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An automobile with a mass of 1240.00 kg has 3.99 m between
thefront and rear axles. Its center of gravity is located 0.622
mbehind the front axle. With the automobile on level
ground,determine the magnitude of the force from the ground
on(a) each front
wheel(assuming equal forces on the front wheels) and (b) each rear wheel (assuming
equalforces on the rear wheels).

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Answer:

[tex]948.15248\ N[/tex]

[tex]5134.04751\ N[/tex]

Explanation:

[tex]F_f[/tex] = Force on front wheels

[tex]F_r[/tex] = Force on rear wheels

Distance between CG and rear wheel = 3.99-0.622 = 3.368 m

[tex]F_r=1240\times 9.81-F_f[/tex]

As the forces are conserved we have

[tex](1240\times 9.81-F_f)3.368=F_f\times 0.622\\\Rightarrow 12164.4-F_f=\dfrac{0.622F_f}{3.368}\\\Rightarrow 12164.4-F_f=0.18467F_f\\\Rightarrow F_f=\dfrac{12164.4}{1.18467}\\\Rightarrow F_f=10268.09503\ N[/tex]

On rear wheels

[tex]F_r=1240\times 9.81-10268.09503\\\Rightarrow F_r=1896.30497\ N[/tex]

The force on each rear whees is [tex]\dfrac{1896.30497}{2}=948.15248\ N[/tex]

Force on each front wheel is [tex]\dfrac{10268.09503}{2}=5134.04751\ N[/tex]

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