Question 7.9:
A
car weighs 1800 kg. The distance between its front and back axles is
1.8 m. Its centre of gravity is 1.05 m behind the front axle.
Determine the force exerted by the level ground on each front wheel
and each back wheel.
Solution:
Mass of the car, m = 1800 kg
Distance
between the front and back axles, d
= 1.8 mMass of the car, m = 1800 kg
Distance between the C.G. (centre of gravity) and the back axle = 1.05 m
The various forces acting on the car are shown in the following figure.

At translational equilibrium:
Rf + Rb = mg
= 1800 X 9.8
= 17640 N ....(i)
For rotational equilibrium, on taking the torque about the C.G., we have:
Rf(1.05) = Rb(1.8 - 1.05)
Rb / Rf = 7 / 5
Rb = 1.4 Rf ......(ii)
Solving equations (i) and (ii), we get:
1.4Rf + Rf = 17640
Rf = 7350 N
∴ Rb = 17640 – 7350 = 10290 N
Therefore, the force exerted on each front wheel = 7350 / 2 = 3675 N, and
The force exerted on each back wheel = 10290 / 2 = 5145 N
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