Book plots position & force on the same graph, still gets right answer

[DrBanana]

Homework Statement

A uniform rod/ladder rests against a smooth wall, with its foot resting on a smooth plane that makes an angle ##\theta## with the wall. The wall is perpendicular to the horizontal x axis, and the ladder is at an angle ##\alpha## with this x axis. Show that ##tan\theta=2tan\alpha##

Relevant Equations

##\sum F = 0##, ##\sum M = 0##

In the picture you can see what I'm talking about. This is a simple problem to do if you can use moments, but alas my education system seems to have something against those.

##AB## is the uniform rod/ladder, G is its centre of mass, BE is the horizontal x axis, AD is the wall, BD is the smooth surface. S is the normal force on the foot of the ladder, so it's perpendicular to BD. R is the normal force from the wall. BC and GO are just geometric constructions, and they are both parallel to the wall.

My book claims that since the net force on the ladder is zero, the line of action of S will have to go through O. I want to know why that should be the case, where does it come from, is it actually a valid way to solve problems, if so then does it actually have a physical meaning, and if it does have a physical meaning, then how. And if it doesn't have a physical meaning, then why does this 'method' still work.


The proof continues by showing ##\angleBOC=\theta##, and then that O is the midpoint of AC (because G is the midpoinnt of AB and OG and BC are parallel). So ##tan\theta=\frac{BC}{\frac{1}{2}AC}=2tan\alpha##.


So in the same picture we represented forces, positions and also arbitrary (as in, with no physical existence) lines, and it all sort of worked out. For context, all of our statics proofs are like this, and it's a predicament that's shared by the bloke who made this post six years ago: https://physics.stackexchange.com/q...which-the-resultant-of-two-unlike-unequa?rq=1

 

[kuruman]

In my opinion this is the way to introduce the condition of equilibrium that the sum of the moments (or torques) must be zero without really saying so. This is why.

1. If the system is in equilibrium, the sum of all the torques acting on it is zero regardless of where you choose the reference point about which you calculate the torques.

2. Here you have three forces acting on the ladder. Given that the ladder is in equilibrium, all there forces, when extended along their line of action, must intersect at a single point. Clearly, the torque generated by each force about that point is zero. The condition that must be true for this to happen is the equation you are asked to show.

 

[DrBanana]

In my opinion this is the way to introduce the condition of equilibrium that the sum of the moments (or torques) must be zero without really saying so. This is why.

1. If the system is in equilibrium, the sum of all the torques acting on it is zero regardless of where you choose the reference point about which you calculate the torques.

2. Here you have three forces acting on the ladder. Given that the ladder is in equilibrium, all there forces, when extended along their line of action, must intersect at a single point. Clearly, the torque generated by each force about that point is zero. The condition that must be true for this to happen is the equation you are asked to show.

Are lines of action 'fixed' in space? Because if not, I can just move the point of intersection which would, I assume, prohibit me from showing the desired relation of the tangents of the angles.

 

[Lnewqban]

Are lines of action 'fixed' in space? Because if not, I can just move the point of intersection which would, I assume, prohibit me from showing the desired relation of the tangents of the angles.

I believe that the condition of static balance and the location of the center of mass of the ladder at its midpoint are what lock everything for you, regardless the value of both angles.

 

[DrBanana]

Ok so putting it all together (for now):

1. I found this: https://www.damtp.cam.ac.uk/user/po242/pdfs/Mech2015/L2.pdf, might be helpful to any viewers.

2. In these sorts of problems, you cannot move around forces willy nilly.

3. As a consequence of (2), any lines of action of any two non-parallel forces must intersect each other at a fixed point, a point that doesn't change throughout the problem. (This is all under the assumption that all the forces are coplanar)

4. The torques of those two forces about that point is zero, whether the body is in equilibrium or not.

5. If the body is in equilibrium, and three forces are acting on it, then you can take the lines of action of any two (non-parallel) forces, find where they intersect, and be sure that the third force's line of action goes through that point. Because if it doesn't, then ##\tau=r \times F \neq 0##, which implies that the net torque isn't zero, which is a contradiction.

6. Because of the above, you can relate lines of action of forces to normal geometric points on your plane, which might then allow you to use tricks from geometry.

