m_1g Sin60^{\circ} = m_1a }$

→$\mathsf{T_1 - 20\times \dfrac{\sqrt{3}}{2}= 2 \times 1.33}$

→$\mathsf{T_1 -10\sqrt{3} =2.66}$

→$\mathsf{T_1 = 2.66 + 17.32}$

→$\mathsf{T_1 = 19.98 N}$

Now,

→$\mathsf{ F - T_2 - m_3g Sin60^{\circ} = m_3a}$

→$\mathsf{120 - T_2 - 6 \times 10 \times \dfrac{\sqrt{3}}{2}= 6 \times 1.33 }$

→$\mathsf{120 - T_2 -30\sqrt{3} = 7.98}$

→$\mathsf{120 - 51.96 -T_2 = 7.98}$

→$\mathsf{T_2 = 120 - 51.96 -7.98}$

→$\mathsf{T_2 = 120 - 59.94 }$

→$\mathsf{T_2= 60.06 N }$

hence,

The tension force is 19.98 N and 60.06 N.

Answer 2

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