Question

A train is moving with a speed of $72km/h$ on a curved railway line of $400m$ radius. A spring balance loaded with a block of weight of $5kg$ is suspended from the roof of the train. What would be the reading of the balance? ($g=10m/s^2$)

Answer 1

If T be the tension in the spring balance, then $\begin{array}{c}T\cos \theta =mg\\ andT\sin \theta =\frac{m{v}^2}{r}\end{array}$

On squaring and then adding,

$\begin{array}{c}{T}^2={\left(mg\right)}^2+{\left(\frac{m{v}^2}{r}\right)}^2\\ Or,T=m\sqrt{g^2+\frac{v^4}{r^2}}\end{array}$

According to question, we have$v=72km/h=20m/s,r=400m,m=5kg,g=10m/s^2$

Putting the above values for $T$ , we get

$T=5\times \sqrt{{10}^2\times \frac{{20}^4}{{400}^2}}=50.25N\left(approx\right)$

Hence Spring balance reading $\frac{50.25}{10}=5.025kg$