# Simple statics: force due to pressure and tension in an equilateral triangle

A rope encloses a compressed two-dimensional gas with pressure $$p$$. First suppose the gas is constrained to be square (e.g. it's in a square container that can shrink and expand). Let $$T_1$$ be the constant tension of the rope. If the square has side length $$x$$, the force due to pressure on the right side is $$px$$. Assuming equilibrium, the tension on the rope equals this force and we have $$2T_1 = px$$.

Similarly for a circular shape, the diameter experiences force $$pd$$ to the right. Similar reasoning yields $$2T_2 = pd$$ where $$T_2$$.

The following diagrams are from Levi - The Mathematical Mechanic

Now let $$z$$ be the side length of an equilateral triangle. The following diagram considers equating the force due to pressure on a corner of the triangle with the tension of the rope. The force due to pressure is downward and the tension is upward. However, it suggests that the force due to pressure on a side of the triangle is $$pz/2$$ and not $$pz$$ which is what I expected.

Can someone help me make sense of why it's $$pz/2$$? I don't believe I left anything out of the problem and this result is presented without much discussion.

At the upper edge the force of the gas on the rope is $$pz$$ upwards. This must be balanced by tension forces $$T$$ at the left and right ends making angles of $$\theta=60^{\circ}$$ with the edge downwards. Therefore $$pz=2T\sin\theta$$ which works out the same as in diagram 3.13.
• For the upper edge, solving for $T$ yields $T = pz\sqrt{3}/3$, which is still a off by a factor of $1/2$. Shouldn't this value be the same? As for the corner, your explanation makes sense, but my intuition seems to be off about pressure. Why doesn't the entire left or right edge contribute to the bottom corner? Sep 17, 2022 at 18:29