This exercise investigates an asymmetrical load on two ropes and is about how to determine geometrical conditions for a force in a concurrent force system.

## Task

A weight load of G = 100 N shall be suspended from two ropes 1 and 2 in such a way that the dimensions a = 2 m and h = 2 m are met and the force in rope 1 reach the value S_{1} = 80 N. How long must the distance between the two supports, hereby referred as l, be?

## Solution

The following video is in german language. Please scroll down for the written solution.

The load G splits into the two rope forces S1 and S2.

At the connection point of the ropes a concurrent force system establishes. All forces act at the same central point. (The forces S_{1} and S_{2} are drawn side by side for readability, both are acting on the same point.)

The equilibria of forces for the x and y directions are established. For this purpose, the rope forces are broken down into their respective components.

\[\tag{1} \sum F_x = 0 = -S_{1x} + S_{2x} \]

\[\tag{2} \sum F_y = 0 = S_{1y} + S_{2y} -- G \]

The angle α can be calculated from the given lengths h and a.

\[\tag{3} \alpha = arctan \left( \frac{h}{a} \right) = 45° \]

With this angle and the specified force for rope 1, the x-components of the force in rope 1 can be calculated.

\[\tag{4} S_{1x} = S_1 cos \alpha \]

\[\tag{5} S_{1x} = 80 N \cdot cos 45° = 56.6 N \]

\[\tag{6} S_{2x} = S_{1x} = 56.6 N \]

And also the y component.

\[\tag{7} S_{1y} = S_1 sin \alpha \]

\[\tag{8} S_{1y} = 80 N \cdot sin 45° = 56.6 N \]

From equation 2 follows

\[\tag{9} S_{2y} = G -- S_{1y} \]

\[\tag{10} S_{2y} = 100 N -- 56.6 N = 43.4 N \]

The force in rope 2 is calculated using the Pythagorean theorem.

\[\tag{11} S_2 = \sqrt{S_{2x}^2 + S_{2y}^2} \]

\[\tag{12} S_2 = \sqrt{56.6^2 N^2 + 43.4^2 N^2} = 71.3 N \]

The angle β results from

\[\tag{13} S_{2x} = S_2 cos \beta \]

\[\tag{14} \beta = arccos \left( \frac{S_{2x}}{S_2} \right) = 37.5° \]

The angle β is related to h, l and a. The required length l can now be calculated from this.

\[\tag{15} \beta = arctan \left( \frac{h}{l-a} \right) \]

\[\tag{16} tan \beta = \frac{h}{l-a} \]

\[\tag{17} l = \frac{h}{tan \beta} + a = 4.6 m \]

So this was the way to determine geometrical conditions for an asymmetrical load on two ropes. Here we have a quite similar exercise about a symmetrical suspended load on two ropes.

Don’t miss the other exercises regarding Engineering Mechanics 1!