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Why a tightrope can never be perfectly horizontal

The person's weight must be balanced by the upward parts of the rope's tension. The flatter the rope, the smaller that upward part, so a truly horizontal rope would need an impossible tension.

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A single force acting at an angle can be split into two parts at right angles, a horizontal component and a vertical component. This resolution of forces lets us handle slanted forces one direction at a time.

When someone stands on a tightrope, the rope sags and pulls up on them from both sides at a slight angle. Only the vertical components of these tensions support the person's weight. When the rope is steeply sloped, those vertical parts are large and easily carry the load. As the rope is pulled flatter, the vertical parts shrink, so the tension must rise sharply to keep supporting the same weight.

A perfectly horizontal rope would have no vertical component at all, so it could never balance the weight, no matter how tight. That is why every loaded cable, washing line or power line always sags a little.

In SPM you should resolve forces into components and apply equilibrium to problems like this.

Common misconceptions

  • A strong enough rope can be pulled perfectly straight under a load -> A horizontal rope has no vertical component to support weight, so some sag is unavoidable.
  • Tension is the same however the rope is angled -> As the rope flattens, tension rises sharply to keep the vertical component large enough.
  • Only the horizontal pull matters -> The vertical component is what balances the weight; both components matter.

Force and Motion II

The physics behind it

The key tool here is the resolution of forces: a single tension acting at an angle can be split into a horizontal component T cos θ and a vertical component T sin θ, where θ is the angle the rope makes with the horizontal. When someone stands at the middle of the rope, the two halves pull up and outward.

The horizontal components point in opposite directions and cancel, while the two vertical components together must support the person's weight, so 2T sin θ = W.

Take a performer of weight W = 600 N on a rope sagging at just θ = 5°. Then T = W ÷ (2 sin θ) = 600 N ÷ (2 × sin 5°) = 3440 N. The tension is almost six times the performer's weight, all because sin 5° is small.

As the rope is pulled flatter, θ approaches 0°, sin θ approaches 0, and the tension needed grows without limit. A perfectly horizontal rope would demand infinite tension, which no real rope can provide, so it must always sag.

See it in daily life

Watch a badminton net in a school hall and you will notice it never hangs dead straight; it always dips slightly in the middle even after being pulled tight, because the weight of the net itself needs a vertical component of tension to hold it up. A festival banner strung between two lamp posts across a kampung road behaves the same way: however hard it is tensioned, it settles into a shallow curve, and on a windy day it sags more as extra load is added.

A hammock tied between two coconut trees makes the effect obvious and comfortable, sagging deeply so that the steeply angled ropes easily carry your weight with a modest tension. If you tried to tie the hammock almost horizontal, the ropes would either snap or the knots would give, because the tension required would be enormous.

The gentle dip you always see in cables, nets and washing lines is not sloppiness; it is the only way the vertical components can add up to the load.

How this comes up in SPM

In Paper 2 this sits in the Force and Motion II chapter and is examined with command words such as explain, state, describe and determine. You may be asked to resolve a tension into components, to determine the tension in a cable supporting a load at a given angle, or to explain why a horizontal rope cannot support a weight.

A neat vector diagram, or the equation 2T sin θ = W together with the cancelling horizontal parts, is usually expected in the working.

The topic sits among close neighbours in the same chapter: the resolution of forces into perpendicular components, forces in equilibrium where a body is acted on by three forces, and the idea of a resultant force being zero for a body at rest. It also relates to the inclined-plane situations analysed by resolving weight along and across the slope.

Practise choosing sensible horizontal and vertical directions and writing one equation for each, so equilibrium problems with angled forces become routine rather than confusing.

Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)

Written by the spmphysics.com.my editorial team.· Updated 5 Sept 2026

Frequently asked questions

How is this examined in SPM?
It can appear in Paper 1 and Paper 2. We do not predict questions.
Why does the tension become so large when the rope is nearly flat?
The vertical components must always add up to the weight, following 2T sin θ = W. When θ is very small, sin θ is tiny, so T has to be very large to keep the product the same. That is why a nearly horizontal cable is under enormous tension.
Do the horizontal pulls just disappear?
No, they are still there on both sides, but they point in opposite directions and are equal in size, so they cancel out and produce no net horizontal force. Only the vertical components are left to balance the weight.
Would a stronger rope let me make it perfectly horizontal?
No. A perfectly horizontal rope has no vertical component at all, so it cannot balance any weight no matter how strong it is. A stronger rope can be pulled tighter and sag less, but some sag is always unavoidable.

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