Tension Calculator

When an elevator accelerates upward, the cable must not only support the weight ($$mg$$) but also provide the upward net force ($$ma$$) for the acceleration. Thus $$T = m(g + a) > mg$$. Conversely, downward acceleration reduces the tension. This is why you feel heavier when an elevator starts going up and lighter when it starts going down.

As the rope angle approaches horizontal ($$\theta \to 0°$$), tension approaches infinity because $$T = mg/\sin\theta$$ and $$\sin\theta \to 0$$. This means it is physically impossible to pull a rope perfectly horizontal under load — there will always be some sag, no matter how much tension is applied.

Yes. In free fall ($$a = -g$$), vertical tension becomes $$T = m(g + (-g)) = 0$$ — this is the weightless condition. Tension is also zero in a slack rope. Ropes and cables can only pull, not push, so tension is always zero or positive.

For an ideal massless rope, tension is the same throughout its length. For a real rope with mass, tension varies along the length — it is greatest at the top (supporting the rope's own weight plus the load) and least at the bottom. For light ropes supporting heavy loads, the massless approximation is very accurate.

An ideal frictionless pulley redirects the tension without changing its magnitude. In a simple Atwood machine (two masses on a string over a pulley), the tension is $$T = \frac{2m_1 m_2 g}{m_1 + m_2}$$. More complex pulley systems (block and tackle) can multiply force by using multiple rope segments to share the load.