Problem — Bar suspended by two ropes

Problem

A uniform bar of mass $5{,}0\;\mathrm{kg}$ and length $2{,}0\;\mathrm{m}$ is suspended horizontally by two vertical ropes at its ends. A $20\;\mathrm{kg}$ weight hangs from the right end. Calculate the tension in the two ropes.

Solution

Two equilibrium conditions: the sum of the vertical forces is zero and the sum of the torques is zero. The bar’s weight $Mg = 5{,}0\cdot 9{,}81 = 49{,}1\;\mathrm{N}$ acts at the centre ($1{,}0\;\mathrm{m}$); the load $mg = 20\cdot 9{,}81 = 196{,}2\;\mathrm{N}$ at the right end.

Torques about the left end (eliminates $T_\text{sx}$): $T_\text{dx}\cdot 2{,}0 = Mg\cdot 1{,}0 + mg\cdot 2{,}0 = 49{,}1 + 392{,}4 = 441{,}5\;\mathrm{N\cdot m}$ $T_\text{dx} = \frac{441{,}5}{2{,}0} \approx \ev{221\;\mathrm{N}}$ Vertical equilibrium: $T_\text{sx} = Mg + mg - T_\text{dx} = 245{,}3 - 220{,}7 \approx \ev{25\;\mathrm{N}}$ The right rope, closer to the load, bears almost all the weight.

Links

Topics: Statics and equilibrium Concepts: Torque · Static equilibrium Skills: Torque balance Objects: Rod