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Two identical uniform rods, each of weight $W$, are hinged together to form a structure which is resting on a rough floor as shown.

enter image description here

If the reaction forces acting on the structure by the floor are $R_1$ and $R_2$, which of the following shows the forces acting on the structure?

enter image description here

The answer given is $D$. The following is the answer provided:

Since the centre of gravity of the structure is at the mid-point of the line joining the centres of the two uniform rods, and the centre of gravity is nearer to the left rod, the vertical component of $R_1$ should be bigger than that of $R_2$ such that the sum of the vertical components of $R_1$ and $R_2$ is equal to $2W$.

I don't understand why the centre of gravity of the structure is nearer to the left rod. First, I find the centre of gravity of the two rods separately, then I join them using a straight line. The midpoint on the line is the centre of gravity of the structure, but it seems to me that it is nearer to the right rod.

What's wrong here?

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3 Answers 3

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The centre of gravity is closest to the side with most mass. Remember, you are only considering the horizontal 1 dimension.

  • What if the left rod was vertical? Then surely all it's mass would be concentrated at one point horizontally. In that case the centre of mass is surely being pulled towards that point. You would have the mass of the right rod spread out and then suddenly the same amount of mass at one single point at the left end. Naturally, the centre of gravity is pulled towards that left end.

  • If the left rod now starts tilting, becoming less and less vertical, its mass gradually spreads out horizontally. The centre of gravity then starts moving more towards the right. But as long as the left rod is still more vertical than the right rod, the mass is still more concentrated at the left end. So the centre og gravity is still more left.

The centre of mass is about where most mass is bundled up. And since you only consider the sideways position, you can forget about how high the beams reach and only consider how far they spread horizontally in this way.


A side note

You can solve this question without considering centre of gravity.

The arrow lengths represent the force magnitudes. Thinking of Newton's 1st law vertically, both weights must be balanced. That only happens on pictures C and D.

Thinking horizontally, you know that friction must be present, ruling out C. And the question is solved.

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  • $\begingroup$ No friction gives an unstable equilibrium situation. $\endgroup$
    – Farcher
    Commented Nov 16, 2016 at 13:36
  • $\begingroup$ That is very true, @Farcher. May I ask what you are commenting on? $\endgroup$
    – Steeven
    Commented Nov 16, 2016 at 15:06
  • $\begingroup$ It was your comment that friction must be present. $\endgroup$
    – Farcher
    Commented Nov 16, 2016 at 15:07
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The diagram can be redrawn as follows with the forces at the bottom of the ladders $R$ replaced by two normal reactions $N$ and two frictional forces $f$ which must be equal and opposite.

enter image description here

By inspection $a<c$.

Moments about $A \Rightarrow N_2(a+b+c) = W(a+b) + Wa = 2Wa + Wb$

Moments about $C \Rightarrow N_1(a+b+c) = W(b+c) + Wc = 2Wc + Wb$

As $a<c$ so $N_1>N_2$ and $R_1>R_2$

So you look for the only diagram showing $R_1>R_2$.

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The formulation of the answer is indeed confusing. What they mean to say is that the center of gravity is nearer to the point where the left rod rests on the floor than to the point where the right rod rests on the floor.

Therefore the vertical component of the reaction force on the left rod is larger.

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