# Physics

Workers have loaded a delivery truck in such a way that its center of gravity is only slightly forward of the rear axle. .63m in front of the rear wheels and 2.3m behind the front wheels. The mass of the truck and its contents is 9155 kg. Find the magnitudes of the forces exerted by the ground on the front wheels and rear wheels?

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1. The total frce on front and rear wheels is

F1 + F2 = Mg = 9155*9.8= 8.9*10^4 Newtons

For the force on the front wheels (F1), set the torque about the rear wheels equal to zero.

8.9*10^4*0.63 - F1*2.3 = 0

Solve for F1. Then subtract that from 8.9*10^4 N to get F2 (the rear wheel force).

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2. This response above is *technically* correct in everything, but the length for this equation
- 8.9*10^4*0.63 - F1*[LENGTH] = 0
would be the total length of the wheels, so 0.63 + 2.3 which would be 2.93,
making the equation:
- 8.9*10^4*0.63 - F1*2.93 = 0
This will give the correct answer.

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