Tension in Two Ropes Holding a Weight

A weight hangs in static equilibrium from two ropes attached at different angles. Each rope pulls along its length, with a tension whose components balance gravity. To find the tensions, you set up two simultaneous equations from the equilibrium conditions $\sum F_x = 0$ and $\sum F_y = 0$.

The Setup

Let the two ropes make angles $\theta_1$ and $\theta_2$ from the vertical. The horizontal components of the two tensions point in opposite directions (one pulls left, the other right) and must cancel; the vertical components both pull up and together must support the weight.

$\sum F_x: \;\;-T_1\sin\theta_1 + T_2\sin\theta_2 = 0$

$\sum F_y: \;\;\;T_1\cos\theta_1 + T_2\cos\theta_2 - W = 0$

Step-by-Step Solution

Given: $W = 100\;\text{N}$, $\theta_1 = 30°$ (left rope), $\theta_2 = 45°$ (right rope), both measured from the vertical.

Find: $T_1$ and $T_2$.

---

Step 1 — Write the horizontal equilibrium equation.

Choose right as positive $x$. Rope 1 pulls left, rope 2 pulls right:

$-T_1\sin 30° + T_2\sin 45° = 0$

$T_2\sin 45° = T_1\sin 30°$

Solve for $T_2$:

$T_2 = T_1 \cdot \dfrac{\sin 30°}{\sin 45°} = T_1 \cdot \dfrac{0.5}{0.7071} \approx 0.7071\,T_1$

Step 2 — Write the vertical equilibrium equation.

$T_1\cos 30° + T_2\cos 45° - 100 = 0$

$T_1(0.8660) + T_2(0.7071) = 100$

Step 3 — Substitute $T_2 = 0.7071\,T_1$ into the vertical equation.

$0.8660\,T_1 + (0.7071)(0.7071\,T_1) = 100$

$0.8660\,T_1 + 0.5000\,T_1 = 100$

$1.3660\,T_1 = 100$

$T_1 = \dfrac{100}{1.3660} \approx 73.21\;\text{N}$

Step 4 — Solve for $T_2$.

$T_2 = 0.7071 \times 73.21 \approx 51.77\;\text{N}$

Hmm — let me double-check this. We had $T_2 = T_1 \sin 30° / \sin 45° = T_1 \cdot 0.7071$. So:

$T_2 = 73.21 \times 0.7071 \approx 51.77\;\text{N}$

Step 5 — Verify the equilibrium.

Horizontal:

$-T_1\sin 30° + T_2\sin 45° = -73.21(0.5) + 51.77(0.7071)$

$= -36.61 + 36.61 = 0 \;\;\checkmark$

Vertical:

$T_1\cos 30° + T_2\cos 45° = 73.21(0.8660) + 51.77(0.7071)$

$= 63.40 + 36.61 = 100.01\;\;\checkmark$

(Off by 0.01 due to rounding — exact within precision.)

---

Answer:

$\boxed{\;T_1 \approx 73.21\;\text{N},\quad T_2 \approx 51.77\;\text{N}\;}$

The rope at the smaller angle from vertical (30°) carries the larger tension, because it's pulling more "straight up" against gravity. The 45° rope pulls more sideways, so a larger fraction of its tension goes into the horizontal balance — and less into supporting the weight.

Try It

• Adjust the two angles with the sliders.
• Try setting both angles equal — by symmetry, the tensions become equal.
• Try setting one angle near 0° (rope nearly vertical) — that rope's tension grows close to $W$ and the other rope's tension shrinks.
• The HUD shows the live tensions and verifies that $\sum F_x = \sum F_y = 0$.