Answer:
a) T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] , b) T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]
c) x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex], d) m₂ = m₁ ( [tex]\frac{ L}{2d} -1[/tex])
Explanation:
After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act
a) The tension of string A is requested
The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive
∑ τ = 0
T_A d - W₂ x -W₁ L/2 = 0
T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]
b) the tension in string B
we write the expression of the translational equilibrium
∑ F = 0
T_A - W₂ - W₁ - T_B = 0
T_B = T_A -W₂ - W₁
T_ B = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] - g m₂ - g m₁
T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]
c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system
T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0
at the point that begins to rotate T_B = 0
g m₂ (d -x) - g m₁ (0.5 L -d) + 0 = 0
m₂ (d-x) = m₁ (0.5 L- d)
m₂ x = m₂ d - m₁ (0.5 L- d)
x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex]
d) The mass of the block for which it is always in equilibrium
this is the mass for which x = 0
0 = d - \frac{m_1}{m_2} \ \frac{L}{2d}
[tex]\frac{m_1}{m_2} \ (0.5L -d) = d[/tex]
[tex]\frac{m_1}{m_2} = \frac{ d}{0.5L-d}[/tex]
m₂ = m₁ [tex]\frac{0.5 L -d}{d}[/tex]
m₂ = m₁ ( [tex]\frac{ L}{2d} -1[/tex])
