Balancing Torques and Forces on this Beam

[Quantum Psi Inverted]

Homework Statement

Prove: If the system is motionless, then F_3a = F_2(a + b). In other words,
the torques (force times distance) around the left end cancel. And you can show
that they cancel around any other point, too.

Relevant Equations

F_3f(a)=F_2f(a+b)

In the given solution, it states that F_3f(a)=F_2f(a+b) is a "reasonable assumption". However, I don't see how we can just assume that. I can very clearly see that F_i for i=1,2,3 is likely proportional to a and b in some kind of way, but I can not clearly establish a line of reasoning for this. Can someone help me solidify that line of reasoning?

 

[Lnewqban]

The area of any of the rectangles formed by [Force] times [Its distance to fulcrum] is equal to each other, when all CW and CCW torques are balanced.
Note that F₃ has not been represented below because its distance to fulcrum is zero.

 

[256bits]

Homework Statement: Prove: If the system is motionless, then F_3a = F_2(a + b). In other words,
the torques (force times distance) around the left end cancel. And you can show
that they cancel around any other point, too.
Relevant Equations: F_3f(a)=F_2f(a+b)

F_3f(a)=F_2f(a+b)

What is said, is that,
The torque clockwise about point 1 is: F₃ a
The torque counter-clockwise about point 1 is : F₂ ( a + b )

Since the system for this problem is motionless ( meaning it is not accelerating ) ie static equilibrium, the torques clockwise and counter-clockwise must add up to 0 ( zero ), which gives,
F₃ a = F₂ ( a + b )

You do understand torques, and levers, and moments?

 

[haruspex]

The area of any of the rectangles formed by [Force] times [Its distance to fulcrum] is equal to each other, when all CW and CCW torques are balanced.
Note that F₃ has not been represented below because its distance to fulcrum is zero.

What is said, is that,
The torque clockwise about point 1 is: F₃ a
The torque counter-clockwise about point 1 is : F₂ ( a + b )

Since the system for this problem is motionless ( meaning it is not accelerating ) ie static equilibrium, the torques clockwise and counter-clockwise must add up to 0 ( zero ), which gives,
F₃ a = F₂ ( a + b )

You do understand torques, and levers, and moments?

I think you have both missed the point of the question.
The author is attempting to prove that F₃ a = F₂ ( a + b ) without assuming the standard torque balance equation. The method starts by assuming there is a relationship of the form $F_3f(a)=F_2f(a+b)$, for some function f. @Quantum Psi Inverted seeks justification of that assumption.

 

[Ibix]

Sum the moments of the three forces around a point some distance $x$ from the left hand end. What is the value of this sum? Differentiate with respect to $x$. What function $df/dx$ satisfies the resulting equation given that $a$, $b$ and $x$ could be any value? So what is $f$?

 

[haruspex]

Sum the moments of the three forces around a point some distance $x$ from the left hand end. What is the value of this sum? Differentiate with respect to $x$. What function $df/dx$ satisfies the resulting equation given that $a$, $b$ and $x$ could be any value? So what is $f$?

See post #4.

 

[Ibix]

See post #4.

I did read that. On re-reading the OP, it seems to me to segue from asking for a proof that "torques, whatever they may be, sum to zero" is reasonable into requesting a proof of $f(x)=x$. I've answered the second, but acknowledge that the question may be the first.

 

[haruspex]

I did read that. On re-reading the OP, it seems to me to segue from asking for a proof that "torques, whatever they may be, sum to zero" is reasonable into requesting a proof of $f(x)=x$. I've answered the second, but acknowledge that the question may be the first.

Unfortunately, we do not know how the author shows f(x)=x.

 

[Quantum Psi Inverted]

Unfortunately, we do not know how the author shows f(x)=x.

In particular the author shows that f(x)=x by first considering F_3f(a)=F_2f(a+b) and F_3f(b)=f(a+b). Combining these two with the fact that F_3=F_2+F_1, we result with f(a)+f(b)=f(a+b) over the positive reals, which is known as Cauchy's Functional Equation. The method that I used here to prove that f(x)=x diverges from the author however, since the author merely assumes that f is linear: It is well known that if f is continuous and bounded that f must be linear. Since f(x) is already known to be continuous and f(x)>0 must be true(truthfully it doesn't really matter as it gets cancelled out so we are able to state this on a case-by-case basis), We let f(x)=kx+q, leading us to f(x)=kx for some k, though this ultimately does not matter.

 

[Ibix]

@Quantum Psi Inverted - can you clarify for me whether you are interested in why $f(x)$ is $x$, or in why $\sum F_if(x_i)$ is a plausible measure of total torque?

 

[haruspex]

@Quantum Psi Inverted - can you clarify for me whether you are interested in why $f(x)$ is $x$, or in why $\sum F_if(x_i)$ is a plausible measure of total torque?

As I understand it, the latter is closer, but still not quite right.
There is no concept of torque at this stage. The question is what relationship between the forces and distances in the setup creates balance.

 

[Quantum Psi Inverted]

I think you have both missed the point of the question.
The author is attempting to prove that F₃ a = F₂ ( a + b ) without assuming the standard torque balance equation. The method starts by assuming there is a relationship of the form $F_3f(a)=F_2f(a+b)$, for some function f. @Quantum Psi Inverted seeks justification of that assumption.

