Question

an 18.6 kg bar is 1.05 m long and is supported by two cables a and b. a 15.2 kg mass is hung from the bar at a distance of 13.5 cm from cable A. the system is in static equilibrium what's the tension in cables a and b

Answer 1

For a bar of mass 18.6 kg, There would be following forces exerted on bar:

1) Tension by a as well as b in upward direction(2T₁)

2) Weight of bar in downward direction(m₁g)

3) Tension of mass 15.2 kg hung from the bar in downward direction(T₂)

For Equilibirium:

2T₁ = m₁g + T₂ .................................................... 1

For mass 15.2 kg hung from the bar, following Forces will be exerted:

1) Weight of mass 15.2 kg in downward direction

2) Tension of rope hung from bar (T₂)

For Equilibirium

m₂g = T₂.................................. 2

From 1 and 2:

So,   2T₁ = 2T₁ = m₁g + T₂ = m₁g + m₂g = 9.8(18.6 + 15.2)

Tension in cable a and b(T₁₎ = 165.62 N