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Exercises · 6.9

Q.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.

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Taking torques about the front axle, the back axle carries 10290 N and the front axle carries 7350 N in total; dividing by two wheels per axle gives 3675 N on each front wheel and 5145 N on each back wheel.

Setting up

The car (mass 1800 kg) is in static equilibrium, so both the net force and net torque on it are zero. Its weight is

W=mg=1800×9.8=17640 NW=mg=1800\times9.8=17640\text{ N}

acting at the centre of gravity, which is 1.05 m behind the front axle (and so 0.75 m in front of the back axle, since the wheelbase is 1.8 m). Let FfF_f be the total upward force from the ground on both front wheels, and FbF_b the total force on both back wheels.

Torque balance about the front axle

Choosing the front axle as the pivot eliminates FfF_f from the equation (its moment arm is zero). The weight, 1.05 m behind the front axle, creates a torque balanced by FbF_b acting at the full wheelbase distance:

Fb×1.8=W×1.05F_b\times1.8 = W\times1.05

Fb=17640×1.051.8=185221.8=10290 NF_b = \frac{17640\times1.05}{1.8} = \frac{18522}{1.8} = 10290\text{ N}

Force balance

Ff+Fb=W  ⇒  Ff=17640−10290=7350 NF_f + F_b = W \;\Rightarrow\; F_f = 17640-10290 = 7350\text{ N}

Force per wheel …

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