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enter image description here

Referring to the image above:

  • The black lines are ONE continuous string whose ends are where the red thing is.
  • The red thing can tension the string and can measure the force used to tension the string. This force is 100 Newton.
  • The white lines are beams (metal) that make up a triangle.
  • The black line is slung around the sides of the triangle pulling the three triangle sides inwards.
  • Assumption: The tension in the string is distributed evenly throughout the string.
  • Assumption: all triangle angles and sides are equal.

Context: I want to calculate the bending stress experienced by each triangle side.

The question: What is the force that is pulling each of the beams (white triangle) sides inwards?

Is it as obvious as it first seems?

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1 Answer 1

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If all the triangles are equilateral triangles, then the symmetry means each side of the larger triangle will experience an equal force at the midpoint of each side and perpendicular to each side pointing inward:

$$F_{side} = 2Tcos\frac{\theta}{2}$$ where T is the tension in the string and \theta = 60°.

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  • $\begingroup$ Thanks for your answer. So if the force by which the black string is tensioned is 100 N then EACH side of the white triangle is experiencing a force equal to 2 * 100 * cos(60 / 2) = 173.2 N? $\endgroup$
    – aehhhhmm
    Commented Apr 10 at 17:07
  • $\begingroup$ If it is indeed 2 * 100 * cos(60 / 2) = 173.2 N per beam then the total force acting on all beams together would be 519.6 N. I thought that the reaction force (beams resisting the tension from the string) has to be exactly equal to the tension in the string (1/3 of tension in string per beam) for the system to be in an equilibrium. I don't understand where the extra force (beyond the 100N that has been put into the string initially) comes from. $\endgroup$
    – aehhhhmm
    Commented Apr 10 at 17:34
  • $\begingroup$ Forces have direction. You can't simply add the Force magnitudes. The sum of the Forces in the x and y directions equal zero. If you draw a free body diagram and sum the component Forces then it becomes clear $\endgroup$ Commented Apr 10 at 18:09

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