Normal reactions with moments A uniform ladder, $$AB$$, is leaning against a smooth vertical wall on rough horizontal ground at an angle of $$70°$$ to the horizontal. The ladder has length $$8\ \rm m$$, and is held in equilibrium by a frictional force of magnitude $$60\ \rm N$$ acting horizontally at $$B$$, as shown in the diagram. Write down the magnitude of the normal.

The normal reaction makes both A and B rotate, so I don't understand how we can ignore it by taking the moment at B OR A?

I have 60*8cos20 on the LHS (clockwise) = X*9.8*4sin20 (anti) but then I remember the normal force and it feels like I should add everything from the LHS to the RHS and I don't understand why that's wrong.

If I'm taking the moment at B, the 60N to the left frictional force still makes point -A rotate clockwise. and the Normal force, makes point A rotate anti-clockwise, so I still can't ignore it?

Similarly, there have been other seesaw questions, where I am told if I take the moment at the pivot then I can ignore the reaction at the pivot. Sure, but isn't there still a normal reaction from the seesaw back up into the (particle) weights on top of them?

Please can someone explain the simple thing I must be doing incorrectly, thanks!

• To the person voting to close as homework (and presumably downvoting the answers given at the time); note that this isn't asking for us to verify their work, or to solve a question for them. They are asking a conceptual question about the forces at a pivot, and only providing the homework question as an example of when this occurs. – JMac Apr 10 at 11:23

In situations like this, it doesn't mean the force no longer matters, it's just saying that a force doesn't have a torque about the point at which it is applied.

The torque of a force $$\mathbf F$$ applied at position $$\mathbf r$$ relative to the point you are calculating the torque about is given by $$\mathbf\tau=\mathbf r\times\mathbf F$$ If you are calculating the torque about the point where the force is applied, then the torque of that force is $$0$$ because $$\mathbf r=0$$

You might be confused because you are thinking "if a force causes rotation about some point then it has a torque." But the issue is you have to specify the point you are calculating the torque about first. You can't talk about torques without first specifying the point you are "looking at". In other words, the statement "A force has a torque of $$\tau\ \rm{N\cdot m}$$" is meaningless.

• Why the down vote? – Aaron Stevens Apr 10 at 11:32
• I think whoever VTC'd as homework decided to downvote us because they thought we were doing someone's homework. It's pretty annoying, since actually reading the questions or the answers show that this is a conceptual question and conceptual answers. – JMac Apr 10 at 11:36
• @JMac oh yeah, I knew that happens all of the time here, but I hadn't even thought of this as a homework problem, so I didn't even consider that possibility. – Aaron Stevens Apr 10 at 11:47
• Yeah, same. I figured if anything, someone would find a good dupe for this. – JMac Apr 10 at 12:03

The forces acting on B without pivot are unable to provide a force that allows the ladder to rotate around point B.

To apply a moment at point B, you need to have both a force, and a moment arm. The forces applied at point B have no moment arm, and therefore cannot contribute to the ladder rotating around that point (or, you could say when you do the math, their moment arm is zero, and therefore the moment they produce is 0).

That does not mean that the forces at point B cannot cause rotation around a different pivot point; just that those forces do not contribute to the pivoting around point B.

Consider a see-saw. At the pivot in the centre, obviously the force of the pivot itself is keeping the see-saw upright; but isn't actually doing anything to move the ends of the see-saw. You need people on the ends adding a force to make it rotate around the pivot. Consider the reference frame of you sitting on one end of the see-saw. The reaction force of the pivot and the other person on the see-saw are causing a moment on your end of the see-saw, which is rotating it relative to the surroundings. It's just also moving up and down, so it's not pure rotation.

• ok, its the normal reaction i dont understand though if im taking moments at point B.. then the normal reaction is pushing back (vertically right or parallel to the ladder?) it is point A that is touching the wall, so regardless, there is still a distance of the walls reaction force between point B, so I thought that this would be considered another moment.. but it is completely ignored. – ramose Apr 10 at 13:41
• @ramose It's completely ignored when checking the rotation around point B, because it is applied directly onto point B. When taking the moments around point A, you completely ignore the forces acting at point A, and you would consider the reaction forces at B. – JMac Apr 10 at 13:43
• I see the normal force being applied directly at point A, becasue it is A that is touching the wall, I cannot see why it is directly being applied to B? – ramose Apr 10 at 13:48
• @ramose Because B is touching the area where the force is applied. Think about it this way; if the two forces at B were the only forces acting on this (also getting rid of pseudo-forces like inertial effects on the centre of mass; which would make it tend to rotate when held at B); then what would the ladder to? It wouldn't rotate, it would only cause translation. This is why you can say it has no moment at point B; because in isolation, any force applied at that point will not cause rotation. – JMac Apr 10 at 13:52
• Ok, so If there is a reaction at two points, shared by the same normal force, I always ignore the force being applied at all points, even though I am taking the moment at an individual point? I could remember by rote memory, I dont understand it though but thanks. – ramose Apr 10 at 14:03

I have included the forces which are acting on the ladder in the left-hand diagram. As the net force acting on the ladder has to be zero in terms of magnitudes $$N_1= F$$ and $$N_2=W$$.

Now look at the torques acting on the ladder in a slightly different way.

The ladder is subjected to two couples $$W\,\frac x 2$$ anticlockwise and $$F\, y$$ clockwise which for static equilibrium must be equal in magnitude.

ie $$W \, \dfrac x 2 = F \,y$$ which is the same as taking moments about $$A$$ and as taking moments about $$B$$ as the magnitudes of $$N_1$$ and $$F$$ are the same.
Indeed taking moments about any point will yield the same result.

Your seesaw, with $$N=W_1+W_2$$ as the net force equals zero condition, is subject to two equal magnitude and opposite direction couples, $$W_1\,x$$ and $$W_2\,y$$, so that $$W_1\, x = W_2\, y$$ which is just the equation you might write when taking moments about the pivot $$C$$?
Again taking moments about any point will yield the same result.

• Thankyou, the problem I have is in this diagram I see N(sub1)*y normal reaction of the wall pushing back at the ladder (anti-clockwise) and I added that in my calculations, I know it's wrong, but i do not see why. – ramose Apr 10 at 13:45