If half of the weight of a flatbed truck is supported by it two Drive wheels, wh

Question

If half of the weight of a flatbed truck is supported by it two Drive wheels, what is the maximum acceleration it can achieve on dry concrete where the coefficient of kinetic friction is 0.7 in the coefficient of static friction is 1. In case the truck has four-wheel-drive in the cabinet is wooden what is the maximum acceleration and will the cabinet slip? If half of the weight of a flatbed truck is supported by it two Drive wheels, what is the maximum acceleration it can achieve on dry concrete where the coefficient of kinetic friction is 0.7 in the coefficient of static friction is 1. In case the truck has four-wheel-drive in the cabinet is wooden what is the maximum acceleration and will the cabinet slip?

Explanation / Answer

Using force balance

Fnet = Ff

m*a = uk*(m*g/2)

a = uk*g/2

a = 0.7*9.81/2 = 3.43 m/sec^2

B.

in case of four wheel drive

amax = uk*g

amax = 0.7*9.81 = 6.87 m/sec^2

since acceleration of flatbed truck is lower, so it will not slip.

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