Question

A horizontal uniform bar of mass 2.7 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 0.65 m from the other end. A monkey of mass 1.35 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:

14.36 N

Explanation:

$T_{1}$ = Tension in string 1

$T_{2}$ = Tension in string 2

$m_b$ = mass of the bar = 2.7 kg

$W_b$ = weight of the bar

weight of the bar is given as

$W_b = m_{b} g = (2.7) (9.8) = 26.46$N

$m_m$ = mass of the bar = 1.35 kg

$W_m$ = weight of the monkey

weight of the monkey is given as

$W_m = m_{m} g = (1.35) (9.8) = 13.23$N

Using equilibrium of torque about left end

$W_{m} (AB) + W_{b} (AB) = T_{2} (AC)\\W_{m} (AB) + W_{b} (AB) = T_{2} (AD - CD)\\(13.23) (1.5) + (26.46)(1.5) = T_{2} (3 - 0.65)\\\\T_{2} = 25.33$ N

Using equilibrium of force in vertical direction

$T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36$ N