Question

18. A turn of radius 20 m is banked for the vehicles going
at a speed of 36 km/h. If the coefficient of static friction
between the road and the tyre is 0.4, what are the
possible speeds of a vehicle so that it neither slips down
nor skids up ?

Answer 1

Answer on Question #60162, Physics / Mechanics | Relativity

18. A turn of radius $20\mathrm{m}$ is banked for the vehicles going at a speed of $36\mathrm{km/h}$ . If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up?

Solution:

Given:

$$\begin{array}{l} v = 3 6 \frac {k m}{h r} = 1 0 m / s, \\ r = 2 0 m, \\ \mu_ {s} = 0. 4 \\ \end{array}$$

The road is banked with an angle,

$$\theta = \tan^ {- 1} \left(\frac {v ^ {2}}{r g}\right) = \tan^ {- 1} \left(\frac {1 0 ^ {2}}{2 0 * 1 0}\right) = 2 6. 5 7 {}^ {\circ}$$

When the car travels at max speed it slips upward

Force equations at maximum speed $\mathbf{v}$ , at threshold of sliding up incline

$$\begin{array}{l} \sum F _ {x} = \frac {m v ^ {2}}{r} = N \sin \theta + \mu_ {s} N \cos \theta \\ \sum F _ {y} = 0 = N \cos \theta - \mu_ {s} N \sin \theta - m g \\ \end{array}$$

Solving the pair of equations for the maximum speed $\mathbf{v}$ gives:

$$v _ {m a x} = \sqrt {\frac {r g (\sin \theta + \mu_ {s} \cos \theta)}{\cos \theta - \mu_ {s} \sin \theta}}$$ $$v _ {m a x} = \sqrt {\frac {2 0 \cdot 1 0 \cdot (\sin 2 6 . 5 7 {}^ {\circ} + 0 . 4 \cdot \cos 2 6 . 5 7 {}^ {\circ})}{\cos 2 6 . 5 7 {}^ {\circ} - 0 . 4 \cdot \sin 2 6 . 5 7 {}^ {\circ}}} = 1 5 \frac {m}{s} = 1 5 \cdot 3. 6 \frac {k m}{h r} = 5 4 \frac {k m}{h r}$$

For the case of sliding down

$$\sum F _ {x} = \frac {m v ^ {2}}{r} = N \sin \theta - \mu_ {s} N \cos \theta$$ $$\sum F _ {y} = 0 = N \cos \theta + \mu_ {s} N \sin \theta - m g$$

Solving the pair of equations for the minimum speed $v$ gives:

$$v _ {m i n} = \sqrt {\frac {r g (\sin \theta - \mu_ {s} \cos \theta)}{\cos \theta + \mu_ {s} \sin \theta}}$$ $$v _ {m i n} = \sqrt {\frac {2 0 \cdot 1 0 \cdot (\sin 2 6 . 5 7 {}^ {\circ} - 0 . 4 \cdot \cos 2 6 . 5 7 {}^ {\circ})}{\cos 2 6 . 5 7 {}^ {\circ} + 0 . 4 \cdot \sin 2 6 . 5 7 {}^ {\circ}}} = 4. 0 8 5 \frac {m}{s} = 4. 0 8 5 \cdot 3. 6 \frac {k m}{h r} = 1 4. 7 \frac {k m}{h r}$$

Answer: $v_{min} = 14.7 \frac{km}{hr}$ , $v_{max} = 54 \frac{km}{hr}$ .

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