Question

Two strings are tied 2 m apart on a rod and on the other end mass 200 g is tied as shown. Each string is 1.25 m long. Find the tension $T_1$ and $T_2$ if the rod is rotated with 60 rmp.

• A. 6 N, 3 N
• B. 6.25 N, 3.75 N
• C. 4.25 N, 5.75 N
• D. 5.25 N, 4.75 N

Answer 1

Answer: (B)

6.25 N, 3.75 N

Resolve $T_1$ and $T_2$:

$T_1\cos \theta +T_2\cos \theta =mr{\omega }^2$

$T_1\sin \theta =T_2\sin \theta +mg$ or $(T_1+T_2)\left(0.6\right)=0.2\times \frac{3}{4}{\left(2\pi \right)}^2$ or $T_1+T_2=10$

$T_1-T_2=\frac{mg}{\sin \theta }=\frac{0.2\times 10}{0.8}=2.5$

On solving we get: $T_1$ $=6.25N$ and $T_2$ $=3.75N$