At what point of the pole should the rope be tied so that the man can turn the pole upside down easily

classical-mechanicsphysicssolution-verificationstaticstrigonometry

Question:

One end of a $20m$ long rope is tied to a telegraph pole standing perpendicularly on the ground and its other end is being pulled by a person with a definite force. At what point of the pole should the rope be tied so that the person can turn the pole upside-down easily?

My book's attempt:

enter image description here

Let the rope be tied at point $A$. Now,

$$BE=BC\sin\theta$$

$$=AC\cos\theta\sin\theta\tag{1}$$

$$=20\sin\theta\cos\theta$$

$$=10\sin2\theta$$

It will be easy for the person to turn the pole upside-down if the moment is maximum about point $B$. Calculating moment about $B$,

$$F\cdot BE$$

$$10F\sin2\theta$$

The moment will be maximum if $\sin2\theta=1$ or $\theta=45^{\circ}$.

If $\theta=45^{\circ}$, then $x=20\cdot\sin45^{\circ}=20\cdot\frac{1}{\sqrt2}=10\sqrt2\ \text{(Ans.)}$

My analysis:

From $(1)$,

$$BE=AC\cos\theta\sin\theta$$

$$=AB\cos\theta$$

Now, calculating the moment about $B$,

$$F\cdot BE$$

$$F\cdot AB\cos\theta$$

The moment will be maximum if $\cos\theta=1\implies\theta=0^{\circ}$.

My question:

  1. According to my book, $\theta=45^{\circ}$; according to me, $\theta=0^{\circ}$. Who is correct?

Best Answer

The book. You make the mistake in forgetting that $AB$ is a function of $\theta$ [because the length of the rope is fixed]; in particular if $\theta$ is $0$ then so must $AB$ be $0$. So $AB\cos \theta$ is not maximized at $\theta=0$ after all, because $AB$ must be $0$ at $\theta=0$.

Your analysis would be correct if you had a rope of infinite length tied to the pole at a fixed point $A$; as in the person picks a point $A$ on the pole to tie the rope around, and then he decides $\theta$ by deciding how far away to stand from the pole with the rope in his hand to pull; he would be able to do with a rope of infinite length. [In that case the person could pick $A$ to be as high as possible, and then start pulling from very far away so that simultaneously $AB$ is large and $\theta$ close to $0$, which would maximize his strength in pulling the pole over.] However here, because the rope has length $20$, the higher $A$ is, the closer to the pole he must be to start pulling, which would mean that $\cos \theta$ would be smaller.

Related Question