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# Calculate the permissible angle

This exercise shows how to calculate the permissible rope angle for a given load.

A load of G = 10 kN has to be attached to two suspension points with ropes. The maximum permissible tensile force of the rope is 20 kN. What value must the angle α have so the rope doesn't break?

## Solution

The following video is in german language.

Due to the symmetrical arrangement of the suspension, the look on one side is sufficient. It is a system of concurrent forces in a plane.

The vertical component of the rope force equals to the half of the load G:

$\tag{1} {F_{\mathit{1y}}}=\frac{G}{2}$

and the relationship to the rope force is

$\tag{2} {F_1} \sin{\left( \alpha \right) }={F_{\mathit{1y}}}$

$\tag{3} F_1 = \frac{F_1y}{\sin\left( \alpha \right) }$

The rope force shall not be greater than 20 kN, so the following applies

$\tag{4} 20 kN \geq \frac{F_1y}{\sin\left( \alpha \right) }$

$\tag{5} \alpha \geq \arcsin \left( \frac{5 kN}{20 kN } \right)$

$\tag{6} \alpha \geq 14.48°$

So the angle α must stay or be equal to 14.48°, or the rope will break. This was the way to calculate it for a symmetrical setup. We have another exercise, where an asymetrical layout is investigated, just take a look into the category Engineering Mechanics I in the menu at the side.

This was the exercise about how to calculate the permissible rope angle if you have a given load and maximum rope force. If you enjoyed it, please share it!