Problem — Picture hung on two wires

Problem

A picture of mass $m = 8{,}0\;\text{kg}$ hangs on the wall by two symmetric wires that each make an angle of $30^\circ$ with the ceiling. Find the tension in each wire. (Use $g = 9{,}8\;\text{m/s}^2$.)

Solution

Step 1 — Forces involved. On the picture act the weight $\vv{P}$ downward and the two tensions $\vv{T}_1$, $\vv{T}_2$ along the wires. By the symmetry of the system the two tensions have the same magnitude: $T_1 = T_2 = T$.

Step 2 — Decomposition. Each wire makes $30^\circ$ with the ceiling (horizontal), hence $30^\circ$ with the horizontal. The vertical component of each tension is thus $T\sin 30^\circ$; the horizontal components are equal and opposite and cancel.

Step 3 — Vertical equilibrium. The sum of the vertical components of the two tensions balances the weight:

$2\,T\sin 30^\circ = mg$

Step 4 — Tension. Solving for $T$ with $\sin 30^\circ = 0{,}5$:

$T = \frac{mg}{2\sin 30^\circ} = \frac{8{,}0\cdot 9{,}8}{2\cdot 0{,}5} = \frac{78{,}4}{1{,}0} = 78{,}4\;\text{N}$

Result.

$\ev{T_1 = T_2 \approx 78\;\text{N}}$

Note on the answer key $T \approx 49\;\text{N}$) is not consistent with the statement. With the angle of $30^\circ$ measured relative to the ceiling, the useful vertical component of each tension is $T\sin 30^\circ = 0{,}5\,T$, so $2\cdot 0{,}5\,T = mg$ gives $T = mg = 78{,}4\;\text{N}$. Wires only slightly inclined from the horizontal must indeed "pull hard" to support the weight. The correct value is $\approx 78\;\text{N}$: the $49\;\text{N}$ key value is a typo.

The answer printed in the margin (

Linked atoms

Argomenti: Statics and equilibrium Concetti: Tension · Weight force Competenze: Vector decomposition Metodi: Decomposition into Cartesian components Oggetti: Ideal rope