Tension in Uniform Rope Shortcut

 

❓ Question

A uniform rope of mass $m$ is supported by two level pin supports as shown in the figure. The rope makes an angle of $30^\circ$ with the horizontal at each support.

Find the tension at the midpoint of the rope.

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✍️ Solution

Because the setup is symmetric, consider only the left half of the rope.

Forces acting on the left half

• Weight of half rope:

$$\frac{mg}{2}$$

acting vertically downward.

• Tension at the left support $T$, making an angle $30^\circ$ with the horizontal.
• Tension at the midpoint $T_{\text{mid}}$, which is horizontal due to symmetry.

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Step 1: Vertical equilibrium

Vertical component of support tension balances the weight of half rope.

$$T\sin30^\circ=\frac{mg}{2}$$

Since

$$\sin30^\circ=\frac12,$$
$$\frac{T}{2}=\frac{mg}{2}$$
$$\boxed{T=mg}$$

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Step 2: Horizontal equilibrium

The horizontal component of the support tension equals the midpoint tension.

$$T_{\text{mid}}=T\cos30^\circ$$

Substitute $T=mg$:

$$T_{\text{mid}} = mg\cos30^\circ$$
$$= mg\cdot\frac{\sqrt3}{2}$$

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✅ Final Answer

$$\boxed{T_{\text{mid}}=\frac{\sqrt3}{2}\,mg}$$

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Key Idea

At the lowest (mid) point of a symmetric hanging rope, the tangent is horizontal. Hence the tension there is purely horizontal, making it equal to the horizontal component of the end tension.

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