Simple Statics Force diagram problem using moments.

Question

I have a statics problem in which my approach differs from the suggested solution, so I was wondering if anyone can point out my flawed logic. Below is the question.

Two identical uniform rods $ab$ and $bc$, each of length $2l$ and weight $W$, are smoothly hinged at $b$. Ends $a$ and $c$ rest on a smooth horizontal floor. The system is kept in equilibrium in a vertical plane by a light inextensible string joining $a$ to the midpoint of $bc$. the and between $ab$ and $bc$ is $2\alpha$. Show the tension of the string is $T = \frac{W}{4} \sqrt{9\tan^2\alpha + 1}$.

So below is the force diagram I have created.

The suggested solution for this problem resolves the $\vec{j}$ components of the system as:

$$ R_a + R_c = 2W $$

without taking into consideration the resolved tension force, $T$, of the light inextensible string.

Also taking moments about the point $a$ of the system abc the solution is:

$R_c \cdot 4l \sin \alpha = l \sin\alpha \cdot W + 3l\sin \alpha \cdot W $$

Once again no mention of the tension.

So I am I wrong to include the tension force for the $\vec{j}$ components of the system and the moments about $a$.

Any help appreciated.

Answer 1

It is true that the tension force $T$ in the string is a single force. But, this tension in the string exerts two equal forces of $T$ on the system at both ends, i.e. at $a$ and at $d$, as shown in the diagram.

If you still have doubts, remove the string and apply two unequal forces, say $T_1$ and $T_2$, but acting along the same straight line at $a$ and $d$. When you resolve the forces acting on the system in $\underline{i}$ and $\underline{j}$ directions assuming that the system is in stable equilibrium, you get the two equation shown below. $$R_a + R_c = 2W +T_2\sin(\alpha) – T_1\sin(\alpha)$$ $$T_1\cos(\alpha) - T_2\cos(\alpha) = 0 \qquad\rightarrow\qquad T_1 = T_2$$

Since $T_1 = T_2$, the first equation reduces to $$R_a + R_c = 2W$$

When you are taking moments about a given point, you can ignore all the forces passing through that point, because such forces generate zero moments. Therefore, you have to consider neither $T_1$ nor $T_2$ as they both pass through $a$.

Answer 2

Calling the forces indexed by point of application

$$ \cases{ T_a = T(\cos\beta,\sin\beta)\\ F_a = (0,R_a)\\ F_b = (0,R_b)\\ F_c = (H_c,V_c)\\ F_{g_1} = (0,-W)\\ F_{g_2} = (0,-W)\\ a = (0,0)\\ b = 2l(\sin\alpha,0)\\ c = 2l(\sin\alpha,\cos\alpha)\\ g_1 = \frac 12 (a+c)\\ g_2 = \frac 12(b+c)\\ } $$ with $\tan\beta = \frac 13\cot\alpha$ we have the equilibrium equations

$$ \cases{ F_a+T_a+F_{g_1}+F_b = 0\\ F_{g_1}\times (g_1-a)+F_c\times(c-a) = 0\\ } $$

and

$$ \cases{ -F_b+F_{g_2}-T_a + F_c = 0\\ (-F_c)\times(b-c)+(F_{g_2}-T_a)\times(g_2-c) = 0 } $$

Solving the corresponding linear system we got

$$ \cases{ R_a = \frac{W (\cos (\alpha -\beta )+6 \cos (\alpha +\beta ))}{6 \cos (\alpha ) \cos (\beta )-2 \sin (\alpha ) \sin (\beta )} \\ R_b = \frac{W (\sin (\alpha ) \sin (\beta )+5 \cos (\alpha ) \cos (\beta ))}{6 \cos (\alpha ) \cos (\beta )-2 \sin (\alpha ) \sin (\beta )} \\ H_c = \frac{2 W \sin (\alpha ) \cos (\beta )}{\sin (\alpha ) \sin (\beta )-3 \cos (\alpha ) \cos (\beta )} \\ V_c = -\frac{W \cos (\alpha -\beta )}{6 \cos (\alpha ) \cos (\beta )-2 \sin (\alpha ) \sin (\beta )} \\ T = \frac{2 W \sin (\alpha )}{3 \cos (\alpha ) \cos (\beta )-\sin (\alpha ) \sin (\beta )} \\ } $$

and using the relationship $\tan\beta = \frac 13\cot\alpha$ we have finally

$$ T = \frac{3}{4} W \tan \alpha \sqrt{\frac{\cot^2\alpha}{9}+1} = \frac W4\sqrt{9\tan^2\alpha+1} $$