A uniform rod BC of length $L$ and weight $W_1$ is hinged to a fixed point at $B$. A particle of weight $W_2$ is attached to the rod at $C$. The system is in equilibrium by a light elastic string in the same vertical plane as the rod. One end of the elastic string is attached to the rod at $C$ and the other end is attached to a fixed point $A$ which is at the same horizontal level as $B$. The rod and the string each make an angle of $30^o$ with the horizontal. (see diagram). Find tension in AC.
If I resolve the forces horizontally at C, I find that $T_1\cos 30=T_2\cos 30$
Then resolving the forces vertically at C, I find that $2T_1\sin 30=W_2$. Using the above equation as well gives $T_1=T_2=W_2$
However, if I take moments about B,
$(L/2)(W_1\cos 30)+(W_2\cos 30)(L)=(L\sin 60)(T_1)$
I find that $T_1=\frac{W_1}{2}+W_2$ which is the correct answer.
Why is the first method of resolving forces at C incorrect?
Best Answer
Resolving the horizontal components at C is correct as only the tensions have horizontal components acting on both C and the bar BC of which point C is a part.
But you can't only consider the vertical force acting directly on C because C is a part of bar BC which must also be in rotational equilibrium for C to be in equilibrium. For the Bar to be in rotational equilibrium it means the sum of the moments about point B must be zero.
In any case it should be obvious that $W_1$ has to influence the tensions, not only $W_2$.
Hope this helps.