Chain on a slope remains still if its end points are of equal height?

[Happiness]

Let $\vec{a}$ be the gravitational acceleration along the curve.
Then $|\vec{a}|=a=-g\sin\theta$ and
$\vec{a}=a_x\,\vec{i}+a_y\,\vec{j}=-g\sin\theta\cos\theta\,\vec{i}-g\sin^2\theta\,\vec{j}$

My question is why does the solution below ignore the direction of $\vec{a}$ and calculate $F$ as $F=\int a\,dm$ instead of calculating $\vec{F}$ as $\vec{F}=\int a_x\,dm\,\vec{i}+\int a_y\,dm\,\vec{j}$?

Solution:

 

[Nugatory]

Can the chain move if its center of mass is not moving? Will the center of mass of the chain move downwards if there is no net vertical force on it? How would you go about calculating that net vertical force?

 

[Happiness]

Can the chain move if its center of mass is not moving? Will the center of mass of the chain move downwards if there is no net vertical force on it? How would you go about calculating that net vertical force?

No and no.

Net vertical force = $\vec{F_y}=\int a_y\,dm\,\vec{j}$
$=\int -g\sin^2\theta\,dm\,\vec{j}$
$=\int\frac{-gf'^2}{1+f'^2}\rho\sqrt{1+f'^2}\,dx\,\vec{j}$
$=-g\rho\int\frac{f'^2}{\sqrt{1+f'^2}}\,dx\,\vec{j}$

$\int\frac{f'^2}{\sqrt{1+f'^2}}\,dx$ is a function of $x$. Let's call it $H(x)$.

Then $\vec{F_y}=-g\rho\big(H(b)-H(a)\big)\,\vec{j}$, which may not be zero.

For example, suppose $f(x)=-x^4-2x^3+11x^2+12x$, which has a root at $x=0$ and another one at $x=3$. Let $u=f'^2=(-4x^3-6x^2+22x+12)^2$.

Then $\int_0^3\frac{f'^2}{\sqrt{1+f'^2}}\,dx=71.9$, calculated using
http://www.wolframalpha.com/input/?i=u=(-4x^3-6x^2+22x+12)^2,+integrate+u/sqrt(1+u)+dx+from+x=0+to+x=3

 

[Doc Al]

All you care about is the component of weight tangential to the curve, given by $g \sin\theta dm$. Any other component will be canceled by the normal force.

$\vec{a}=a_x\,\vec{i}+a_y\,\vec{j}=-g\sin\theta\cos\theta\,\vec{i}-g\sin^2\theta\,\vec{j}$

By breaking that tangential component into x and y components you are just making it harder to calculate. I haven't tried it, but done properly you should get the same result.

 

[Happiness]

All you care about is the component of weight tangential to the curve, given by $g \sin\theta dm$. Any other component will be canceled by the normal force.

By breaking that tangential component into x and y components you are just making it harder to calculate. I haven't tried it, but done properly you should get the same result.

As you said, the net force $dF$ on a mass element $dm$ is $g\sin\theta\,dm$. But force is a vector. So we should be considering $\vec{dF}$ together with its direction. The direction of $\vec{dF}$ is different for every mass element because the slope of the curve is different in different places in general.

I don't get why the solution treats force as a scalar and calculates $F=\int dF$ instead of $\vec{F}=\int\vec{dF}=\int dF_x\,\vec{i}+\int dF_y\,\vec{j}$.

 

[Happiness]

I found out, I believe, the reason why sometimes we can "treat force as a scalar". I then argued that we can't do so in the first post and hence the answer given is wrong. Please correct me if I am wrong.

The example below (attached) also treats force, in this case tension, as a scalar instead of as a vector, when it writes the equation
$T(\theta+d\theta)\leq T(\theta)+\mu N_{d\theta}$. --- (1)

I believe the correct equation (considering components in the tangential direction) should be
$T(\theta+d\theta)\cos\frac{d\theta}{2}\leq T(\theta)\cos\frac{d\theta}{2}+\mu N_{d\theta}$. --- (2)

However, since $d\theta$ is small, $\cos\,d\theta\approx 1$. And so equation (2) reduces to equation (1).

This means that we could "treat force as a scalar" if we consider the tensions between adjacent mass elements.

Let Figure 1.3 (attached below) represents the mass element of the motionless chain in the first post. (The mass element is shown as the arc in the figure.) The curve on which the chain lies is arbitrary. So Figure 1.3 should be rotated accordingly when analyzing different mass elements in various parts of the chain.

For each mass element to be at rest, the normal reaction force would adjust itself so that it balances the radial component (the one perpendicular to the slope) of the weight of the mass element $(mg\cos\theta$, where $\theta$ is the angle the slope makes with the horizontal$)$ and the radial components of the two tensions $(T\sin\frac{d\theta}{2})$. The radial components of the two tensions change according to their tangential components as they are related by a factor of $\tan\frac{d\theta}{2}$, while their tangential components would adjust themselves to balance $mg\sin\theta$, the tangential component of the weight of the mass element.

