Question

In the given figure find the acceleration of the system and tension T1 and t2 in the strings assume that the table and the police are frictionless and the two strings are massless take gis equal to 9.8 M per second square

Answer 1

The acceleration of the system is 1.4m/s², T₁=22.4 N and T₂=33.6 N

The system moves with an acceleration a.

Write the force equation for the 2 kg mass. (m₁).

$T_1-m_1g=m_1a........(1)$

Write the force equation for the 8 kg mass (M) on the table.

$T_2-T_1=Ma........(2)$

Write the force equation for the 4 kg mass (m₂)

$m_2g-T_2=m_2a.......(3)$

Add equations (1), (2) and (3)

$(m_2-m_1)g=(m_1+M+m_2)a\\ a=\frac{m_2-m_1}{m_1+M+m_2} g$

Substitute 2 kg for m₁, 8 kg for M , 4 kg for m₂ and 9.8 m/s² for g.

$a=\frac{m_2-m_1}{m_1+M+m_2} g\\ =\frac{(4kg)-(2kg)}{(2 kg)+(8kg)+(4kg)} (9.8m/s^2)\\ =1.4m/s^2$

Calculate T₁ from equation (1).

$T_1-m_1g=m_1a\\ T_1=m_1(g+a)\\ =(2 kg)[(9.8m/s^2)+(1.4m/s^2)]\\ =22.4N$

Calculate T₂ from equation (3).

$m_2g-T_2=m_2a\\ T_2=m_2(g-a)\\ =(4kg)[(9.8m/s^2)-(1.4m/s^2)]\\ =33.6 N$

Thus the acceleration of the system is 1.4m/s², T₁=22.4 N and T₂=33.6 N

Answer 2

Answer:-

$a = 1.4 m/s^2$

$T_1 = 22.4 N$

$T_2 = 33.6 N$

Given :-

g = 9.8 m/s²

$m_1 = 2kg\\ M = 8kg \\ m_2 = 4kg$

To find :-

The acceleration of the whole system and the tension force.

Solution:-

Let the mass be $m_1 , M, m_2$ of each block respectively.

Let$T_1 and T_2$ be the tension force on string.

For bock $m_1$

acceleration is in upward.

→$\mathsf{T_1 - m_1 g = F_{Net} }$

→$\mathsf{T_1 -m_1g = m_1a}$

→$\mathsf {T_1 = m_1a + m_1 g}-----1$

For block $M$

acceleration is in right side.

→$\mathsf{T_2 - T_1 = F_{Net}}$

→$\mathsf{T_2 -T_1 = Ma}------2$

For block $m_2$

acceleration is in downward direction.

→$\mathsf{m_2g -T_2 = m_2a}$

→$\mathsf{-T_2 = m_2a - m_2 g}-----3$

• Adding equation 1 , 2 and 3.

We get,

→$\mathsf{ T_1 + T_2 -T_1 -T_2 = m_1a +m_1g + Ma + m_2a -m_2g }$

→$\mathsf{0 = m_1g -m_2g +m_1a + m_2a + Ma}$

→$\mathsf{m_2g -m_1g = a(m_1 + m_2 + M) }$

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

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

Put the given value,

→$\mathsf {a = \left(\dfrac{g\left(4-2\right)}{2+4+8}\right) }$

→$\mathsf{a = \left(\dfrac{2g}{14}\right)}$

→$\mathsf{ a = \dfrac{2 \times 9.8}{14}}$

→$\mathsf{a = \dfrac{19.6}{14}}$

→$\mathsf{a = 1.4 m/s^2}$

Now,

Tension forces on string will be :-

→$\mathsf{ T_1 = m_1 (a +g) }$

→$\mathsf{T_1 = 2 (1.4+ 9.8)}$

→$\mathsf{T_1 = 2 \times 11.2}$

→$\mathsf{ T_1 = 22.4 N}$

→$\mathsf{T_2 = m_2g - m_2 a}$

→$\mathsf{ T_2 = m_2 (g-a) }$

→$\mathsf {T_2 = 4 (9.8 -1. 4) }$

→$\mathsf{T_2 = 4 \times 8.4}$

→$\mathsf{ T_2 = 33.6 N }$

hence,

Accelerate of the system will be 1.4 m/s².

And tension forces on string will be 22.4 N and 33.6 N respectively.