[Physics] Distribution of weight of an inclined object

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We have a rod, 1m long and negligible height compared to the length, mass 10Kg uniformly distributed (center of gravity in the center). It is tilted theta degrees to the horizontal, and rests on two supports at two extreme ends, with one support higher than the other. What is the weight distribution (forces) on both supports?

If calculating, we find that although tilted, the lines of action of the upward reaction forces from both supports are equidistant from the center of gravity, so the forces on both supports should be equal, so that there is no net rotational moment. So, both supports should bear 50% of the weight, regardless of angle of incline.

However, from experience we know that the lower support bears greater weight than the upper support, and as the angle of incline increases to 90 degrees, all of the weight shifts to the lower support, with none on the upper support. This seems to defy logic because if the weight on both supports is unequal and their lines of action are equidistant from the center while tilted, then there should be a rotational moment and the object should rotate.

So what is the correct method of calculation of distribution of weight depending on the angle of incline?
Note that I am considering the case where the height is negligible, so the supports are at equal distance from the center, regardless of the angle of incline.

Best Answer

A rod fixed at the centre of two points

We refer to the figure above. Provided the rod is firmly fixed at ‘A’ and ‘B’, i.e. there is no sliding at ‘A’ and ‘B’, although it may swivel as on hinges, and gravity acts downwards in the figure (effective as force $F_C$), the supporting forces at ‘A’ and ‘B’ will always be $F_C/2$. Changes in θ do not change the relationship. Even in the vertical state when ‘A’, ‘B’ and ‘C’ lie on a vertical line, the rod is equally supported at ‘A’ and ‘B’. We apply some statics.

Setting A as the centre of rotation, the vertical forces are: $$F_BL\cos\theta = 0.5F_CL\cos\theta$$ $$therefore\qquad F_B = 0.5F_C$$

Similarly, setting B as the centre of rotation, the vertical forces will are: $$F_AL\cos\theta = 0.5F_CL\cos\theta$$ $$F_A = 0.5F_C$$

Now if, for example the fix point at A is a slot which allows the rod to not only swivel, but also to move freely in the longitudinal direction, then:

  • Nothing changes at $\theta$ = $0^{\circ}$ since the rod will experience the full normal (upward) force at ‘A’.
  • At $\theta$ = $90^{\circ}$ however, Fix point ‘A’ would merely keep the rod from rotating in case of a slight imbalance in sideways forces on the rod, which is common in the real world, and point ‘B’ would bear the full weight of the rod.
  • At $\theta$ between $0^{\circ}$ and $90^{\circ}$, provided friction at 'A' was not very high, 'B' would bear more than half the weight of the rod, since point 'A' would merely be required to prevent the rod from rotating left and falling towards the horizontal. The vertical component borne at 'A' would be less. As $\theta$ increased, 'B' would bear an increasingly greater burden, until it bore all the weight at $\theta$ = $90^{\circ}$. The ratio of the weight it bore would be proportional to the friction coefficient of the slot at 'A'. A high enough frictional coefficient, like the feet of those lizards that walk on walls, at 'A' would cause 'A' to bear half the weight even at $90^{\circ}$, just as in the original condition above.
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