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:

Question

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:

TuteeHUB · Accepted Answer

NA

TuteeHUB · Answer

$\frac{W}{4} + \frac{P}{4} + \frac{Q}{4}$

TuteeHUB · Answer

$\frac{W}{4} + \frac{P}{4} - \frac{Q}{4}$

TuteeHUB · Answer

$\frac{W}{4} - \frac{P}{4} + \frac{Q}{4}$

TuteeHUB · Answer

$\frac{W}{4} - \frac{P}{4} - \frac{Q}{4}$

1.	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:
A.	\(\frac{W}{4} + \frac{P}{4} + \frac{Q}{4}\)
B.	\(\frac{W}{4} + \frac{P}{4} - \frac{Q}{4}\)
C.	\(\frac{W}{4} - \frac{P}{4} + \frac{Q}{4}\)
D.	\(\frac{W}{4} - \frac{P}{4} - \frac{Q}{4}\)
Answer» E.

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