Question

A horizontal uniform bar of mass 2.5 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 0.74 m from the other end. A monkey of mass 1.25 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

The tension T₁ in string 1 at the moment that the monkey is halfway between the ends of the bar is 34.68 N

Step-by-step explanation:

Given;

mass of the uniform horizontal bar, m₁ = 2.5 kg

length of the bar, L = 3.0 m

mass of the monkey, m₂ = 1.25 kg

distance from the right end of the second string, d = 0.74 m

For a body to remain in rotational equilibrium, the net external torque acting on it due to applied external forces must be equal.

∑τ = 0

For vertical equilibrium of bar-string system, in which T₁ and T₂ are the tension on both ends of the string;

T₁ + T₂ = (m₁ + m₂)g

T₁ + T₂ = (2.5 + 1.25) 9.8

T₁ + T₂ = 36.75 N

For rotational equilibrium when the monkey is halfway between the ends of the bar, take moment about the left end of the string.

(m₁ + m₂) L /2 = T₂(L - d)

(m₁ + m₂)0.5L = T₂( L - d)

(2.5 + 1.25)0.5 x 3 = T₂ ( 3 - 0.74)

4.6875 = T₂ (2.26)

T₂ = (4.6875) / (2.26)

T₂ = 2.074 N

Thus, T₁ = 36.75 N - T₂

T₁ = 36.75 N - 2.074 N

T₁ = 34.68 N