Jointed rods

Suppose that three identical uniform rods of mass $M$ and length $l$ are joined together to form an equilateral triangle, and are then suspended from a cable, as shown in Fig. 94. What is the tension in the cable, and what are the reactions at the joints?

Figure 94: Three identical jointed rods.

Let $X_1$, $X_2$, and $X_3$ be the horizontal reactions at the three joints, and let $Y_1$, $Y_2$, and $Y_3$ be the corresponding vertical reactions, as shown in Fig. 94. In drawing this diagram, we have made use of the fact that the rods exert equal and opposite reactions on one another, in accordance with Newton's third law. Let $T$ be the tension in the cable.

Setting the horizontal and vertical forces acting on rod $AB$ to zero, we obtain

$$X_1 - X_3 = 0,$$ (491)

$$T + Y_1 + Y_3 - M g = 0,$$ (492)

respectively. Setting the horizontal and vertical forces acting on rod $AC$ to zero, we obtain

$$X_2 - X_1 = 0,$$ (493)

$$Y_2 - Y_1 - M g = 0,$$ (494)

respectively. Finally, setting the horizontal and vertical forces acting on rod $BC$ to zero, we obtain

$$X_3 - X_2 = 0,$$ (495)

$$-Y_2 - Y_3 - M g = 0,$$ (496)

respectively. Incidentally, it is clear, from symmetry, that $X_1 = X_3$ and $Y_1=Y_3$. Thus, the above equations can be solved to give

$$T = 3 M g,$$ (497)

$$Y_2 = 0,$$ (498)

$$X_1=X_2=X_3 = X,$$ (499)

$$Y_1=Y_3 = -M g.$$ (500)

There now remains only one unknown, $X$.

Now, it is clear, from symmetry, that there is zero net torque acting on rod $AB$. Let us evaluate the torque acting on rod $AC$ about point $A$. (By symmetry, this is the same as the torque acting on rod $BC$ about point $B$). The two forces which contribute to this torque are the weight, $M g$, and the reaction $X_2=X$. (Recall that the reaction $Y_2$ is zero). The lever arms associated with these two torques (which act in the same direction) are $(l/2) \cos\theta$ and $l \sin\theta$, respectively. Thus, setting the net torque to zero, we obtain

$$M g (l/2) \cos\theta + X l \sin\theta = 0,$$ (501)

which yields

$$X = -\frac{M g}{2 \tan\theta} = -\frac{M g}{2 \sqrt{3}},$$ (502)

since $\theta = 60^\circ$, and $\tan 60^\circ = \sqrt{3}$. We have now fully determined the tension in the cable, and all the reactions at the joints.