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I am trying to determine whether the ground exerts a horizontal frictional force On the ladder (in addition to the normal contact force) when the surface is not frictionless and came up with the above thought experiment. If there exists such a force, then in its absence, the ladder should also slide in addition to falling.

If such a force does exist, what is causing it? I understand that friction on the ladder is the horizontal component of the reaction force to that exerted by the ladder in the ground , but there is only a vertical gravitational force on the ladder pushing it vertically into the ground, so shouldnt the reaction force be vertically upward only?

I have a feeling that this has something to do with moment, of which i have only the basic knowledge

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4 Answers 4

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I am trying to determine whether the ground exerts a horizontal frictional force On the ladder (in addition to the normal contact force) when the surface is not frictionless and came up with the above thought experiment. If there exists such a force, then in its absence, the ladder should also slide in addition to falling.

You are correct that, in the absence of friction on the ground, the ladder will slide on the ground in addition to falling. Since there would be no external horizontal forces acting upon the ladder, the center of mass (COM) can only have vertical motion. For the motion to be strictly vertical, the foot of the ladder has to slide. See Fig 1 below.

But we also know that if friction is present it will oppose the relative motion between the foot of the ladder and the ground. Friction will either prevent relative motion between the surfaces (static friction), or act in opposition to the sliding motion (kinetic friction). If the maximum possible static friction force is not exceeded, horizontal motion (sliding) will be prevented and the motion of the COM will follow a circular path, as shown in Fig 2 below.

I understand that friction on the ladder is the horizontal component of the reaction force to that exerted by the ladder in the ground, but there is only a vertical gravitational force on the ladder pushing it vertically into the ground, so shouldn’t the reaction force be vertically upward only?

Although the gravitational force acts vertically on the COM, the gravitational force can be resolved into components acting parallel and perpendicular to the ladder at the COM. See Fig 3 below. The component that acts parallel to and down the ladder can be resolved into normal and horizontal (friction) reaction forces.

By the way I came across the following answer by @ja72 to a similar question: Motion of the center of mass of a falling rod. Note the graphic provided by ja72.

Hope this helps.

enter image description here

enter image description here

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  • $\begingroup$ Thankyou for the answer. Cleared a lot of things up. i think my main problem here is that I am ignoring the forces actimg on each specific point and generalising he forces the force(s) acting on the centre of mass to every point. If understood this correctly, we can consider the gravitational force to act as a single downward force, Mg, which is resolved into 2 components. Now, since the line of action of Mgsin θ passes through the point of contact, it acts on it but not Mgcos θ. The force perpendicular to the ladder at the point of contact is actually mgcos θ $\endgroup$ Jun 10, 2020 at 4:38
  • $\begingroup$ Where m is the infinitismal mass of that point. As a result, the only force on the POC is actually Mgsin θ, which is again resolved and results in two components, one along the x axis which produces friction, and one along the y axis, which produces the normal contact force. If this is correct, whats the origin of F| at the point of contact. Shouldnt it be the reaction force to that infinitismal mgcos θ and thus be effectively 0? $\endgroup$ Jun 10, 2020 at 4:44
  • $\begingroup$ @OVERWOOTCH Oops. I was working on a couple of drafts and used the wrong Fig 3. See updated figure and text. Sorry for the confusion. $\endgroup$
    – Bob D
    Jun 10, 2020 at 15:40
  • $\begingroup$ Oh ok so “N” and “F” are just the Mgsinx resolved back right? Is my understanding about the rest as mentioned in the comments correct? $\endgroup$ Jun 10, 2020 at 16:28
  • $\begingroup$ @OVERWOOTCH Let’s just say your understanding and mine are the same and, hopefully, are correct. If my trig is correct, the reactions to mg sin θ are a normal force of N= mg sin$^2$θ and a required horizontal friction reaction of $F_{f} = \frac {mg}{2}$ sin 2θ. The interesting thing is if this is correct, then when θ is less than 45$^0$ the coefficient of static friction would need to be greater than 1 to prevent sliding. At 10$^0$ or less it would have to be greater than 5.67! $\endgroup$
    – Bob D
    Jun 11, 2020 at 13:47
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Ladder FBD

I have considered the vertical wall as frictionless for the sake of simplicity.

