Worked exercise — Beam with two loads hung from two ropes

Problem

A rigid beam $17\;\text{m}$ long and of mass $10\;\text{kg}$ hangs horizontally from two vertical ropes $R_1$ (on the left) and $R_2$ (on the right). A mass of $20\;\text{kg}$ rests on the beam $3\;\text{m}$ from the left rope, and another mass of $10\;\text{kg}$ rests at the right end. Find the tensions $R_1$ and $R_2$.

Solution

Step 1 — The weights involved. Weight of the $20\;\text{kg}$ mass: $P_A = 20\cdot 9{,}81 = 196{,}2\;\text{N}$; weight of the beam and of the $10\;\text{kg}$ mass: $98{,}1\;\text{N}$ each. The beam’s weight is applied at its centre, $8{,}5\;\text{m}$ from the left end.

Step 2 — Translational equilibrium (vertical $y$ axis):

$R_1 + R_2 - 196{,}2 - 98{,}1 - 98{,}1 = 0 \;\Rightarrow\; R_1 + R_2 = 392{,}4\;\text{N}$

Step 3 — Rotational equilibrium (pole at the left rope). Choosing the pole at the attachment point of the left rope, $R_1$ has zero lever arm and vanishes:

$-196{,}2\cdot 3 - 98{,}1\cdot 8{,}5 - 98{,}1\cdot 17 + R_2\cdot 17 = 0$

Step 4 — Calculating $R_2$ and $R_1$.

$R_2 = \frac{588{,}6 + 833{,}85 + 1667{,}7}{17} \approx 181{,}8\;\text{N}$

$R_1 = 392{,}4 - 181{,}8 \approx 210{,}6\;\text{N}$

$\ev{R_1 \approx 210{,}6\;\text{N}, \qquad R_2 \approx 181{,}8\;\text{N}}$

Note how the load placed closer to the left is supported mainly by $R_1$, consistent with intuition.

Linked atoms

Argomenti: Statics and equilibrium Concetti: Static equilibrium · Moment of a force · Tension · Weight force Competenze: Moment balance · Choice of pole Oggetti: Rigid body · Rod