7. Also, let the resultant of all the forces acting on a body be ##F## and if P and O are arbitrary points, then ##\tau_P=\tau_O + \overrightarrow{PO} \times F##, where ##\overrightarrow{PO}## is the vector going from P to O, and ##\tau_P## and ##tau_O## are the torques about points P and O respectively. If the body is at equilibrium, then ##F## is zero. This means ##\tau_P=\tau_O##. Let P be the point that was mentioned in (5), and O is any other point. This means that for a body in equilibrium, if the net torque is zero about any point, it is zero about all points.

 

[haruspex]

Ok so putting it all together (for now):

1. I found this: https://www.damtp.cam.ac.uk/user/po242/pdfs/Mech2015/L2.pdf, might be helpful to any viewers.

2. In these sorts of problems, you cannot move around forces willy nilly.

3. As a consequence of (2), any lines of action of any two non-parallel forces must intersect each other at a fixed point, a point that doesn't change throughout the problem. (This is all under the assumption that all the forces are coplanar)

4. The torques of those two forces about that point is zero, whether the body is in equilibrium or not.

5. If the body is in equilibrium, and three forces are acting on it, then you can take the lines of action of any two (non-parallel) forces, find where they intersect, and be sure that the third force's line of action goes through that point. Because if it doesn't, then ##\tau=r \times F \neq 0##, which implies that the net torque isn't zero, which is a contradiction.

6. Because of the above, you can relate lines of action of forces to normal geometric points on your plane, which might then allow you to use tricks from geometry.

7. Also, let the resultant of all the forces acting on a body be ##F## and if P and O are arbitrary points, then ##\tau_P=\tau_O + \overrightarrow{PO} \times F##, where ##\overrightarrow{PO}## is the vector going from P to O, and ##\tau_P## and ##tau_O## are the torques about points P and O respectively. If the body is at equilibrium, then ##F## is zero. This means ##\tau_P=\tau_O##. Let P be the point that was mentioned in (5), and O is any other point. This means that for a body in equilibrium, if the net torque is zero about any point, it is zero about all points.

All looks right, except I'm not sure what your point 2 means. Do you mean that in the linear balance of forces you can shift any force sideways, and since its vector doesn’t change the balance still works; but for balance of torques you can't do that?

 

[DrBanana]

All looks right, except I'm not sure what your point 2 means. Do you mean that in the linear balance of forces you can shift any force sideways, and since its vector doesn’t change the balance still works; but for balance of torques you can't do that?

Sorry it was late at night and I didn't expand on it. I've thought about it a bit more and it comes down to this: You can move around forces (not changing their direction of course) as long as they still act on the body, and then the body's translational motion won't change. However moving around forces can change the torque about a point. However from my last point I think you can say that in a statics problem you can move them around since the torque is going to be zero anyway.

But even though moving them around might be allowed, it might mean locking yourself out of the solution to the given problem. So practically, you can only move them around if it is useful to do so. That's my full thought process on this.

 

[haruspex]

I think you can say that in a statics problem you can move them around since the torque is going to be zero anyway.

Not so.

 

[Lnewqban]

My book claims that since the net force on the ladder is zero, the line of action of S will have to go through O. I want to know why that should be the case, where does it come from, is it actually a valid way to solve problems, if so then does it actually have a physical meaning, and if it does have a physical meaning, then how. And if it doesn't have a physical meaning, then why does this 'method' still work.

I would like to clarify that manipulation of the points of application of the forces is not the way to work this type of problems.
Perhaps you already understand it, but I would describe the physical phenomenon that the diagram is representing.

My understanding is that the ladder is in a slippery situation, because both, the vertical and the sloped surfaces that contact its ends have zero friction.
Based on that, remaining in a precarious non-moving position of equilibrium is very difficult to achieve (just imagine wheels attached to both ends of our ladder).

Our diagram shows that, if the surface BD were horizontal, our ladder would fall down while doing a clockwise rotation.
If instead, the slope of the surface BD were greater than represented, our ladder would also fall down, but while doing a counter-clockwise rotation.

There is only one exact slope for BD for which all the forces and moments acting on our ladder perfectly cancel each other, and the mentioned precarious static balance is achieved.

That balance happens only when the angle formed by force S, and its magnitude, are just right.
For this specific slippery problem, with one vertical and another sloped surfaces, each time the body in question is uniform, making the location of its CM midway its length, and no other external forces are acting on that body, that geometrical equation will be satisfied.