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.

  • $\begingroup$ you have to take friction into account. From the center of gravity of the rod, there are 4 forces bringing about torque, 2 normal reactions and 2 frictions(one for each surface) $\endgroup$ Commented Aug 3, 2017 at 6:09
  • $\begingroup$ @PrabhdeepSingh Only the normal reactions "bring about torque". The lines of action of the frictional forces act through the centre of mass? $\endgroup$
    – Farcher
    Commented Aug 3, 2017 at 9:49
  • $\begingroup$ Depends on the support mentioned. Think of the floor on which a slanted ladder rests $\endgroup$ Commented Aug 3, 2017 at 10:02
  • $\begingroup$ Related: Question about holding a couch at an angle $\endgroup$
    – Floris
    Commented Feb 5, 2018 at 12:26

2 Answers 2


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.

You correct that the normal reactions at each of the rod are going to be equal but you have neglected the fact that there are also frictional forces present.

Let the bar $AB$ have its centre of mass in the middle at $C$ and it has a weight $mg$.

The general diagram with friction forces acting at both ends of the rod is difficult to draw so I have drawn diagrams with friction only acting at one of the ends.

The left hand diagram shows a possible configuration with no friction at the point of contact $B$ and the right hand diagram with no friction at the point of contact $A$.

Hopefully this gives you an indication of how you might analyse a more general situation?

enter image description here

Referring to the left hand diagram.
The bar is subjected to three forces, its weight $mg$, a force $X_1$ at end $A$ and a normal reaction force $R_2$ at the end $B$ where there is no friction.

Since the bar is in static equilibrium the lines of action of those three forces must pass through a point $D$.

The force at end $A$ can be resolved into two components as shown in blue.
The normal reaction force $R_1$ and the frictional force $F_1$.
It so happens in this diagram that the magnitudes of these two forces are the same so the coefficient of static friction must be at least one.

Considering torques about the centre of mass $C$ shows that $R_1 = R_2$ and this will always be so irrespective of where the frictional forces act.
The weight $mg$ has also been resolved into two components shown in green to illustrate the equilibrium of the forces.

The right hand diagram can be similarly analysed but this time the lines of action of the three forces acting on the rod, $mg$, $R_1$ and $X_2$ meet at point $E$.

  • $\begingroup$ If there are 4 forces on the rod, and it is in equilibrium, then the 4 forces should all meet at one point. You cannot say that 3 meet at one point and 3 meet at another point. $\endgroup$ Commented Aug 3, 2017 at 9:43
  • $\begingroup$ @user1648764 There are three forces acting on the rod for one of which it is convenient to split into two components - normal reaction and friction. $\endgroup$
    – Farcher
    Commented Aug 3, 2017 at 9:46
  • $\begingroup$ I would assume that the forces X1 and X2 shown in the figure are vertically upwards, and add up to mg. That is the only way the rod can be in equilibrium. If X1 and X2 were both inclined to the left as shown, then there would be a net force to the left on the rod. Is my assumption correct? If correct, then we have 2 vertical upward forces X1, X2 equidistant from the centre, so they must be equal or they generate a net torque. So why is there more weight on the lower end? That was the question. $\endgroup$ Commented Aug 3, 2017 at 9:52
  • $\begingroup$ @user1648764 The diagrams represent two different situations. One situation has X1, R2 and mg acting as the three forces and the other situation has X2, R1 and mg acting. By more weight at the end I assume that by that you mean the support pushes harder. If that is the case it all depends on where and the magnitude of the frictional forces acting at the end. In the left hand diagram the support at the top (A) is "pushing" harder. $\endgroup$
    – Farcher
    Commented Aug 3, 2017 at 9:55
  • $\begingroup$ If the 3 forces on the rod are X1, X2 and mg, then X1, X2 and mg should meet at one point (or should be parallel). You cannot say that X1, a component of X2 and mg meet at one point and X2, a component of X1 and mg meet at another point. Perhaps this is the reason why your X1 and X2 are tilting to the left when they should be vertically upwards. $\endgroup$ Commented Aug 3, 2017 at 10:00

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