While the direction of the tensions are fixed by the slope of the curve, the magnitude of the tensions can be freely adjusted so as to balance $mg\sin\theta$. For the next mass element, the magnitude of one of the tension is fixed because of the action-reaction law, but we can still freely adjust the magnitude of the other tension to balance $mg\sin\theta$. This is true until we reach the mass element at the end of the chain, which is only experiencing one tension. Thus, this "treating force as a scalar" method fails, I believe, and the solution given in post #1 is not correct.

The "treating force as a scalar" method holds for the example below because all mass elements, including the mass elements at the ends of the rope, experience two tensions. (The ends of the rope are the points where the rope first touches the pole and where it last leaves the pole, and need not be the actual physical ends of the rope.)

We may make the method work for the chain in post #1 by changing the question such that a tension is applied to each end of the chain. Let's call the tensions $T_1$ and $T_2$. Following the given solution in post #1 (and also that in the example below), $T_1$ and $T_2$ must have the same magnitude. The chain would remain motionless if we remove $T_1$ and $T_2$ provided they are pointing in opposite directions. This is achieved if the endpoints have the same slope. So we expect the solution given in post #1 to be correct if the endpoints have the same slope (and the same height).

Consider $f=3x^5-5x^4+x^3+x^2$ with the endpoints $(0, 0)$ and $(1, 0)$, which have zero slope. However, it's calculated that $\vec{F_x}$ and $\vec{F_y}$ are not zero. What's wrong?

$\vec{F_x}=-g\rho\int_0^1\frac{f'}{\sqrt{1+f'^2}}\,dx\,\vec{i}=-0.00589g\rho\,\vec{i}$
http://www.wolframalpha.com/input/?i=v=(15x^4-20x^3+3x^2+2x),+integrate+v/sqrt(1+v^2)+dx+from+x=0+to+x=1

$\vec{F_y}=-g\rho\int_0^1\frac{f'^2}{\sqrt{1+f'^2}}\,dx\,\vec{j}=-0.121g\rho\,\vec{j}$
http://www.wolframalpha.com/input/?i=v=(15x^4-20x^3+3x^2+2x),+integrate+v^2/sqrt(1+v^2)+dx+from+x=0+to+x=1

But nonetheless they are values close to zero (compared to $\vec{F_y}=-71.9g\rho\,\vec{j}$ in post #3 where the endpoints do not have the same slope).

Is this because the "treating force as a scalar" method (and hence the conclusion that the chain is motionless) is only an approximation as the method approximates $\cos\,d\theta\approx 1$ in equation (2)?

For the case where the curve has the zero slope at its endpoints, in order for the mass element at each endpoint (which is horizontal) to be at rest when $T_1$ and $T_2$ (which are the applied forces at each ends of the chain) are equal in magnitude and opposite in direction, we would require that $F_1$ and $F_2$ (the internal forces) are also equal in magnitude and opposite in direction, where $F_1$ is the tension exerted on the mass element at the left endpoint by its adjacent mass element and $F_2$ is the tension similarly defined for the mass element at the right endpoint. But this seems unlikely to be satisfied for an arbitrary curve (arbitrary except for condition of having zero slope at its endpoints). Nonetheless, $F_1=F_2$ when the curve is symmetrical. Thus, could it be that the "treating force as a scalar" method is an approximation that becomes exact (for $\vec{F_x}$) when the curve becomes symmetrical?

Note that $\theta$ has a different meaning in the example above vs that in the motionless chain, where $\theta$ is the angle the slope makes with the horizontal. In the example above, integration is with respect to $\theta$ but in the motionless chain, it is done with respect to $x$.

The following is not needed for my question, but it's included for completeness.

 

[versine]

You can't break up the force into $$g\sin\theta\cos\theta$$ and $$g\sin\theta\cos\theta$$ because there is tension gradient. By Rolle's theorem there is a portion of the chain with zero slope and if you consider a mass element the tension gradient has a net vertical component.
You can integrate along the chain to show that the tension at the beginning = tension at the end in the x direction I think.

 

[N1206]

"The direction of dF→ is different for every mass element because the slope of the curve is different in different places in general."

That may be true, but it is also true that in order for the endpoints to be at the same height, all of the forces each mass element may have pulling to the left, will have corresponding ones pulling to the right. Think of it this way: If you broke the arbitrary curve down to infinitesimal bits and then re-arranged the bits by height, smoothly rising to the apex and then back to the origin, you'd have a symmetric curve--and any arbitrary curve with endpoints at the same height and the same distance apart would yield that same curve.

Instead of chain, picture a ball launched at a certain height and blown by a gusty wind over the time of its flight, and then caught at the same height it was launched at. Its vertical launch speed will equal its downward speed at catch time. That is straight ballistics. The ball will have some horizontal speed as an unequal force (the wind) acted on it in the horizontal plane. The chain, on the other hand, has no unequal forces acting on it other than gravity pulling it downward. Any force acting to pull it to the right because of the slope at any infinitesimal spot, will have some corresponding spot causing a leftward pull--and they will cancel out.

Only if the endpoints were NOT at the same height would you have any net horizontal forces.