Question

A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the end of the bar, and string 2 is attached a distance 0.6 m from the other end. A monkey of mass 1.5 kg walks from one end of the bar to the other. Find the tension T1 in string 1 at the moment that the monkey is halfway between the ends of the bar.

Answer 1

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