A horizontal uniform bar of mass 2.9 kgkg and length 3.0 mm 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 mm from the other end. A monkey of mass 1.45 kgkg walks from one end of the bar to the other. Find the tension T1T1T_1 in string 1 at the moment that the monkey is halfway between the ends of the bar.

T₁= 14.62 N , T₂  = 28.87 N

Explanation:

Given that

M = 2.9 kg

L=3 mm

x=0.74 mm

m = 1.45 kg

Lets take tension in the string 1 is T₁ and tension in the string two is T₂

T₁  +  T₂ = (M+m) g

T₁  +  T₂ = 2.9 x 10 + 1.45 x 10

T₁  +  T₂ =43.5 N

 Taking moment about T₁

T₂  x 2.26 =(M+m) g x 1.5

T₂  x 2.26 = 43.5 x 1.5

T₂  = 28.87 N

T₁ = 43.5 - 28.87

T₁= 14.62 N

Therefore the tension will be  T₁= 14.62 N , T₂  = 28.87 N

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