Rope tension holding a 20 kg sign with ropes at 45°

Answer: tension in each rope = 138.687 N (horizontal pull at each anchor 98.066 N).

What we know

Hanging mass: 20 kg
Rope angle above horizontal: 45°

Step by step

Given: m = 20 kg, θ = 45°
Formula: T = m·g / (2·sin θ)
Substitute: T = 20 × 9.80665 / (2 × sin 45)
Result: T = 138.687 N

All results

Tension in each rope	138.687 N
Horizontal pull at each anchor	98.066 N

Open the two-rope tension calculator with these numbers → and change any of them.

The physics

A weight hung from two symmetric ropes — a sign between two posts, a lamp on a cable, a load on a sling — is a static equilibrium problem. The two rope tensions must together support the full weight vertically while their horizontal components cancel, because the mass is not going anywhere.

Standard acceleration of gravity →