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A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the centerline at a point 1.15 m behind the front axle. Find the force exerted by the ground on each wheel.

Respuesta :

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

[tex]F_1 + F_2 = W[/tex]

[tex]F_1 + F_2 = (1530\times 9.81)[/tex]

[tex]F_1 + F_2 = 15009.3 N[/tex]

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

[tex]F_1(1.15) = F_2(2.70 - 1.15)[/tex]

[tex]F_1 = 1.35 F_2[/tex]

now from above equation

[tex]F_2 + 1.35F_2 = 15009.3[/tex]

now we have

[tex]F_2 = 6392.85 N[/tex]

now the other force is given as

[tex]F_1 = 8616.45 N[/tex]

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