Question

The figure below shows a uniform rod of length $30\text{cm}$ having a mass of $3.0\text{kg}.$ The strings as shown in the figure are pulled by constant forces of $20\text{N}$ and $32\text{N}$. Find the force exerted by the $20\text{cm}$ part of the rod on the $10\text{cm}$ part. All the surfaces are smooth, and the strings are light.

• A. $36\text{N}$
• B. $12\text{N}$
• C. $64\text{N}$
• D. $24\text{N}$

Answer 1

The correct option is D $24\text{N}$

The $20\text{cm}$ part of rod will pull the $10\text{cm}$ part.

Let the force exerted be $F$.

Considering rod as system

Using second law of motion

$F_{net}=ma$

$\Rightarrow (32-20)=3a$

$\therefore a=\frac{12}{3}=4{\text{m/s}}^2$

Now from the FBD of $10\text{c}m$ part of rod.


Mass of rod is $3kg$, thus mass per unit length $\lambda$ is given by

$\lambda =\frac{3}{30}\text{kg/cm}$

$\therefore$ mass of $10\text{cm}$ part will be

$m_1=\frac{3}{30}\times 10=1\text{kg}$

Applying second law,

$\Rightarrow F-20=ma$

$\Rightarrow F-20=1\times 4\Rightarrow F=24\text{N}$

$\begin{array}{c}Ans.D\end{array}$

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: In such problems, always try to visualize}\\ \text{the whole body as two sections, then one}\\ \text{section will be exerting electromagnetic}\\ \text{force of attraction at the another section.}\end{array}$