Engineer's Notebook §1.2

Equilibrium

A body stays at rest only if the forces on it cancel and their moments about any point cancel too:

ΣF = 0,ΣM = 0
F
force
N
M
moment
N·m
W
weight
N
T1
left cable tension
N
T2
right cable tension
N
θ1
left cable angle from horizontal
°
θ2
right cable angle from horizontal
°
H
horizontal pull on each anchor
N
A weight hangs from a knot held by two cables running up to a ceiling at different angles. Each cable is labelled with its tension. θ₁ θ₂ W = 500 N T₁ = 254 N T₂ = 389 N
Fig. 1.2-1Both tensions follow from ΣF = 0 at the knot.
T1 = W cos θ2 / sin(θ1 + θ2)254 N
T2 = W cos θ1 / sin(θ1 + θ2)389 N
H = T1 cos θ1194 N
Tmax389 N

Worked example

Uses the values in Fig. 1.2-1. Change them and the working updates.

Given
  • Whanging weight500 N
  • θ1left cable, from horizontal40°
  • θ2right cable, from horizontal60°
Find

The tension in each cable.

Solution
ΣFx = 0: T1 cos θ1 = T2 cos θ2 ΣFy = 0: T1 sin θ1 + T2 sin θ2 = W T1 = W cos θ2 / sin(θ1 + θ2) = 500 N × cos 60° / sin(40° + 60°) = 254 N T2 = W cos θ1 / sin(θ1 + θ2) = 500 N × cos 40° / sin(40° + 60°) = 389 N
Answer
T₁ = 254 N, T₂ = 389 N

Check: T₁ sin θ₁ + T₂ sin θ₂ = 500 N, the weight.

Related cases

  • Lever on a pivot

    F1a = F2b

  • Two equal cables at θ

    T = W2 sin θ

  • Beam on two supports, load at a

    RA = PbLRB = PaL

Assumptions
  • The cables are weightless, do not stretch and can only pull.
  • All three forces meet at the knot, so their moments about it are zero and ΣM = 0 holds automatically. ΣF = 0 does the work.
  • The anchors are rigid.

Derivation

Draw the free body of the knot: T1 pulls up and to the left at θ1, T2 up and to the right at θ2, and the weight W pulls straight down.

  1. Horizontal: T1 cos θ1 = T2 cos θ2.
  2. Vertical: T1 sin θ1 + T2 sin θ2 = W.
  3. Put T2 = T1 cos θ1 / cos θ2 into the vertical equation and multiply by cos θ2: T1(sin θ1 cos θ2 + cos θ1 sin θ2) = W cos θ2.
  4. The bracket is sin(θ1 + θ2), so T1 = W cos θ2 / sin(θ1 + θ2).

Swapping the subscripts gives T2. As both cables flatten, the sine in the denominator shrinks and the tensions grow without limit, which is why a loaded line can never be pulled perfectly straight.