Question

Two strings are tied $2\text{m}$ apart on a rod and on the other end a mass $200\text{g}$ is tied as shown in figure. Each string is $1.25\text{m}$ long. FInd the tensions $T_1$ and $T_2$ if the rod is rotated with $60\text{rpm}$. Consider ${\pi }^2=10$

• A. $6\text{N},3\text{N}$
• B. $6.25\text{N},3.75\text{N}$
• C. $4.25\text{N},5.75\text{N}$
• D. $5.25\text{N},4.75\text{N}$

Answer 1

The correct option is B $6.25\text{N},3.75\text{N}$

From triangle $\sin \theta =\frac{1}{1.25}=0.8\Rightarrow \cos \theta =0.6$
Given $\omega =60rpm=2\pi rad/s$
Resolve $T_1$ and $T_2$.
$T_1\cos \theta +T_2\cos \theta =mr\omega 2$
$T_1\sin \theta =T_2\sin \theta +mg$
$(T_1+T_2)\left(0.6\right)=0.2\times \left(\frac{3}{4}\right)(2\pi {)}^2$
$T_1+T_2=10$
$T_1-T_2=\frac{mg}{\sin \theta }=\frac{0.2\times 10}{0.8}=2.5$
solving $T_1=6.25N$ and $T_2=3.75N$