(a) If half of the weight of a flatbed truck is supported by its two drive wheel

Question

(a) If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve on dry concrete where the coefficient of kinetic friction is 0.7 and the coefficient of static friction is 1.
m/s2

(b) Will a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate where the coefficient of kinetic friction is 0.3 and the coefficient of static friction is 0.55?

(c) Answer both of these questions for the case that the truck has four-wheel drive, and the cabinet is wooden.
maximum acceleration
m/s2

Explanation / Answer

(A) on drive wheels,

N = M g / 2

f_max = us N = M g / 2

Fnet = M a

M g / 2 = M a

a = 4.9 m/s^2

B) f_max = us N = us m g

f_max = 0.55 m g

f_max = m a_max

a_max = 0.55 x 9.8 = 5.39 m/s^2

it will slip if acceleration is more than 5.39 m/s^2.

so it will not slip.

(c) if four wheel drive then

f_max = Mg

a = g = 9.8 m/s^2

(d) now it will slip .

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