Statics with Trigonometry!

Classical Mechanics Level 3

As shown in the figure, the non-stretch light rope $AO$ and $BO$ are attached to a node $O$ , $O$ is attached to an object $P$ , $BO>AO$ , where point $A$ and $B$ are fixed on the same horizontal line. At the beginning, $AO$ is vertical and the two ropes are just stretched. The tensions of $AO$ and $BO$ are denoted as $F_{A}$ and $F_{B}$ respectively. Now, keeping $A$ and $B$ on the same horizontal line, if we slowly move point $B$ to the left , what can we say about $F_{A}$ and $F_{B}$ as $AB$ increases?

Note: Can you rigorously prove it?

F_{B}

decreases.

F_{A}

decreases, and then increases.

F_{B}

decreases, and then increases.

F_{A}

increases.

1 solution

Zhengxi Gao
Jul 7, 2019

Let's make a force analysis of the object. And it gives us
Fb * cosθ = Fa * cosα
and
Fb*sinθ +Fa *sinα =G
So we will get Fa =G * $\frac{cosθ}{sin(θ+α)}$
Then its first order derivation of θ is Fa' {θ} =G * $\frac{-cos(α)}{ sin^{2}(θ+α)}$ which 0<2θ+α<1.5π and its first order derivation of α is Fa' {α} = G * $\frac{-cosθ*cos(θ+α)}{sin^{2}(θ+α)}$ 0<θ<0.5π 0<θ+α<π
So Fa' {θ}<0 all the time
We can know that at first 0.5π<θ+α.hence, Fa' {α}>0. And | Fa' {θ}|>|Fa' {α}|
Gradually,it becomes 0<θ+α<0.5π,which makes Fa'_{α}<0.
As a result , we can know that Fa decreases ,and then increases

With respect to what variable did you take the derivative? It seems you took wrt $\theta$ , but $\alpha$ is also changing as $\theta$ changes. Why did you ignore that? Thanks

Harsh Poonia - 1 year, 10 months ago

I'm posting this here as the app doesn't seem to let me post a solution.

I'll call the vertical rope's length a and the other one's b. The angles that these make with the horizontal will be α and β respectively. The length AB will be d and the angle between the ropes θ. The forces will be F and G respectively, as I can't write subscripts. W is the weight of the object.

The angles satisfy: [1] α+β+θ=π.

The horizontal tensions need to sum 0 and the vertical forces upwards hold the weight W: Fcosα-Gcosβ=0 Fsinα+Gsinβ=W.

Solving for F and using [1] gives: [2] F=Wcosβ/sin(α+β)=Wcosβ/sin(π-θ)=Wcosβ/sinθ.

For any triangle: cosβ=(b^2+d^2-a^2)/(2bd) cosθ=(a^2+b^2-d^2)/(2ab).

Defining γ=b/a δ=d/a

these relations become cosβ=(γ^2+δ^2-1)/(2δγ) cosθ=(γ^2-δ^2+1)/(2γ).

So, sinθ=sqrt(1-cos^2(θ))=sqrt(-γ^4-δ^4+2γ^2*δ^2+2γ^2+2δ^2-1)/(2γ).

Using [2] and these relations: F=W/δ(γ^2+δ^2-1)/sqrt(2γ^2δ^2+2γ^2+2δ^2-γ^4-δ^4-1).

a and b are constants, so γ is constant. I'll define the constants x=γ^2-1.

As γ>1, x is positive. Then F=W/δ*(δ^2+x)/sqrt(-δ^4+2(x+2)δ^2-x^2).

Taking the derivative of F with respect to d we get: [3] dF/dd=dF/dδdδ/dd=W/(aδ^2)(δ^6+3xδ^4-(5x+8)xδ^2+x^3)/sqrt(-δ^4+2(x+2)δ^2-x^2)^3.

Note that this is only a function of δ^2, not δ. In the initial position the triangle is right, so it satisfies: a^2+d^2=b^2,

and dividing by a^2, δ^2=γ^2-1=x.

Then, dF/dd(δ^2=x)=-W/(a*sqrtx)<0,

meaning that the tension decreases as d increases at first. There greatest possible value of d, which I'll call D, is: D=a+b,

and dividing by a, Δ=γ+1

is the greatest possible value of δ. This corresponds to Δ^2=x+2+2sqrt(x+1).

Therefore the value δ^2=x+2 always lies within the limit and is a valid configuration, while at the same time being greater than x. Evaluating [3] there gives: dF/dd(δ^2=x+2)=W/(a(x+2)*sqrt(x+1))>0.

This means that at this separation the tension always increases.

Therefore the tension always decreases first and then increases.

Augusto Boglione - 1 year, 6 months ago

Statics with Trigonometry!

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