How strong does a cable need to be on a crane to lift a load without breaking? Let’s say the boom (AC) weighs 400 lbs, with center of gravity at its center, and supports a 2400 lb. load at C. Distances: AB = 10 ft AD = 14 ft DC = 6 ft BD changes as the boom is raised and lowered How strong does cable BD need to be to hold up the 400 lb. boom plus the 2400 lb. load?
Solution to the Cable Problem: It turns out that the force on the cable depends on the angle θ so we will have to leave θ as an unknown. (1) BD = √𝟏𝟎𝟐 + 𝟏𝟒𝟐 − 𝟐(𝟏𝟎)(𝟏𝟒)𝒄𝒐𝒔𝜽 or √𝟐𝟗𝟔 − 𝟐𝟖𝟎𝒄𝒐𝒔𝜽… law of cosines Let’s name angle BDA α (2) α = 𝐬𝐢𝐧−𝟏 (
𝟏𝟎𝒔𝒊𝒏𝜽
√𝟐𝟗𝟔−𝟐𝟖𝟎𝒄𝒐𝒔𝜽
) … law of sines
In order for the cable not to snap the moment (desire to rotate) that the weight of the load and boom cause clockwise around point A must be equal to the moment caused counter clockwise by the cable … engineers would call this a state of equilibrium. Moment = force x distance … the distance must be perpendicular to the force Note that the angle between the boom and the load will also be θ Moment of load L: The perpendicular distance between the force of load L and A would be 20sin(θ) and its force is 2400 lbs. (3) ML = 48000 sin(θ) Moment of boom G: The perpendicular distance between the force of the boom at G and A would be 10sin(θ) and its force is 400 lbs. (4) MG = 4000 sin(θ) Moment of cable BD: In this case it will be easier to find the component of the cables force that is perpendicular to the boom. Since we don’t know what it is let’s call it F. Its force perpendicular to the boom would then be Fsinα, but since α = 10𝑠𝑖𝑛𝜃
sin−1 (
√296−280𝑐𝑜𝑠𝜃
(5) MF =
) and its distance from A would be 14 ft.
𝟏𝟒𝟎𝑭𝒔𝒊𝒏𝜽 √𝟐𝟗𝟔−𝟐𝟖𝟎𝒄𝒐𝒔𝜽
Finally, MF = ML + MG ... or (6) F =
140𝐹𝑠𝑖𝑛𝜃 √296−280𝑐𝑜𝑠𝜃
𝟐𝟔𝟎𝟎√𝟐𝟗𝟔−𝟐𝟖𝟎𝒄𝒐𝒔𝜽 … 𝟕
where 0 < Ɵ < 180 Maximum tension in the cable is at Ɵ = 180 so F =
𝟔𝟐𝟒𝟎𝟎 ≈ 𝟕
8914 lbs.
= 48000 sin(θ) + 4000 sin(θ) … dividing sin(θ) from both sides and solving for F:
Here is a chart I hastily found online that lists breaking strengths for different diameter wire ropes sold by Ingersoll Rand: https://www.ingersollrandproducts.com/en-us/lifting-equipment-material-handling/Support/winch-selectionsupport/wire-rope-selection.html Notice that 8mm diameter would hold the weight but the chart also shows it would take 15mm diameter for a safe working load at 3.5:1 and even 19mm diameter for a safe working load at 5:1
