Suppose I have a rigid beam attached to a wall, how can the reaction force on the point the beam is attached to generate a static equilibrium on the beam?


In particular, how can a force on the point of contact generate a moment that can cancel the moment due to the weight of the beam (that I imagine acting on the center of mass of the beam). If I sum the moments about that point any force there will have zero moment...

  • 3
    $\begingroup$ The wall must exert a moment on the beam; no single force is sufficient to keep the beam from rotating. Is this what you're asking about? $\endgroup$ Aug 27, 2018 at 19:08
  • $\begingroup$ @Chemomechanics Yes $\endgroup$
    – Julian
    Aug 27, 2018 at 19:27
  • $\begingroup$ The end of the beam must be loaded by the wall such that the turning moment is being balanced. This can be checked by putting different loads on the end strip embedded in the wall.. $\endgroup$
    – drvrm
    Aug 27, 2018 at 20:11

2 Answers 2


There is no "point" of contact. It is not possible for a horizontal beam with non-zero mass to be supported by a wall at a single point. The beam must have finite thickness and/or extend inside the wall. In both cases contact is made over an area. This enables forces to be exerted on the beam in opposite directions a finite distance apart, thus creating an anti-clockwise turning moment to balance the clockwise moment provided by the weight of the beam.

For example :

  1. If the beam does not penetrate into the wall but is attached to it by a bracket at the top then this bracket must pull the beam towards the wall while the wall pushes the beam away at the bottom.

  2. If the beam is embedded in the wall then on the underside more upward force is provided close to the face of the wall while on the upperside more downward force is provided far from the face of the wall.

  3. The beam is supported like a shelf from below, on a bracket which is fixed to the wall in more than one place and extends some distance from the wall. The bracket provides a downward force close to the wall and an upward force far from the wall.

  4. The beam is supported from above by a wire or rope attached to some point further up the wall, as in Horizontal rod attached to a wall. The wire provides a force upwards and inwards towards the wall, while the wall provides reaction on the beam away from the wall.

  • $\begingroup$ I see, thank you. In analogy, do you confirm that the classical example of two springs at the same distance from center, like in this question link is not possible? Either they must have different distance to balance torque or the weight will not be flat (and the springs will result in different elongations, thus making the usual Keff = K1+K2 wrong). Am I right? $\endgroup$
    – Julian
    Aug 28, 2018 at 4:51
  • $\begingroup$ @Julian I am not sure what restrictions you are imposing on the other question, the problem statement is very vague. If this is a classic example, do you have a link to a complete statement of the problem? eg Are the springs attached to the same upper point or different points? Do the springs have the same relaxed length? What do you mean by the weight being "flat"? $\endgroup$ Aug 28, 2018 at 5:42

This are the two option I thought: option 1) you exert from the external a force $\vec{f}$ that press the beam to the wall in order to cause a normal force from the wall to the beam ($\vec{F_{12}} = -\vec{F_{21}} = \vec{N}$) and to cause a static friction force $f_s = \mu_s \cdot \vec{N}$ or 2) the beam is not ideally 1d but 3d and the wall applies a torque $\vec{L}$ (the image in the circle is the zooming) that is against the torque of the gravitational force (applied in the center of gravity of the beam) that prevent the beam to fall even in absence of support ropes (the falling movement-trajectory is represented in green). It works if the wall has some glue.


  • $\begingroup$ Sorry, discard the option 1): it would fall because there is no support on the right side where the force is applied. It would have been correct if the force $\vec{F}$ was not parallel to the beam but with an angle such that the component of the force F parallel to the wall could have put to zero the whole torque of the bar. $\endgroup$
    – Costantino
    Aug 27, 2018 at 20:06

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