Yes, thank you for poi

@Quantum Psi Inverted - can you clarify for me whether you are interested in why $f(x)$ is $x$, or in why $\sum F_if(x_i)$ is a plausible measure of total torque?

I would like to understand how the idea that F_3f(a)=F_2f(a+b) came to be. I'm very confused about what motivates this, as it's neither stated in the book, nor anywhere else. I'm not quite interested in why f(x)=x nor the sum of f_if(x_i); I can see those working out. However, it still stands that the conception of the F_3f(a)=F_2f(a+b) is still not very clear. What might lead one to see this relationship?

 

[haruspex]

I would like to understand how the idea that F_3f(a)=F_2f(a+b) came to be.

Maybe thus…
Let the leftmost point be constrained by a hinge (instead of by a given force).
Balance can occur with two forces applied normally to the rod.
It will occur when there is a certain relationship between the magnitudes of the two forces and their distances from the hinge.
The relationship is scale-invariant: doubling both forces maintains balance. (Likewise, doubling both distances.) You could derive that from dimensional analysis.

Of course, each of these steps is open to challenge, but perhaps each is reasonable.

 

[SammyS]

In particular the author shows that f(x)=x by first considering F_3f(a)=F_2f(a+b) and F_3f(b)=f(a+b).
. . .

Is there a typo in the last equation?

Should that be $F_3 \, f(b) = F_1 \, f(a+b)$ ?

 

[Ibix]

See what others come up with, but here's my stab at an argument.

Hopefully we can agree that the only available physical variables are the $n$ forces $F_1,\ldots,F_n$ and their positions $x_1,\ldots,x_n$. Hopefully we can agree that under different conditions the rod accelerates clockwise or counter clockwise or stays at a constant spin rate, and it would be useful if there were some function of the forces and positions that were correspondingly positive, negative, or zero.

What do we want such a function to do?

We'd like it to be linear in the forces. This way we don't have philosophical issues if we place two coins each of weight $W$ on a lever - does that count as two forces of magnitude $W$ or one of $2W$? If the function is linear in the forces we don't care, but if it is, say, quadratic then $(2W)^2\neq 2W^2$ and we have a while can of worms there. So we've argued that our function ought to be $F_1f_1(x_1)+F_2f_2(x_2)+\ldots+F_nf_n(x_n)$, where the functions $f_1,\ldots,f_n$ are not known and may differ.

We want our function to be zero when we have equal and opposite forces applied at the same point. For this, we need the same function of distance associated with for each independent force, so $f_1=f_2=\ldots=f_n=f$.

With that, I think I've made a reasonable argument that the quantity in question ought to be $\sum F_if(x_i)$, and with the additional assumption that $f(0)=0$, that's what you started with.

Does that make sense?

 

[Quantum Psi Inverted]

Is there a typo in the last equation?

Should that be $F_3 \, f(b) = F_1 \, f(a+b)$ ?

Oops, yes.

See what others come up with, but here's my stab at an argument.

Hopefully we can agree that the only available physical variables are the $n$ forces $F_1,\ldots,F_n$ and their positions $x_1,\ldots,x_n$. Hopefully we can agree that under different conditions the rod accelerates clockwise or counter clockwise or stays at a constant spin rate, and it would be useful if there were some function of the forces and positions that were correspondingly positive, negative, or zero.

What do we want such a function to do?

We'd like it to be linear in the forces. This way we don't have philosophical issues if we place two coins each of weight $W$ on a lever - does that count as two forces of magnitude $W$ or one of $2W$? If the function is linear in the forces we don't care, but if it is, say, quadratic then $(2W)^2\neq 2W^2$ and we have a while can of worms there. So we've argued that our function ought to be $F_1f_1(x_1)+F_2f_2(x_2)+\ldots+F_nf_n(x_n)$, where the functions $f_1,\ldots,f_n$ are not known and may differ.

We want our function to be zero when we have equal and opposite forces applied at the same point. For this, we need the same function of distance associated with for each independent force, so $f_1=f_2=\ldots=f_n=f$.

With that, I think I've made a reasonable argument that the quantity in question ought to be $\sum F_if(x_i)$, and with the additional assumption that $f(0)=0$, that's what you started with.

Does that make sense?

Yes, it does. Thank you!

 

[PeroK]

Homework Statement: Prove: If the system is motionless, then F_3a = F_2(a + b). In other words,
the torques (force times distance) around the left end cancel. And you can show
that they cancel around any other point, too.
Relevant Equations: F_3f(a)=F_2f(a+b)

View attachment 338042
In the given solution, it states that F_3f(a)=F_2f(a+b) is a "reasonable assumption". However, I don't see how we can just assume that. I can very clearly see that F_i for i=1,2,3 is likely proportional to a and b in some kind of way, but I can not clearly establish a line of reasoning for this. Can someone help me solidify that line of reasoning?

You could look at how to balance things if each force acts separately over a small angle. Consider a rod hinged between the ends with distances $a$ and $b$ to either end. The opposing forces act through a distance $a\theta$ and $b\theta$. The total KE gained by the rod is $F_aa\theta$ and $F_b b\theta$. In order to maintain a dynamic equilibrium you would need each force to be inversely proportional to the distance.

If the forces were inversely proportional, therefore, and the rod moved, then it would have no kinetic energy, which is something of a contraction. Hence in this case, and this case only, it must remain in static equilibrium.

Note that this argument applies independent of the mass distribution of the bar.