Question

A light string is warped around a cylindrical log of wood which is placed on a horizontal surface with its axis vertical and it is pulled with a constant force $F$ as shown in the figure. (Friction is absent everywhere).

• A. Tension $T$ in string will increase with increase in $\theta$.
• B. Tension $T$ in string will decrease with increase in $\theta$.
• C. Tension $T>F$ if $\theta >\frac{\pi }{3}$
• D. Tension $T>F$ if $\theta >\frac{\pi }{4}$

Answer 1

Answer: (A), (C)

Tension $T>F$ if $\theta >\frac{\pi }{3}$

Correct Option: Option A

Step 1: Given;

Step 2: Formula used;

$F_{net}=0$

Step 3: Find the tension, $T$ in the string;

Since, the system is in Equilibrium then $F$(right) = $F$ (left).

$$\begin{gathered} \Rightarrow T\cos \theta +T\cos \theta =F \\ \Rightarrow 2T\cos \theta =F \\ \Rightarrow T=\frac{F}{2\cos \theta } \end{gathered}$$

Step 4: Find the nature of $T$;

Since, $\cos \theta$ is decreasing function.

So, $T$ will increase with an increase in $\theta$.

Option C:

When $\theta <\frac{\mathrm{\pi }}{3}$, then $\cos \theta <\cos \frac{\mathrm{\pi }}{3}$ and $T=\frac{F}{2\cos \theta }$. Using these establish relation between $T$ and $F$.

$$\begin{gathered} \theta <\cos \theta <\cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow \cos \theta <\frac{1}{2} \\ \Rightarrow \frac{F}{2T}<\frac{1}{2} \\ \Rightarrow F<T \end{gathered}$$

So, Tension $T>F$ if $\theta >\frac{\pi }{3}$.

Incorrect Options

Option B:

As it is proved, that $T$ will increase with an increase in $\theta$, so this option is incorrect.

Option D:

When $\theta <\frac{\mathrm{\pi }}{4}$, then $\cos \theta <\cos \frac{\mathrm{\pi }}{4}$ and $T=\frac{F}{2\cos \theta }$. Using these establish relation between $T$ and $F$.

$$\begin{gathered} \theta >\cos \theta <\cos \frac{\mathrm{\pi }}{4} \\ \Rightarrow \cos \theta <\frac{1}{\sqrt{2}} \\ \Rightarrow \frac{F}{2T}<\frac{1}{\sqrt{2}} \\ \Rightarrow F<\sqrt{2}T \end{gathered}$$

So, Tension $T>F$ if $\theta >\frac{\pi }{4}$ is incorrect.

Thus, options A and C are correct.