An old streetcar rounds a flat corner of radius $\mathbf{9}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{m}$, at$\mathbf{16}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}$. What angle with the vertical will be made by the loosely hanging hand straps?

• $\text{Radius=9.1\hspace{0.17em}m}$
  
• $\text{Velocity=16\hspace{0.17em}km/hr}$
  

The problem deals with Newton’s laws of motion which describe the relations between the forces acting on a body and the motion of the body.

Formula:

$\left|\mathrm{a}\right|={\mathrm{v}}^2/\mathrm{R}$

The free-body diagram (for the hand straps of mass m) is the view that a passenger might see if she was looking forward and the streetcar was curving towards the right (so it points rightwards in the figure).

We note that $\left|\mathrm{a}\right|={\mathrm{v}}^2/\mathrm{R}$

Where,

$\begin{array}{c}\mathrm{v}=\text{16\hspace{0.17em}km/hr}\\ =4.4\text{\hspace{0.17em}m/s}\end{array}$

Applying Newton’s law to the axes of the problem (+x is rightward and +y is upward), we obtain,

$\mathrm{T}\text{sin}\mathrm{\theta }={\mathrm{mv}}^2/\mathrm{R}$

$\mathrm{T}\mathrm{cos\theta }=\mathrm{mg}$

Solving these equations for the angle,

$\mathrm{\theta }={\text{tan}}^{-1}\left(\frac{{\mathrm{v}}^2}{\mathrm{Rg}}\right)$

Substitute the values in the above equation, and we get,

$\begin{array}{c}\mathrm{\theta }={\text{tan}}^{-1}\left(\frac{{\left(4.4\text{\hspace{0.17em}m/s}\right)}^2}{\left(\text{9.1\hspace{0.17em}m}\right)\left(9.8{\text{\hspace{0.17em}m/s}}^2\right)}\right)\\ \mathrm{\theta }={\text{tan}}^{-1}\left(\frac{{\left(4.4\text{\hspace{0.17em}m/s}\right)}^2}{\left(\text{9.1\hspace{0.17em}m}\right)\left(9.8{\text{\hspace{0.17em}m/s}}^2\right)}\right)\\ \mathrm{\theta }={\text{tan}}^{-1}\left(0.2170\right)\\ \mathrm{\theta }=12°\end{array}$

Thus,the vertical angle will be made by the loosely hanging hand straps is $12°$.