Answer:
T₁ = 2.8125 N
Explanation:
The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:
∑M₂ = 0
M₂ = F*d
Where:
∑M₂ : Algebraic sum of moments in the the point (2) of the bar
M₂ : moment in the point 2 ( N*m)
F : Force ( N)
d : Horizontal distance of the force to the point 2 ( N*m
Data
mb = 3 kg : mass of the bar
mm = 1.5 kg : mass of the monkey
L = 3m : lengt of the bar
g = 9.8 m/s²: acceleration due to gravity
Forces acting on the bar
T₁ : Tension in string 1 (vertical upward)
T₂ : Tension in string 2 (vertical upward)
Wb :Weihgt of the bar (vertical downward)
Wm: Weihgt of the monkey (vertical downward)
Calculation of the weight of the bar (Wb) and of the monkey(Wm)
Wb = m*g = 3 kg*9.8 m/s² = 29.4 N
Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N
Calculation of the distances from forces the point 2
d₁₂ = (3-0.6) m = 2.4m : Distance from T1 to the point 2
db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2
dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2
Equilibrium of moments at the point 2 on the bar
∑M₂ = 0
T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0
T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0
T₁(2.4) =3*(1.5) + 1.5*(1.5)
T₁(2.4) =6.75
T₁ = 6.75 / (2.4)
T₁ = 2.8125 N
⇒ 
