Question:
One end of a rope of length $l$ is attached to a vertical wall, the other end being attached to a point on the surface of a uniform sphere of radius $a$. If the weight of the sphere is $W$, then show that the tension of the rope is $T=\frac{W(l+a)}{\sqrt{2al+l^2}}$.
My tutor's attempt:
$$\text{We shift $\vec{W}$ to the line segment CA}\tag{1}$$
Now,
$$OC=OB+BC$$
$$OC=a+l$$
And,
$$OA=a$$
Now,
$$OC^2=OA^2+AC^2$$
$$AC^2=(a+l)^2-a^2$$
$$AC^2=a^2+2al+l^2-a^2$$
$$AC^2=2al+l^2$$
$$AC=\sqrt{2al+l^2}$$
Now, comparing $\vec{N}$, $\vec{T}$ and the shifted $\vec{W}$ with $\triangle AOC$,
$$\frac{N}{a}=\frac{T}{l+a}=\frac{W}{\sqrt{2al+l^2}}$$
$$[\text{If three forces acting at a point are in equilibrium then the sides of any triangle taken in order parallel to the forces will be proportional to the forces}]$$
Using 2nd and 3rd fractions,
$$\frac{T}{l+a}=\frac{W}{\sqrt{2al+l^2}}$$
$$T=\frac{W(l+a)}{\sqrt{2al+l^2}}\ \text{(showed)}$$
My book's attempt:
Using Lami's theorem,
$$\frac{T}{\sin(90^{\circ})}=\frac{W}{\sin(180^{\circ}-\angle COA)}$$
$$T=\frac{W}{\sin(\angle COA)}$$
$$T=\frac{W}{\frac{CA}{CO}}$$
$$T=\frac{W}{\frac{\sqrt{2al+l^2}}{a+l}}$$
$$T=\frac{W(l+a)}{\sqrt{2al+l^2}}\ \text{(showed)}$$
My comments:
I think my tutor's attempt is wrong (no offense intended to him) and my book's is right, even though both were able to prove the question statement. My tutor in $(1)$ shifted $\vec{W}$ to $CA$. How can he do that? Isn't force only a sliding vector or a bound/localized/fixed vector? As my tutor shifted $\vec{W}$, I think my tutor is wrong.
On the contrary, my book did the math easily and accurately in my opinion.
My question:
- Isn't my tutor's attempt wrong as he shifted $\vec{W}$?
Best Answer
Notice that Lami's theorem is derived by basically performing your tutor's steps; in other words, the latter is the mechanics under the hood of the former.
Free Vector vs. Sliding Vector vs. Bound Vector may be a relevant classification system for when torque and position are relevant features of the system; however, the given exercise is mathematically analysing just vectors, which are indeed portable by translation in space, i.e., “free”.
Addendum
Notice that both your tutor and book started by (fully) mathematically modelling the given physical setup this way:
Notice also that the sphere has been modelled as a particle, that torque and position are not features of the model, and that the blue triangle supplies only direction information about the vectors.
Since the system is in equilibrium, the three forces' resultant is $\mathbf 0;$ so, following the usual laws of vector addition, the vector diagram above can be rearranged into a closed triangle (in the process freely translating the vector $\mathbf W):$
Then we can use this triangle to complete the given task: expressing magnitude $T$ in terms of magnitude $W:$ $$T=\frac W{\sin{\measuredangle COA}}=\cdots=\frac{W(l+a)}{\sqrt{2al+l^2}}.$$
P.S. Distangling the blue triangle (the concrete physical scenario) from the vector diagram (an abstraction) is to make clear that the notion of “free” vs. “sliding” vs. “bound” vectors doesn't interfere with our usual process of geometrically adding/manipulating vectors.