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When talking about roller supports in a statics problem, we say that the roller prevents only the vertical translation of the beam, but not the rotation or the horizontal translation of it. So there is only one vertical reaction applied on the beam by the roller.

Also, we know a beam is connected to the roller using a pin, and the reaction force is the result of the pin that acts on the beam, as shown in this picture:

Now, suppose that I have a beam that is connected to a roller and I am going to push the beam in the horizontal direction (rollers don't prevent translation in the horizontal direction). What reactions should I expect?

I think that the pin is going to apply a force in the horizontal direction (due to the contact between the pin and the beam), which means that there is a reaction force in the horizontal direction. Is that true? If it is true, then why don't we consider horizontal reactions when analyzing rollers?

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  • $\begingroup$ I'm pretty sure rollers aren't pinned, that's pinned supports. This link shows the different supports and what they do (this was a first result google search, you may want to look at sources with more citation, but they should all still say this). Roller supports are able to move horizontally because they aren't pinned. $\endgroup$
    – JMac
    Jan 26 '17 at 17:52
  • $\begingroup$ @JMac: As you can see in the link you posted, "roller" really just means that it allows movement in one direction. Rollers can be either fixed or pinned. $\endgroup$
    – Wasabi ♦
    Jan 26 '17 at 18:00
  • $\begingroup$ I don't understand what you're imagining. You have a beam with how many supports? Are all of them rollers allowing horizontal movement? Or is one of them a pinned (non-roller) support, which allows for rotations but not displacements? $\endgroup$
    – Wasabi ♦
    Jan 26 '17 at 18:02
  • $\begingroup$ This link contains a picture ,,it explains what I am thinking about .. images.google.com.sa/… $\endgroup$
    – Eman.suradi
    Jan 26 '17 at 18:33
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    $\begingroup$ @Eman.suradi There would be very minimal force from that. There is nothing to resist the movement, so it would just move instead of creating a horizontal reaction force. The pin is just how the allow it to rotate instead of it being fixed directly to the plate. $\endgroup$
    – JMac
    Jan 26 '17 at 19:32
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If a bar is only supported by roller supports (which allow for horizontal displacements) and a horizontal force is applied, the bar will not suffer any internal forces because the entire structure (bar and supports) will suffer a rigid body motion sliding along the ground. At most, there'll be a slight force due to the friction of the rollers on the ground, but that's usually assumed to be zero.

Once you get into a dynamic system under a constant force (and therefore constant acceleration), then sure, you might get some internal inertial forces, but I don't think that's what you're asking about.

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If you have a horizontal beam with a roller on the bottom and you give it a push in the horizontal direction, it will accelerate in that direction. That is no longer a statics problem and you can't analyze it using a static model.

If you are told, "here is a statics problem," then what that means is that you should assume that the sum of forces and moments are zero in all directions. Can you make this assumption? Sure, we usually hope that the beams in our lives are not accelerating.

You might ask, "What if I have already pushed the beam and it's moving at a constant rate? Then it could still be a statics problem because there is no acceleration (sum of forces and moments can be zero)."

Well, if you stopped pushing it and it's moving at a constant rate, then the net horizontal force is zero. The only way that is possible is if there is no friction in the roller, which brings us back to the static model with zero horizontal reaction force.

Now if you add a horizontal reaction (friction force, in this scenario), then your beam has to accelerate opposite the direction of motion, and you're back to having a dynamics problem (not statics)!

Well, unless you keep pushing it, to balance out the friction. Now you're chasing this beam, pushing it down the road for who knows what reason, and in this case we can probably still use the principles of statics to model this problem but we probably wouldn't model it with a roller support when there is friction. Why? Because it would be even simpler to model it as a beam sliding down a flat surface, and that would give us the same result.

We care less about whether the actual beam has wheels or bearings or anything that rolls, than we care about choosing the simplest model that will allow us to make a correct decision.

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Step one should always be to define what your Free Body is and what question you are trying to answer. If you are looking at the bar pictured as a whole then you aren't concerned with how the roller is attached to the bar (pin). You could draw the free body diagram to include the pin, but everything is going to balance out and muck up your solution with unnecessary forces. Since the pin is internal to the system, no need to include it. It's simply a bar attached the the ground or support with a roller (no horizontal reaction or moment). If you do care about the pin, draw that free body diagram seperately to understand the forces. Simplify!

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  • $\begingroup$ "Since the pin is internal to the system, no need to include it. " I didn't understand what do you mean by that ? .. I understood from your answer -in general- that we draw the FBD for the bar pictured as a whole in order to find the reactions for example , Did I understand you ? $\endgroup$
    – Eman.suradi
    Jan 30 '17 at 12:26
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In a rigid system, the roller transfers lateral forces to the pinned fixings and can only resist perpendicular forces.

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