If we consider this ladder in equilibrium then the vertical wall exerts a reaction force $\vec{N_{1}}$ whereas the floor exerts a rection force $\vec{N_{2}}$.Earth exerts weight $m\vec{g}$ on the ladder on its centre of gravity.

If this ladder is in translation equilibrium then there must be a force directed opposite to $\vec{N_{1}}$ that's why floor must exert a horizontal force $\vec{f}$ on the ladder which we call as friction.

In terms of rotational equilibrium,let's consider the net torque about the centre of ladder then $N_{1}$ provides a torque directed outside the plane of screen whereas $N_{2}$ also provides a torque outside the plane then there must be a force which provides torque directed inside the plane of screen.This force is friction which is exerted by the floor on the wall.

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Yes, it has to do with moments.

For equilibrium the sum of the forces has to be zero and the sum of the moments about any point must be zero.

Assume the ladder is leaning against a frictionless wall at the left and on a floor with friction at the right. See the free body diagram of the ladder below. Take the point of contact with the floor (point B). For simplicity assume the total gravitational force on the ladder of W equals the weight of the ladder plus load acts at the center of the ladder.The gravitational force W on the ladder causes a counterclockwise moment about the floor at A. Assuming a frictionless wall a normal reaction force $R_W$ horizontal to the wall at A is required to provide an equal clockwise moment about the floor so that the sum of the moments about B is zero. Then, in order for the sum of the horizontal forces to be zero, you need a horizontal static friction force $F_f$ equal to the horizontal reaction force of the wall $R_W$.

Hope this helps.

enter image description here

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  • $\begingroup$ I already understand this but there is no wall in the thought experiment. I am just holding it an an incline over a frictionless surface and then releasing it from rest. I now realise that I should have Still pointed That out $\endgroup$ Jun 2, 2020 at 15:56
  • $\begingroup$ In that case, remove $R_W$ from the FBD. You will now have a net counter-clockwise moment about B, whether or not friction is present. The ladder will fall. Whether or not it slides will depend on if static friction is present. No static friction, the ladder will fall and slide to the right. With static friction, it will not slide to the right. The top of the ladder will not fall vertically downward, but move to the left of its original horizontal position. $\endgroup$
    – Bob D
    Jun 2, 2020 at 22:27
  • $\begingroup$ How can static friction exist regardless of the type of surface if there is only a vertical gravitational force pushing the ladder vertically into the ground? $\endgroup$ Jun 3, 2020 at 8:07
  • $\begingroup$ Good point. But let me ask you this. If there is no friction, do you think the bottom of the ladder will slide to the right? $\endgroup$
    – Bob D
    Jun 3, 2020 at 13:11
  • $\begingroup$ Yes, That is exactly what my question is. Now I FEEL that it should (maybe from everyday experiences) but when i think about it in terms of physics, I don't think it should since the net force on it is vertical $\endgroup$ Jun 3, 2020 at 18:13
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Frictionless surface can produce only reaction force in the direction perpendicular to the surface in the point of contact. Any tangential force produced by the surface is by definition force of friction. No force of friction can be perpendicular to the surface.

So the ground exerts only horizontal frictional force. The point of contact of ladder should be sticked to the surface, which means the net vertical force acting on the point of contact is supposed to be zero, otherwise ladder at the point of contact would jump upwards or fall under the ground. The vertical force is the force of constraint - this force is such that the point of contact remains the same - which is your requirement based on the physical situation. Because of its direction, it is called normal force.

So we fixed vertical force by our constraint that point of contact must remain at contact. The only component we are left with is horizontal component, which is by definition force of friction.

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  • $\begingroup$ If the ladder is falling, its center of mass is accelerating toward the ground, in which case the net vertical force is nonzero. This can occur whether there is friction or not. $\endgroup$ Jun 2, 2020 at 16:24
  • $\begingroup$ @NuclearWang I was not talking about ladder as a whole $\endgroup$
    – Umaxo
    Jun 2, 2020 at 20:38

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