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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. Diagram given in question

Here is my diagram of forces

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?

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  • $\begingroup$ But at C, $W1$ is not acting right? Shouldn't we ignore it when resolving forces at C? $\endgroup$
    – creme332
    Commented Jan 25, 2022 at 12:10
  • $\begingroup$ The particle at C is in equilibrium so the vector sum of forces at that point should be 0. There are only 3 forces acting at C and since $W1$ does not act at C, I ignored it. $\endgroup$
    – creme332
    Commented Jan 25, 2022 at 12:16
  • $\begingroup$ Imagine a tug of war between A and B. You want to find the force at B. Do you ignore the force at A because it is not applied at B? $\endgroup$
    – garyp
    Commented Jan 25, 2022 at 13:02
  • $\begingroup$ Ok, I understand why the $W_1$ cannot be ignored. This would mean that the previous method of resolving forces horizontally and vertically at C will not work. Can we somehow modify the equations to include $W_1$ or should we completely give up on this method? The equation for resolving forces horizontally at C is unchanged but the equation for resolving vertically at C now becomes $T_1+W_1=W_2$. Is this correct? $\endgroup$
    – creme332
    Commented Jan 25, 2022 at 13:27

2 Answers 2

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Why is the first method of resolving forces at C incorrect?

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.

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  • $\begingroup$ By saying that the tensions at C can be resolved horizontally, you're also implying that $T_1=T_2$ right? (since the horizontal components must be equal) $\endgroup$
    – creme332
    Commented Jan 25, 2022 at 17:48
  • $\begingroup$ @Bunny Yes they would be equal. $\endgroup$
    – Bob D
    Commented Jan 25, 2022 at 18:17
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From your first two equations: $T_2 = T_1$ but $T_1$ = $(W_1)/(2 sin(30^o))$. The reason this does not work is that because of the weight $W_2$, the rod also exerts a force at (C) which is perpendicular to $T_2$.

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  • $\begingroup$ In that case, the method of taking moments at B is also wrong because it did not account for the additional force perpendicular to $T_2$. The additional force you have mentioned will also have a moment about B, right? $\endgroup$
    – creme332
    Commented Jan 25, 2022 at 17:17

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