Free · Ideal mechanics
Ideal Rope Tension in Static and Accelerating Systems
Calculate supporting-rope forces or the acceleration and internal tensions of one to three frictionless connected objects.
- Formula & worked example
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Calculator inputs
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How to use this calculator
- Enter the known values in the units shown. Results update as you type.
- Where results are editable, change one to solve backwards. Lock a value to hold it fixed.
- Use the worked example to check the method. Reset restores the starting fields.
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Formula and method
A 10 kg stationary mass with g=9.80665 m/s² requires 98.0665 N tension in one vertical rope.
ΣF=ma; static rope components balance weight and each other.
Worked example
Enter these known values and leave the other values blank.
- Tension scenario
- Object is hanging
- Number of ropes
- 1
- Object's mass (m)
- 10 kg
- Gravitational acceleration (g)
- 9.80665 m/s²
- Weight (W)
- 98.07 N
- Tension in the rope (T)
- 98.07 N
Assumptions and limitations
- Ropes are massless, inextensible and taut. Objects accelerate together on a frictionless horizontal support or remain static when hanging.
- Support-rope angles are above horizontal. The exact two-vertical-rope case assumes equal load sharing, which statics alone cannot uniquely establish.
- For an angled pull, only its horizontal component causes horizontal acceleration. The model assumes contact with the support remains intact; it does not calculate lift-off or allowable rope loads.
- Default g is a reference acceleration, not a measurement of local gravity. Enter a known value when appropriate.
Common questions
Why are internal tensions different?
Each connecting rope accelerates the mass behind it, while the applied pulling rope carries the full applied force.
References
- OpenStax: tension and force components
- NIST Handbook 44: exact unit definitions
- Calculation definition and unit reference
Replace explanatory external images with original accessible descriptions preserving each scenario, rope or machine arrangement and every variant condition. Feet per second squared use exactly 0.3048 meters per second squared, replacing a rounded reciprocal factor. The pound-force uses the exact international pound and standard gravity instead of the rounded captured factor. The poundal uses exactly one international pound-foot per second squared. One kip is exactly one thousand pound-force, using the international pound and standard gravity.