A four-wheeler automobile is taking a turn towards left. Consider W = Weight of the vehicle,\(\frac{Q}{2}\) = Magnitude of the reaction due to gyroscopic effect on inner or outer wheels, \(\frac{P}{2}\) = Magnitude of the reaction due to centrifugal effect on inner or outer wheels. The total vertical reaction at each inner wheel is:
1
\(\frac{W}{4} + \frac{P}{4} + \frac{Q}{4}\)
2
\(\frac{W}{4} + \frac{P}{4} - \frac{Q}{4}\)
3
\(\frac{W}{4} - \frac{P}{4} + \frac{Q}{4}\)
4
\(\frac{W}{4} - \frac{P}{4} - \frac{Q}{4}\)