Consider a train that rounds a curve with a radius of 570 m at a speed of 160 km/h (approximately 100 mi/h). (a) Calculate the friction force needed on a train passenger of mass 55 kg if the track is not banked and the train does not tilt. (b) Calculate the fritction force on the passanger if the train tilts at an angle of \({\bf{8}}{\bf{.0}}^\circ \) toward the centre of the curve.

When there is no angle of tilt for the train, then the friction force is equal to the centripetal force. When there is a definite angle of tilt, then using Newton’s law, the friction force can be calculated.

The radius of curvature is \(r = 570\;m\).

The speed of the train is \(v = 160\;km/h\).

The mass of the passanger is \(m = 55\;kg\).

(a)

The figure below represents the free body diagram, with representation of force.

The friction force can be calculated as follows:

\(\begin{aligned}{F_R} &= m\left( {\frac{{{v^2}}}{R}} \right)\\ &= \left( {55\;{\rm{kg}}} \right)\left( {\frac{{160\;{\rm{km/h}}}}{{570\,\,{\rm{m}}}}} \right)\left( {\frac{{1000\;{\rm{m}}}}{{\left( {60\,{\rm{s}} \times 60\;{\rm{s}}} \right)\,1\;{\rm{km/h}}}}} \right)\\ &= 190.6\;{\rm{N}}\\ \approx {\rm{1}}{\rm{.9}} \times {\rm{1}}{{\rm{0}}^2}\;{\rm{N}}\end{aligned}\)

Thus, the friction force needed at the train passanger is \(1.9 \times {10^2}\,{\rm{N}}\).

\({\bf{8}}{\bf{.0^\circ }}\)

(b)

Summation of forces along y-axis is given as,

\(\begin{aligned}\sum {{F_y}} &= 0\\{F_N}\cos \theta - mg - {F_{fr}}\sin \theta &= 0\\{F_N} &= \frac{{mg + {F_{fr}}\sin \theta }}{{\cos \theta }}\end{aligned}\)

Here, \({F_N}\) is the normal force and \({F_{fr}}\) is the friction force.

Summation of forces along x-axis is given as,

\(\begin{aligned}\sum {{F_x}} &= m\left( {\frac{{{v^2}}}{r}} \right)\\{F_N}\sin \theta + {F_{fr}}\cos \theta &= m\left( {\frac{{{v^2}}}{r}} \right)\end{aligned}\)

Substitute value of \({F_N}\) in the above equation,

\(\begin{aligned}\left( {\frac{{mg + {F_{fr}}\sin \theta }}{{\cos \theta }}} \right)\sin \theta + {F_{fr}}\cos \theta &= m\left( {\frac{{{v^2}}}{r}} \right)\\{F_{fr}} &= m\left( {\frac{{{v^2}}}{r}\cos \theta - \sin \theta } \right)\end{aligned}\)

Substitute all the value in the above equation,

\(\begin{aligned}{F_{fr}} &= \left( {55\;{\rm{kg}}} \right)\left( {\frac{1}{{570\;{\rm{m}}}}{{\left( {160\;{\rm{km/h}} \times \left( {\frac{{1000\;{\rm{m}}}}{{\left( {60\,{\rm{s}} \times 60\;{\rm{s}}} \right)\,1\;{\rm{km/h}}}}} \right)} \right)}^2}\cos 8^\circ - \sin 8^\circ } \right)\\ &= \left( {55} \right)\left( {\frac{{{{\left( {44.44} \right)}^2}}}{{570}}\left( {0.99} \right) - \left( {9.8} \right)\left( {0.14} \right)} \right)\;{\rm{N}}\\ &= 113.7\;{\rm{N}}\\ &= {\rm{1}}{\rm{.1}} \times {\rm{1}}{{\rm{0}}^2}\;{\rm{N}}\end{aligned}\)

Thus, the friction force is \(1.1 \times {10^2}\;{\rm{N}}\).