Problem 2B-17***1/2: A massive rope

A uniform rope of mass M and length L is suspended from the ceiling as shown in Fig. 1. The rope forms an angle φ with the ceiling at end points A and B respectively. The length of the portion from the end point A to a point P on the rope is denoted as x. We want to find an equation to describe the shape of the rope that involves x. What is the tension on the rope at the lowest point of the rope?

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Discussion after the solution:

The force acting at the end point of the rope can be computed from Eq. (4) by putting θ = φ. However, the force diverges to infinity as φ aproaches 0. This means that the rope cannot be pulled to a perfect horizontal position. The reason of this suprising conclusion is easy to explain. Suppose that we can pull the rope to a perfectly horizontal line as shown in Fig. 5, then the forces acting on the end points of the rope must be horizontal, too, since the tensions of the rope can only be horizontal as the rope itself. Thus the forces of vertical drection can not be balanced, and the rope cannot be stationary.

Also it is interesting to note that Eq. (6) does not depend on the mass of the rope. Therefore, the shape of the rope is the same as long as its length and the angle φ are the same, no matter the rope is just a thin thread or a very heavy chain.

Solution:

The mass density of the rope, ρ (kg/m), is

ρ = M/L.

We consider a small piece of the rope of length Δx as shown in Fig. 2. At the end point C of this small piece of the rope, a tension force forms an angle θ with the horizontal line, and has a magnitude of T. At he other end point D of this small piece of the rope, a tension force forms an angle θ + Δθ with the horizontal line, and has a magnitude of T + ΔT. The gravitational pull on this small piece of the rope is ρg·Δx. The balance of forces at the horizontal direction says that

T·cosθ = (T + ΔT)·cos(θ + Δθ).                                 (1)

The balance of forces at the vertical direction says that

T·sinθ = ρg·Δx + (T + ΔT)·sin(θ + Δθ).                                 (2)

In this case the consideration of the balance of torques does not help since, in doing so, we need to introduce another unknown parameter, that is, the angle that the small piece of the rope forms with the horizontal line.

We consider first Eq. (1). We have

cos(θ + Δθ) = cosθ·cos(Δθ) - sinθ·sin(Δθ).

Keeping only the first order small terms, that is, terms that are linear to Δθ, ΔT or Δx, we have

cos(Δθ) ≅ 1,     and     sin(Δθ) ≅ Δθ.

Thus

cos(θ + Δθ) ≅ cosθ - sinθ·Δθ .

Substituting this into Eq. (1), and keeping again only the first order small terms, we have

T·cosθ ≅ (T + ΔT)·(cosθ + sinθ·Δθ)
≅ T·cosθ - (T·sinθ)·Δθ + cosθ·ΔT.

This can be simplified as

ΔT/T ≅ (sinθ/cosθ)·Δθ .                                 (3)

Replacing ΔT and Δθ by dT and dθ in Eq. (3) respectively, and integrating both sides of the equation, we get

dT = (sinθ/cosθ)dθ.

Using the relations

(1/T)dT = lnT,     and     (cosθ/sinθ)dθ = -ln(cosθ),

we get

lnT = -ln(cosθ) + B,

where B is an arbitrary constant. This equation can be rewritten as

T(θ) = G/cosθ,                                 (3A)

where G is an arbitrary constant. To determine the constant G, we go to the end point A where θ = φ as shown in Fig. 3. From Fig. 3 we can see that the balance of forces at the vertical direction implies that

2sinφ·T(φ) = Mg,

or

T(φ) = Mg/(2sinφ).

From Eq. (3A) with θ = φ, we have

T(φ) = G/cosφ.

Equating two forms of T(φ), we get

G/cosφ = Mg/(2sinφ),

or

G = (Mg/2)·cotφ.

With this G, Eq. (3A) becomes

T(θ) = {(Mg/2)·cotφ}/cosθ .                                 (4)

Using the first order approximations of cos(Δθ) ≅ 1 and sin(Δθ) ≅ Δθ, Eq. (2) can be written as

T·sinθ = ρg·Δx + (T + ΔT)·{sinθ·cos(Δθ) + cosθ·sin(Δθ)}
≅ ρg·Δx + (T + ΔT)·(sinθ + cosθ·Δθ)
≅ ρg·Δx + T·sinθ + T·cosθ·Δθ + sinθ·ΔT.

The terms T·sinθ on both sides of the equation are cancelled, and we get

- ρg·Δx ≅ T·cosθ·Δθ + sinθ·ΔT.

Substituting Eq. (4) into the above equation, we get

-ρg·Δx ≅ {(Mg/2)·cotφ}·Δθ + sinθ·ΔT.                                 (5)

By differentiating both sides of Eq. (4) by θ, it becomes

dT/dθ = (Mg/2)·cotφ·sinθ/cos2θ,

so we can write

ΔT ≅ (Mg/2)·cotφ·(sinθ/cos2θ)·Δ θ .

Eq. (5) now becomes

-ρg·Δx ≅ {(Mg/2)·cotφ}·(1/cos2θ)·Δθ .

Replacing Δx and Δθ by dx and dθ respectively, and then integrating both sides of the above equation, we get

-ρg·dx = (Mg/2)·cotφ·(1/ cos2θ)·dθ.

Thus we arrive at

-ρg·x = {(Mg/2)·cotφ}·tanθ + Q,                                 (5A)

where Q is an arbitrary constant to be determined. At the end point A, we have x = 0, and θ = φ, so the above equation becomes

0 = {(Mg/2)·cotφ}·tanφ + Q,

or

Q = -Mg/2.

Inserting the above Q and ρ = M/L into Eq.(5A), we finally arrive at the equation

2x/L = -cotφ·tanθ + 1.                                 (6)

Eq. (6) defines the shape of the rope. The meaning of the angle θ should be noted here. If we take a point P on the rope such that the length of the portion of the rope from point A to point P is x, we can draw a line that contacts the rope at point P; let us call this contact line as line RS as shown in Fig. 4. θ is the angle that the line RS forms with the horizontal line.

At the bottom of the rope, we must have θ = 0. Eq. (4) says

T(0) = (Mg/2)·cotφ,

and that is the tension on the rope at the lowest point of the rope.